hypothesis testing z test examples

• Skip to secondary menu
• Skip to main content
• Skip to primary sidebar

Statistics By Jim

Making statistics intuitive

Z Test: Uses, Formula & Examples

By Jim Frost Leave a Comment

What is a Z Test?

Use a Z test when you need to compare group means. Use the 1-sample analysis to determine whether a population mean is different from a hypothesized value. Or use the 2-sample version to determine whether two population means differ.

A Z test is a form of inferential statistics . It uses samples to draw conclusions about populations.

For example, use Z tests to assess the following:

• One sample : Do students in an honors program have an average IQ score different than a hypothesized value of 100?
• Two sample : Do two IQ boosting programs have different mean scores?

In this post, learn about when to use a Z test vs T test. Then we’ll review the Z test’s hypotheses, assumptions, interpretation, and formula. Finally, we’ll use the formula in a worked example.

Related post : Difference between Descriptive and Inferential Statistics

Z test vs T test

Z tests and t tests are similar. They both assess the means of one or two groups, have similar assumptions, and allow you to draw the same conclusions about population means.

However, there is one critical difference.

Z tests require you to know the population standard deviation, while t tests use a sample estimate of the standard deviation. Learn more about Population Parameters vs. Sample Statistics .

In practice, analysts rarely use Z tests because it’s rare that they’ll know the population standard deviation. It’s even rarer that they’ll know it and yet need to assess an unknown population mean!

A Z test is often the first hypothesis test students learn because its results are easier to calculate by hand and it builds on the standard normal distribution that they probably already understand. Additionally, students don’t need to know about the degrees of freedom .

Z and T test results converge as the sample size approaches infinity. Indeed, for sample sizes greater than 30, the differences between the two analyses become small.

William Sealy Gosset developed the t test specifically to account for the additional uncertainty associated with smaller samples. Conversely, Z tests are too sensitive to mean differences in smaller samples and can produce statistically significant results incorrectly (i.e., false positives).

When to use a T Test vs Z Test

Let’s put a button on it.

When you know the population standard deviation, use a Z test.

When you have a sample estimate of the standard deviation, which will be the vast majority of the time, the best statistical practice is to use a t test regardless of the sample size.

However, the difference between the two analyses becomes trivial when the sample size exceeds 30.

Learn more about a T-Test Overview: How to Use & Examples and How T-Tests Work .

Z Test Hypotheses

This analysis uses sample data to evaluate hypotheses that refer to population means (µ). The hypotheses depend on whether you’re assessing one or two samples.

One-Sample Z Test Hypotheses

• Null hypothesis (H 0 ): The population mean equals a hypothesized value (µ = µ 0 ).
• Alternative hypothesis (H A ): The population mean DOES NOT equal a hypothesized value (µ ≠ µ 0 ).

When the p-value is less or equal to your significance level (e.g., 0.05), reject the null hypothesis. The difference between your sample mean and the hypothesized value is statistically significant. Your sample data support the notion that the population mean does not equal the hypothesized value.

Related posts : Null Hypothesis: Definition, Rejecting & Examples and Understanding Significance Levels

Two-Sample Z Test Hypotheses

• Null hypothesis (H 0 ): Two population means are equal (µ 1 = µ 2 ).
• Alternative hypothesis (H A ): Two population means are not equal (µ 1 ≠ µ 2 ).

Again, when the p-value is less than or equal to your significance level, reject the null hypothesis. The difference between the two means is statistically significant. Your sample data support the idea that the two population means are different.

These hypotheses are for two-sided analyses. You can use one-sided, directional hypotheses instead. Learn more in my post, One-Tailed and Two-Tailed Hypothesis Tests Explained .

Related posts : How to Interpret P Values and Statistical Significance

Z Test Assumptions

For reliable results, your data should satisfy the following assumptions:

You have a random sample

Drawing a random sample from your target population helps ensure that the sample represents the population. Representative samples are crucial for accurately inferring population properties. The Z test results won’t be valid if your data do not reflect the population.

Related posts : Random Sampling and Representative Samples

Continuous data

Z tests require continuous data . Continuous variables can assume any numeric value, and the scale can be divided meaningfully into smaller increments, such as fractional and decimal values. For example, weight, height, and temperature are continuous.

Other analyses can assess additional data types. For more information, read Comparing Hypothesis Tests for Continuous, Binary, and Count Data .

Your sample data follow a normal distribution, or you have a large sample size

All Z tests assume your data follow a normal distribution . However, due to the central limit theorem, you can ignore this assumption when your sample is large enough.

The following sample size guidelines indicate when normality becomes less of a concern:

• One-Sample : 20 or more observations.
• Two-Sample : At least 15 in each group.

Related posts : Central Limit Theorem and Skewed Distributions

Independent samples

For the two-sample analysis, the groups must contain different sets of items. This analysis compares two distinct samples.

Related post : Independent and Dependent Samples

Population standard deviation is known

As I mention in the Z test vs T test section, use a Z test when you know the population standard deviation. However, when n > 30, the difference between the analyses becomes trivial.

Related post : Standard Deviations

Z Test Formula

These Z test formulas allow you to calculate the test statistic. Use the Z statistic to determine statistical significance by comparing it to the appropriate critical values and use it to find p-values.

The correct formula depends on whether you’re performing a one- or two-sample analysis. Both formulas require sample means (x̅) and sample sizes (n) from your sample. Additionally, you specify the population standard deviation (σ) or variance (σ 2 ), which does not come from your sample.

I present a worked example using the Z test formula at the end of this post.

Learn more about Z-Scores and Test Statistics .

One Sample Z Test Formula

The one sample Z test formula is a ratio.

The numerator is the difference between your sample mean and a hypothesized value for the population mean (µ 0 ). This value is often a strawman argument that you hope to disprove.

The denominator is the standard error of the mean. It represents the uncertainty in how well the sample mean estimates the population mean.

Learn more about the Standard Error of the Mean .

Two Sample Z Test Formula

The two sample Z test formula is also a ratio.

The numerator is the difference between your two sample means.

The denominator calculates the pooled standard error of the mean by combining both samples. In this Z test formula, enter the population variances (σ 2 ) for each sample.

Z Test Critical Values

As I mentioned in the Z vs T test section, a Z test does not use degrees of freedom. It evaluates Z-scores in the context of the standard normal distribution. Unlike the t-distribution , the standard normal distribution doesn’t change shape as the sample size changes. Consequently, the critical values don’t change with the sample size.

To find the critical value for a Z test, you need to know the significance level and whether it is one- or two-tailed.

Learn more about Critical Values: Definition, Finding & Calculator .

Z Test Worked Example

Let’s close this post by calculating the results for a Z test by hand!

Suppose we randomly sampled subjects from an honors program. We want to determine whether their mean IQ score differs from the general population. The general population’s IQ scores are defined as having a mean of 100 and a standard deviation of 15.

We’ll determine whether the difference between our sample mean and the hypothesized population mean of 100 is statistically significant.

Specifically, we’ll use a two-tailed analysis with a significance level of 0.05. Looking at the table above, you’ll see that this Z test has critical values of ± 1.960. Our results are statistically significant if our Z statistic is below –1.960 or above +1.960.

The hypotheses are the following:

• Null (H 0 ): µ = 100
• Alternative (H A ): µ ≠ 100

Entering Our Results into the Formula

Here are the values from our study that we need to enter into the Z test formula:

• IQ score sample mean (x̅): 107
• Sample size (n): 25
• Hypothesized population mean (µ 0 ): 100
• Population standard deviation (σ): 15

The Z-score is 2.333. This value is greater than the critical value of 1.960, making the results statistically significant. Below is a graphical representation of our Z test results showing how the Z statistic falls within the critical region.

We can reject the null and conclude that the mean IQ score for the population of honors students does not equal 100. Based on the sample mean of 107, we know their mean IQ score is higher.

Now let’s find the p-value. We could use technology to do that, such as an online calculator. However, let’s go old school and use a Z table.

To find the p-value that corresponds to a Z-score from a two-tailed analysis, we need to find the negative value of our Z-score (even when it’s positive) and double it.

In the truncated Z-table below, I highlight the cell corresponding to a Z-score of -2.33.

The cell value of 0.00990 represents the area or probability to the left of the Z-score -2.33. We need to double it to include the area > +2.33 to obtain the p-value for a two-tailed analysis.

P-value = 0.00990 * 2 = 0.0198

That p-value is an approximation because it uses a Z-score of 2.33 rather than 2.333. Using an online calculator, the p-value for our Z test is a more precise 0.0196. This p-value is less than our significance level of 0.05, which reconfirms the statistically significant results.

See my full Z-table , which explains how to use it to solve other types of problems.

Share this:

Reader Interactions

Comments and questions cancel reply.

• Data Science
• Data Analysis
• Data Visualization
• Machine Learning
• Deep Learning
• Computer Vision
• Artificial Intelligence
• AI ML DS Interview Series
• AI ML DS Projects series
• Data Engineering
• Web Scrapping

Z-test : Formula, Types, Examples

Z-test is especially useful when you have a large sample size and know the population’s standard deviation. Different tests are used in statistics to compare distinct samples or groups and make conclusions about populations. These tests, also referred to as statistical tests, concentrate on examining the probability or possibility of acquiring the observed data under particular premises or hypotheses. They offer a framework for evaluating the evidence for or against a given hypothesis.

Table of Content

What is Z-Test?

Z-test formula, when to use z-test, hypothesis testing, steps to perform z-test, type of z-test, practice problems.

Z-test is a statistical test that is used to determine whether the mean of a sample is significantly different from a known population mean when the population standard deviation is known. It is particularly useful when the sample size is large (>30).

Z-test can also be defined as a statistical method that is used to determine whether the distribution of the test statistics can be approximated using the normal distribution or not. It is the method to determine whether two sample means are approximately the same or different when their variance is known and the sample size is large (should be >= 30).

The Z-test compares the difference between the sample mean and the population means by considering the standard deviation of the sampling distribution. The resulting Z-score represents the number of standard deviations that the sample mean deviates from the population mean. This Z-Score is also known as Z-Statistics, and can be formulated as:

[Tex]\text{Z-Score} = \frac{\bar{x}-\mu}{\sigma} [/Tex]

• [Tex]\bar{x}  [/Tex] : mean of the sample.
• [Tex]\mu  [/Tex] : mean of the population.
• [Tex]\sigma  [/Tex] : Standard deviation of the population.

z-test assumes that the test statistic (z-score) follows a standard normal distribution.

The average family annual income in India is 200k, with a standard deviation of 5k, and the average family annual income in Delhi is 300k.

Then Z-Score for Delhi will be.

[Tex]\begin{aligned} \text{Z-Score}&=\frac{\bar{x}-\mu}{\sigma} \\&=\frac{300-200}{5} \\&=20 \end{aligned} [/Tex]

This indicates that the average family’s annual income in Delhi is 20 standard deviations above the mean of the population (India).

• The sample size should be greater than 30. Otherwise, we should use the t-test.
• Samples should be drawn at random from the population.
• The standard deviation of the population should be known.
• Samples that are drawn from the population should be independent of each other.
• The data should be normally distributed , however, for a large sample size, it is assumed to have a normal distribution because central limit theorem

A hypothesis is an educated guess/claim about a particular property of an object. Hypothesis testing is a way to validate the claim of an experiment.

• Null Hypothesis: The null hypothesis is a statement that the value of a population parameter (such as proportion, mean, or standard deviation) is equal to some claimed value. We either reject or fail to reject the null hypothesis. The null hypothesis is denoted by H 0 .
• Alternate Hypothesis: The alternative hypothesis is the statement that the parameter has a value that is different from the claimed value. It is denoted by H A .
• Level of significance: It means the degree of significance in which we accept or reject the null hypothesis. Since in most of the experiments 100% accuracy is not possible for accepting or rejecting a hypothesis, we, therefore, select a level of significance. It is denoted by alpha (∝).
• First, identify the null and alternate hypotheses.
• Determine the level of significance (∝).
• Find the critical value of z in the z-test using
• n: sample size.
• Now compare with the hypothesis and decide whether to reject or not reject the null hypothesis

Left-tailed Test

In this test, our region of rejection is located to the extreme left of the distribution. Here our null hypothesis is that the claimed value is less than or equal to the mean population value.

Right-tailed Test

In this test, our region of rejection is located to the extreme right of the distribution. Here our null hypothesis is that the claimed value is less than or equal to the mean population value.

One-Tailed Test

 A school claimed that the students who study that are more intelligent than the average school. On calculating the IQ scores of 50 students, the average turns out to be 110. The mean of the population IQ is 100 and the standard deviation is 15. State whether the claim of the principal is right or not at a 5% significance level.

• First, we define the null hypothesis and the alternate hypothesis. Our null hypothesis will be: [Tex]H_0 : \mu  = 100        [/Tex] and our alternate hypothesis. [Tex]H_A : \mu > 100 [/Tex]
• State the level of significance. Here, our level of significance is given in this question ( [Tex]\alpha [/Tex]  =0.05), if not given then we take ∝=0.05 in general.
• Now, we compute the Z-Score: X = 110 Mean = 100 Standard Deviation = 15 Number of samples = 50 [Tex]\begin{aligned} \text{Z-Score}&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\&=\frac{110-100}{15/\sqrt{50}} \\&=\frac{10}{2.12} \\&=4.71 \end{aligned} [/Tex]
• Now, we look up to the z-table. For the value of ∝=0.05, the z-score for the right-tailed test is 1.645.
• Here 4.71 >1.645, so we reject the null hypothesis. 
• If the z-test statistics are less than the z-score, then we will not reject the null hypothesis.

Code Implementations of One-Tailed Z-Test

# Import the necessary libraries import numpy as np import scipy.stats as stats # Given information sample_mean = 110 population_mean = 100 population_std = 15 sample_size = 50 alpha = 0.05 # compute the z-score z_score = ( sample_mean - population_mean ) / ( population_std / np . sqrt ( 50 )) print ( 'Z-Score :' , z_score ) # Approach 1: Using Critical Z-Score # Critical Z-Score z_critical = stats . norm . ppf ( 1 - alpha ) print ( 'Critical Z-Score :' , z_critical ) # Hypothesis if z_score > z_critical : print ( "Reject Null Hypothesis" ) else : print ( "Fail to Reject Null Hypothesis" ) # Approach 2: Using P-value # P-Value : Probability of getting less than a Z-score p_value = 1 - stats . norm . cdf ( z_score ) print ( 'p-value :' , p_value ) # Hypothesis if p_value < alpha : print ( "Reject Null Hypothesis" ) else : print ( "Fail to Reject Null Hypothesis" )

Z-Score : 4.714045207910317Critical Z-Score : 1.6448536269514722Reject Null Hypothesisp-value : 1.2142337364462463e-06Reject Null Hypothesis

Two-tailed test

In this test, our region of rejection is located to both extremes of the distribution. Here our null hypothesis is that the claimed value is equal to the mean population value.

Below is an example of performing the z-test:

Two-sampled z-test

In this test, we have provided 2 normally distributed and independent populations, and we have drawn samples at random from both populations. Here, we consider u 1 and u 2 to be the population mean, and X 1 and X 2 to be the observed sample mean. Here, our null hypothesis could be like this:

[Tex]H_{0} : \mu_{1} -\mu_{2} = 0    [/Tex]

and alternative hypothesis

[Tex]H_{1} :  \mu_{1} – \mu_{2} \ne 0    [/Tex]

and the formula for calculating the z-test score:

[Tex]Z = \frac{\left ( \overline{X_{1}} – \overline{X_{2}} \right ) – \left ( \mu_{1} – \mu_{2} \right )}{\sqrt{\frac{\sigma_{1}^2}{n_{1}} + \frac{\sigma_{2}^2}{n_{2}}}}    [/Tex]

where  [Tex]\sigma_1 [/Tex]   and  [Tex]\sigma_2 [/Tex]   are the standard deviation and n 1 and n 2 are the sample size of population corresponding to u 1 and u 2 .  

There are two groups of students preparing for a competition: Group A and Group B. Group A has studied offline classes, while Group B has studied online classes. After the examination, the score of each student comes. Now we want to determine whether the online or offline classes are better.

Group A: Sample size = 50, Sample mean = 75, Sample standard deviation = 10 Group B: Sample size = 60, Sample mean = 80, Sample standard deviation = 12

Assuming a 5% significance level, perform a two-sample z-test to determine if there is a significant difference between the online and offline classes.

Step 1: Null & Alternate Hypothesis

• Null Hypothesis: There is no significant difference between the mean score between the online and offline classes [Tex] \mu_1 -\mu_2 = 0 [/Tex]
• Alternate Hypothesis: There is a significant difference in the mean scores between the online and offline classes. [Tex] \mu_1 -\mu_2 \neq 0 [/Tex]

Step 2: Significance Label

• Significance Label: 5%  [Tex]\alpha = 0.05 [/Tex]

Step 3: Z-Score

[Tex]\begin{aligned} \text{Z-score} &= \frac{(x_1-x_2)-(\mu_1 -\mu_2)} {\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_1}}} \\ &= \frac{(75-80)-0} {\sqrt{\frac{10^2}{50}+\frac{12^2}{60}}} \\ &= \frac{-5} {\sqrt{2+2.4}} \\ &= \frac{-5} {2.0976} \\&=-2.384 \end{aligned} [/Tex]

Step 4: Check to Critical Z-Score value in the Z-Table for apha/2 = 0.025

•  Critical Z-Score = 1.96

Step 5: Compare with the absolute Z-Score value

• absolute(Z-Score) > Critical Z-Score
• Reject the null hypothesis. There is a significant difference between the online and offline classes.

Code Implementations on Two-sampled Z-test

import numpy as np import scipy.stats as stats # Group A (Offline Classes) n1 = 50 x1 = 75 s1 = 10 # Group B (Online Classes) n2 = 60 x2 = 80 s2 = 12 # Null Hypothesis = mu_1-mu_2 = 0 # Hypothesized difference (under the null hypothesis) D = 0 # Set the significance level alpha = 0.05 # Calculate the test statistic (z-score) z_score = (( x1 - x2 ) - D ) / np . sqrt (( s1 ** 2 / n1 ) + ( s2 ** 2 / n2 )) print ( 'Z-Score:' , np . abs ( z_score )) # Calculate the critical value z_critical = stats . norm . ppf ( 1 - alpha / 2 ) print ( 'Critical Z-Score:' , z_critical ) # Compare the test statistic with the critical value if np . abs ( z_score ) > z_critical : print ( """Reject the null hypothesis. There is a significant difference between the online and offline classes.""" ) else : print ( """Fail to reject the null hypothesis. There is not enough evidence to suggest a significant difference between the online and offline classes.""" ) # Approach 2: Using P-value # P-Value : Probability of getting less than a Z-score p_value = 2 * ( 1 - stats . norm . cdf ( np . abs ( z_score ))) print ( 'P-Value :' , p_value ) # Compare the p-value with the significance level if p_value < alpha : print ( """Reject the null hypothesis. There is a significant difference between the online and offline classes.""" ) else : print ( """Fail to reject the null hypothesis. There is not enough evidence to suggest significant difference between the online and offline classes.""" )

Z-Score: 2.3836564731139807 Critical Z-Score: 1.959963984540054 Reject the null hypothesis. There is a significant difference between the online and offline classes. P-Value : 0.01714159544079563 Reject the null hypothesis. There is a significant difference between the online and offline classes.

Solved examples :

Example 1: One-sample Z-test

Problem: A company claims that the average battery life of their new smartphone is 12 hours. A consumer group tests 100 phones and finds the average battery life to be 11.8 hours with a population standard deviation of 0.5 hours. At a 5% significance level, is there evidence to refute the company’s claim?

Solution: Step 1: State the hypotheses H₀: μ = 12 (null hypothesis) H₁: μ ≠ 12 (alternative hypothesis) Step 2: Calculate the Z-score Z = (x̄ – μ) / (σ / √n) = (11.8 – 12) / (0.5 / √100) = -0.2 / 0.05 = -4 Step 3: Find the critical value (two-tailed test at 5% significance) Z₀.₀₂₅ = ±1.96 Step 4: Compare Z-score with critical value |-4| > 1.96, so we reject the null hypothesis. Conclusion: There is sufficient evidence to refute the company’s claim about battery life.

Problem: A researcher wants to compare the effectiveness of two different medications for reducing blood pressure. Medication A is tested on 50 patients, resulting in a mean reduction of 15 mmHg with a standard deviation of 3 mmHg. Medication B is tested on 60 patients, resulting in a mean reduction of 13 mmHg with a standard deviation of 4 mmHg. At a 1% significance level, is there a significant difference between the two medications?

Step 1: State the hypotheses H₀: μ₁ – μ₂ = 0 (null hypothesis) H₁: μ₁ – μ₂ ≠ 0 (alternative hypothesis) Step 2: Calculate the Z-score Z = (x̄₁ – x̄₂) / √((σ₁²/n₁) + (σ₂²/n₂)) = (15 – 13) / √((3²/50) + (4²/60)) = 2 / √(0.18 + 0.2667) = 2 / 0.6455 = 3.10 Step 3: Find the critical value (two-tailed test at 1% significance) Z₀.₀₀₅ = ±2.576 Step 4: Compare Z-score with critical value 3.10 > 2.576, so we reject the null hypothesis. Conclusion: There is a significant difference between the effectiveness of the two medications at the 1% significance level.

Problem 3 : A polling company claims that 60% of voters support a new policy. In a sample of 1000 voters, 570 support the policy. At a 5% significance level, is there evidence to support the company’s claim?

Step 1: State the hypotheses H₀: p = 0.60 (null hypothesis) H₁: p ≠ 0.60 (alternative hypothesis) Step 2: Calculate the Z-score p̂ = 570/1000 = 0.57 (sample proportion) Z = (p̂ – p) / √(p(1-p)/n) = (0.57 – 0.60) / √(0.60(1-0.60)/1000) = -0.03 / √(0.24/1000) = -0.03 / 0.0155 = -1.94 Step 3: Find the critical value (two-tailed test at 5% significance) Z₀.₀₂₅ = ±1.96 Step 4: Compare Z-score with critical value |-1.94| < 1.96, so we fail to reject the null hypothesis. Conclusion: There is not enough evidence to refute the polling company’s claim at the 5% significance level.

Problem 4 : A manufacturer claims that their light bulbs last an average of 1000 hours. A sample of 100 bulbs has a mean life of 985 hours. The population standard deviation is known to be 50 hours. At a 5% significance level, is there evidence to reject the manufacturer’s claim?

Solution: H₀: μ = 1000 H₁: μ ≠ 1000 Z = (x̄ – μ) / (σ / √n) = (985 – 1000) / (50 / √100) = -15 / 5 = -3 Critical value (α = 0.05, two-tailed): ±1.96 |-3| > 1.96, so reject H₀. Conclusion: There is sufficient evidence to reject the manufacturer’s claim at the 5% significance level.

Example 5 : Two factories produce semiconductors. Factory A’s chips have a mean resistance of 100 ohms with a standard deviation of 5 ohms. Factory B’s chips have a mean resistance of 98 ohms with a standard deviation of 4 ohms. Samples of 50 chips from each factory are tested. At a 1% significance level, is there a difference in mean resistance between the two factories?

H₀: μA – μB = 0 H₁: μA – μB ≠ 0 Z = (x̄A – x̄B) / √((σA²/nA) + (σB²/nB)) = (100 – 98) / √((5²/50) + (4²/50)) = 2 / √(0.5 + 0.32) = 2 / 0.872 = 2.29 Critical value (α = 0.01, two-tailed): ±2.576 |2.29| < 2.576, so fail to reject H₀. Conclusion: There is not enough evidence to conclude a difference in mean resistance at the 1% significance level.

Problem 6 : A political analyst claims that 40% of voters in a certain district support a new tax policy. In a random sample of 500 voters, 220 support the policy. At a 5% significance level, is there evidence to reject the analyst’s claim?

H₀: p = 0.40 H₁: p ≠ 0.40 p̂ = 220/500 = 0.44 Z = (p̂ – p) / √(p(1-p)/n) = (0.44 – 0.40) / √(0.40(1-0.40)/500) = 0.04 / 0.0219 = 1.83 Critical value (α = 0.05, two-tailed): ±1.96 |1.83| < 1.96, so fail to reject H₀. Conclusion: There is not enough evidence to reject the analyst’s claim at the 5% significance level.

Problem 7 : Two advertising methods are compared. Method A results in 150 sales out of 1000 contacts. Method B results in 180 sales out of 1200 contacts. At a 5% significance level, is there a difference in the effectiveness of the two methods?

H₀: pA – pB = 0 H₁: pA – pB ≠ 0 p̂A = 150/1000 = 0.15 p̂B = 180/1200 = 0.15 p̂ = (150 + 180) / (1000 + 1200) = 0.15 Z = (p̂A – p̂B) / √(p̂(1-p̂)(1/nA + 1/nB)) = (0.15 – 0.15) / √(0.15(1-0.15)(1/1000 + 1/1200)) = 0 / 0.0149 = 0 Critical value (α = 0.05, two-tailed): ±1.96 |0| < 1.96, so fail to reject H₀. Conclusion: There is no significant difference in the effectiveness of the two advertising methods at the 5% significance level.

Problem 8 : A new treatment for a disease is tested in two cities. In City A, 120 out of 400 patients recover. In City B, 140 out of 500 patients recover. At a 5% significance level, is there a difference in the recovery rates between the two cities?

H₀: pA – pB = 0 H₁: pA – pB ≠ 0 p̂A = 120/400 = 0.30 p̂B = 140/500 = 0.28 p̂ = (120 + 140) / (400 + 500) = 0.2889 Z = (p̂A – p̂B) / √(p̂(1-p̂)(1/nA + 1/nB)) = (0.30 – 0.28) / √(0.2889(1-0.2889)(1/400 + 1/500)) = 0.02 / 0.0316 = 0.633 Critical value (α = 0.05, two-tailed): ±1.96 |0.633| < 1.96, so fail to reject H₀. Conclusion: There is not enough evidence to conclude a difference in recovery rates between the two cities at the 5% significance level.

Problem 9 : Two advertising methods are compared. Method A results in 150 sales out of 1000 contacts. Method B results in 180 sales out of 1200 contacts. At a 5% significance level, is there a difference in the effectiveness of the two methods?

Problem 10 : A company claims that their product weighs 500 grams on average. A sample of 64 products has a mean weight of 498 grams. The population standard deviation is known to be 8 grams. At a 1% significance level, is there evidence to reject the company’s claim?

H₀: μ = 500 H₁: μ ≠ 500 Z = (x̄ – μ) / (σ / √n) = (498 – 500) / (8 / √64) = -2 / 1 = -2 Critical value (α = 0.01, two-tailed): ±2.576 |-2| < 2.576, so fail to reject H₀. Conclusion: There is not enough evidence to reject the company’s claim at the 1% significance level.

1).A cereal company claims that their boxes contain an average of 350 grams of cereal. A consumer group tests 100 boxes and finds a mean weight of 345 grams with a known population standard deviation of 15 grams. At a 5% significance level, is there evidence to refute the company’s claim?

2).A study compares the effect of two different diets on cholesterol levels. Diet A is tested on 50 people, resulting in a mean reduction of 25 mg/dL with a standard deviation of 8 mg/dL. Diet B is tested on 60 people, resulting in a mean reduction of 22 mg/dL with a standard deviation of 7 mg/dL. At a 1% significance level, is there a significant difference between the two diets?

3).A politician claims that 60% of voters in her district support her re-election. In a random sample of 1000 voters, 570 support her. At a 5% significance level, is there evidence to reject the politician’s claim?

4).Two different teaching methods are compared. Method A results in 80 students passing out of 120 students. Method B results in 90 students passing out of 150 students. At a 5% significance level, is there a difference in the effectiveness of the two methods?

5).A company claims that their new energy-saving light bulbs last an average of 10,000 hours. A sample of 64 bulbs has a mean life of 9,800 hours. The population standard deviation is known to be 500 hours. At a 1% significance level, is there evidence to reject the company’s claim?

6).The mean salary of employees in a large corporation is said to be $75,000 per year. A union representative suspects this is too high and surveys 100 randomly selected employees, finding a mean salary of $72,500. The population standard deviation is known to be $8,000. At a 5% significance level, is there evidence to support the union representative’s suspicion?

7).Two factories produce computer chips. Factory A’s chips have a mean processing speed of 3.2 GHz with a standard deviation of 0.2 GHz. Factory B’s chips have a mean processing speed of 3.3 GHz with a standard deviation of 0.25 GHz. Samples of 100 chips from each factory are tested. At a 5% significance level, is there a difference in mean processing speed between the two factories?

8).A new vaccine is claimed to be 90% effective. In a clinical trial with 500 participants, 440 develop immunity. At a 1% significance level, is there evidence to reject the claim about the vaccine’s effectiveness?

9).Two different advertising campaigns are tested. Campaign A results in 250 sales out of 2000 views. Campaign B results in 300 sales out of 2500 views. At a 5% significance level, is there a difference in the effectiveness of the two campaigns?

10).A quality control manager claims that the defect rate in a production line is 5%. In a sample of 1000 items, 65 are found to be defective. At a 5% significance level, is there evidence to suggest that the actual defect rate is different from the claimed 5%?

Type 1 error and Type II error

• Type I error: Type 1 error has occurred when we reject the null hypothesis, even when the hypothesis is true. This error is denoted by alpha.
• Type II error: Type II error occurred when we didn’t reject the null hypothesis, even when the hypothesis is false. This error is denoted by beta.

Z-tests are used to determine whether there is a statistically significant difference between a sample statistic and a population parameter, or between two population parameters.Z-tests are statistical tools used to determine if there’s a significant difference between a sample statistic and a population parameter, or between two population parameters. They’re applicable when dealing with large sample sizes (typically n > 30) and known population standard deviations. Z-tests can be used for analyzing means or proportions in both one-sample and two-sample scenarios. The process involves stating hypotheses, calculating a Z-score, comparing it to a critical value based on the chosen significance level (often 5% or 1%), and then making a decision to reject or fail to reject the null hypothesis.

What is the main limitation of the z-test?

The limitation of Z-Tests is that we don’t usually know the population standard deviation. What we do is: When we don’t know the population’s variability, we assume that the sample’s variability is a good basis for estimating the population’s variability.

What is the minimum sample for z-test?

A z-test can only be used if the population standard deviation is known and the sample size is 30 data points or larger. Otherwise, a t-test should be employed.

What is the application of z-test?

It is also used to determine if there is a significant difference between the mean of two independent samples. The z-test can also be used to compare the population proportion to an assumed proportion or to determine the difference between the population proportion of two samples.

What is the theory of the z-test?

The z test is a commonly used hypothesis test in inferential statistics that allows us to compare two populations using the mean values of samples from those populations, or to compare the mean of one population to a hypothesized value, when what we are interested in comparing is a continuous variable.

Please Login to comment...

Similar reads.

• Engineering Mathematics
• Mathematical
• How to Delete Discord Servers: Step by Step Guide
• Google increases YouTube Premium price in India: Check our the latest plans
• California Lawmakers Pass Bill to Limit AI Replicas
• Best 10 IPTV Service Providers in Germany
• 15 Most Important Aptitude Topics For Placements [2024]

Improve your Coding Skills with Practice

What kind of Experience do you want to share?

Introduction to Statistics and Data Analysis

Chapter 6 hypothesis testing: the z-test.

We’ve all had the experience of standing at a crosswalk waiting staring at a pedestrian traffic light showing the little red man. You’re waiting for the little green man so you can cross. After a little while you’re still waiting and there aren’t any cars around. You might think ‘this light is really taking a long time’, but you continue waiting. Minutes pass and there’s still no little green man. At some point you come to the conclusion that the light is broken and you’ll never see that little green man. You cross on the little red man when it’s clear.

You may not have known this but you just conducted a hypothesis test. When you arrived at the crosswalk, you assumed that the light was functioning properly, although you will always entertain the possibility that it’s broken. In terms of hypothesis testing, your ‘null hypothesis’ is that the light is working and your ‘alternative hypothesis’ is that it’s broken. As time passes, it seems less and less likely that light is working properly. Eventually, the probability of the light working given how long you’ve been waiting becomes so low that you reject the null hypothesis in favor of the alternative hypothesis.

This sort of reasoning is the backbone of hypothesis testing and inferential statistics. It’s also the point in the course where we turn the corner from descriptive statistics to inferential statistics. Rather than describing our data in terms of means and plots, we will now start using our data to make inferences, or generalizations, about the population that our samples are drawn from. In this course we’ll focus on standard hypothesis testing where we set up a null hypothesis and determine the probability of our observed data under the assumption that the null hypothesis is true (the much maligned p-value). If this probability is small enough, then we conclude that our data suggests that the null hypothesis is false, so we reject it.

In this chapter, we’ll introduce hypothesis testing with examples from a ‘z-test’, when we’re comparing a single mean to what we’d expect from a population with known mean and standard deviation. In this case, we can convert our observed mean into a z-score for the standard normal distribution. Hence the name z-test.

It’s time to introduce the hypothesis test flow chart . It’s pretty self explanatory, even if you’re not familiar with all of these hypothesis tests. The z-test is (1) based on means, (2) with only one mean, and (3) where we know \(\sigma\) , the standard deviation of the population. Here’s how to find the z-test in the flow chart:

6.1 Women’s height example

Let’s work with the example from the end of the last chapter where we started with the fact that the heights of US women has a mean of 63 and a standard deviation of 2.5 inches. We calculated that the average height of the 122 women in Psych 315 is 64.7 inches. We then used the central limit theorem and calculated the probability of a random sample 122 heights from this population having a mean of 64.7 or greater is 2.4868996^{-14}. This is a very, very small number.

Here’s how we do it using R:

Let’s think of our sample as a random sample of UW psychology students, which is a reasonable assumption since all psychology students have to take a statistics class. What does this sample say about the psychology students that are women at UW compared to the US population? It could be that these psychology students at UW have the same mean and standard deviation as the US population, but our sample just happens to have an unusual number of tall women, but we calculated that the probability of this happening is really low. Instead, it makes more sense that the population that we’re drawing from has a mean that’s greater than the US population mean. Notice that we’re making a conclusion about the whole population of women psychology students based on our one sample.

Using the terminology of hypothesis testing, we first assumed the null hypothesis that UW women psych students have the same mean (and standard deviation) as the US population. The null hypothesis is written as:

\[ H_{0}: \mu = 63 \] In this example, our alternative hypothesis is that the mean of our population is larger than the mean of null hypothesis population. We write this as:

\[ H_{A}: \mu > 63 \]

Next, after obtaining a random sample and calculate the mean, we calculate the probability of drawing a mean this large (or larger) from the null hypothesis distribution.

If this probability is low enough, we reject the null hypothesis in favor of the alternative hypothesis. When our probability allows us to reject the null hypothesis, we say that our observed results are ‘statistically significant’.

In statistics terms, we never say we ‘accept that alternative hypothesis’ as true. All we can say is that we don’t think the null hypothesis is true. I know it’s subtle, but in science can never prove that a hypothesis is true or not. There’s always the possibility that we just happened to grab an unusual sample from the null hypothesis distribution.

6.2 The hated p<.05

The probability that we obtain our observed mean or greater given that the null hypothesis is true is called the p-value. How improbable is improbable enough to reject the null hypothesis? The p-value for our example above on women’s heights is astronomically low, so it’s clear that we should reject \(H_{0}\) .

The p-value that’s on the border of rejection is called the alpha ( \(\alpha\) ) value. We reject \(H_{0}\) when our p-value is less than \(\alpha\) .

You probably know that the most common value of alpha is \(\alpha = .05\) .

The first publication of this value dates back to Sir Ronald Fisher, in his seminal 1925 book Statistical Methods for Research Workers where he states:

“It is convenient to take this point as a limit in judging whether a deviation is considered significant or not. Deviations exceeding twice the standard deviation are thus formally regarded as significant.” (p. 47)

If you read the chapter on the normal distribution, then you should know that 95% of the area under the normal distribution lies within \(\pm\) two standard deviations of the mean. So the probability of obtaining a sample that exceeds two standard deviations from the mean (in either direction) is .05.

6.3 IQ example

Let’s do an example using IQ scores. IQ scores are normalized to have a mean of 100 and a standard deviation of 15 points. Because they’re normalized, they are a rare example of a population which has a known mean and standard deviation. In the next chapter we’ll discuss the t-test, which is used in the more common situation when we don’t know the population standard deviation.

Suppose you have the suspicion that graduate students have higher IQ’s than the general population. You have enough time to go and measure the IQ’s of 25 randomly sampled grad students and obtain a mean of 105. Is this difference between our this observed mean and 100 statistically significant using an alpha value of \(\alpha = 0.05\) ?

Here the null hypothesis is:

\[ H_{0}: \mu = 100\]

And the alternative hypothesis is:

\[ H_{A}: \mu > 100 \]

We know that the parameters for the null hypothesis are:

\[ \mu = 100 \] and \[ \sigma = 15 \]

From this, we can calculate the probability of observing our mean of 105 or higher using the central limit theorem and what we know about the normal distribution:

\[ \sigma_{\bar{x}} = \frac{\sigma_{x}}{\sqrt{n}} = \frac{15}{\sqrt{25}} = 3 \] From this, we can calculate the probability of our observed mean using R’s ‘pnorm’ function. Here’s how to do the whole thing in R.

Since our p-value of 0.0478 is (just barely) less than our chosen value of \(\alpha = 0.05\) as our criterion, we reject \(H_{0}\) for this (contrived) example and conclude that our observed mean of 105 is significantly greater than 100, so our study suggests that the average graduate student has a higher IQ than the overall population.

You should feel uncomfortable making such a hard, binary decision for such a borderline case. After all, if we had chosen our second favorite value of alpha, \(\alpha = .01\) , we would have failed to reject \(H_{0}\) . This discomfort is a primary reason why statisticians are moving away from this discrete decision making process. Later on we’ll discuss where things are going, including reporting effect sizes, and using confidence intervals.

6.4 Alpha values vs. critical values

Using R’s qnorm function, we can find the z-score for which only 5% of the area lies above:

So the probability of a randomly sampled z-score exceeding 1.644854 is less than 5%. It follows that if we convert our observed mean into z-score values, we will reject \(H_{0}\) if and only if our z-score is greater than 1.644854. This value is called the ‘critical value’ because it lies on the boundary between rejecting and failing to reject \(H_{0}\) .

In our last example, the z-score for our observed mean is:

\[ z = \frac{X-\mu}{\frac{\sigma}{\sqrt{n}}} = \frac{105 - 100}{3} = 1.67 \] Our z-score is just barely greater than the critical value of 1.644854, which makes sense because our p-value is just barely less than 0.05.

Sometimes you’ll see textbooks will compare critical values to observed scores for the decision making process in hypothesis testing. This dates back to days were computers were less available and we had to rely on tables instead. There wasn’t enough space in a book to hold complete tables which prohibited the ability to look up a p-value for any observed value. Instead only critical values for specific values of alpha were included. If you look at really old papers, you’ll see statistics reported as \(p<.05\) or \(p<.01\) instead of actual p-values for this reason.

It may help to visualize the relationship between p-values, alpha values and critical values like this:

The red shaded region is the upper 5% of the standard normal distribution which starts at the critical value of z=1.644854. This is sometimes called the ‘rejection region’. The blue vertical line is drawn at our observed value of z=1.67. You can see that the red line falls just inside the rejection region, so we Reject \(H_{0}\) !

6.5 One vs. two-tailed tests

Recall that our alternative hypothesis was to reject if our mean IQ was significantly greater than the null hypothesis mean: \(H_{A}: \mu > 100\) . This implies that the situation where \(\mu < 100\) is never even in consideration, which is weird. In science, we’re trying to understand the true state of the world. Although we have a hunch that grad student IQ’s are higher than average, there is always the possibility that they are lower than average. If our sample came up with an IQ well below 100, we’d simply fail to reject \(H_{0}\) and move on. This feels like throwing out important information.

The test we just ran is called a ‘one-tailed’ test because we only reject \(H_{0}\) if our results fall in one of the two tails of the population distribution.

Instead, it might make more sense to reject \(H_{0}\) if we get either an unusually large or small score. This means we need two critical values - one above and one below zero. At first thought you might think we just duplicate our critical value from a one-tailed test to the other side. But will double the area of the rejection region. That’s not a good thing because if \(H_{0}\) is true, there’s actually a \(2\alpha\) probability that we’ll draw a score in the rejection region.

Instead, we divide the area into two tails, each containing an area of \(\frac{\alpha}{2}\) . For \(\alpha\) = 0.05, we can find the critical value of z with qnorm:

So with a two-tailed test at \(\alpha = 0.05\) we reject \(H_{0}\) if our observed z-score is either above z = 1.96 or less than -1.96. This is that value around 2 that Sir Ronald Fischer was talking about!

Here’s what the critical regions and observed value of z looks like for our example with a two-tailed test:

You can see that splitting the area of \(\alpha = 0.05\) into two halves increased the critical value in the positive direction from 1.64 to 1.96, making it harder to reject \(H_{0}\) . For our example, this changes our decision: our observed value of z = 1.67 no longer falls into the rejection region, so now we fail to reject \(H_{0}\) .

If we now fail to reject \(H_{0}\) , what about the p-value? Remember, for a one-tailed test, p = \(\alpha\) if our observed z-score lands right on the critical value of z. The same is true for a two-tailed test. But the z-score moved so that the area above that score is \(\frac{\alpha}{2}\) . So for a two-tailed test, in order to have a p-value of \(\alpha\) when our z-score lands right on the critical value, we need to double p-value hat we’d get for a one-tailed test.

For our example, the p-value for the one tailed test was \(p=0.0478\) . So if we use a two-tailed test, our p-value is \((2)(0.0478) = 0.0956\) . This value is greater than \(\alpha\) = 0.05, which makes sense because we just showed above that we fail to reject \(H_{0}\) with a two tailed test.

Which is the right test, one-tailed or two-tailed? Ideally, as scientists, we should be agnostic about the results of our experiment. But in reality, we all know that the results are more interesting if they are statistically significant. So you can imagine that for this example, given a choice between one and two-tailed, we’d choose a one-tailed test so that we can reject \(H_{0}\) .

There are two problems with this. First, we should never adjust our choice of hypothesis test after we observe the data. That would be an example of ‘p-hacking’, a topic we’ll discuss later. Second, most statisticians these days strongly recommend against one-tailed tests. The only reason for a one-tailed test is if there is no logical or physical possibility for a population mean to fall below the null hypothesis mean.

Z-Test for Statistical Hypothesis Testing Explained

The Z-test is a statistical hypothesis test that determines where the distribution of the statistic we are measuring, like the mean, is part of the normal distribution.

The Z-test is a statistical hypothesis test used to determine where the distribution of the test statistic we are measuring, like the mean , is part of the normal distribution .

There are multiple types of Z-tests, however, we’ll focus on the easiest and most well known one, the one sample mean test. This is used to determine if the difference between the mean of a sample and the mean of a population is statistically significant.

What Is a Z-Test?

A Z-test is a type of statistical hypothesis test where the test-statistic follows a normal distribution.  

The name Z-test comes from the Z-score of the normal distribution. This is a measure of how many standard deviations away a raw score or sample statistics is from the populations’ mean.

Z-tests are the most common statistical tests conducted in fields such as healthcare and data science . Therefore, it’s an essential concept to understand.

Requirements for a Z-Test

In order to conduct a Z-test, your statistics need to meet a few requirements, including:

• A Sample size that’s greater than 30. This is because we want to ensure our sample mean comes from a distribution that is normal. As stated by the c entral limit theorem , any distribution can be approximated as normally distributed if it contains more than 30 data points.
• The standard deviation and mean of the population is known .
• The sample data is collected/acquired randomly .

More on Data Science:   What Is Bootstrapping Statistics?

Z-Test Steps

There are four steps to complete a Z-test. Let’s examine each one.

4 Steps to a Z-Test

• State the null hypothesis.
• State the alternate hypothesis.
• Choose your critical value.
• Calculate your Z-test statistics. 

1. State the Null Hypothesis

The first step in a Z-test is to state the null hypothesis, H_0 . This what you believe to be true from the population, which could be the mean of the population, μ_0 :

2. State the Alternate Hypothesis

Next, state the alternate hypothesis, H_1 . This is what you observe from your sample. If the sample mean is different from the population’s mean, then we say the mean is not equal to μ_0:

3. Choose Your Critical Value

Then, choose your critical value, α , which determines whether you accept or reject the null hypothesis. Typically for a Z-test we would use a statistical significance of 5 percent which is z = +/- 1.96 standard deviations from the population’s mean in the normal distribution:

This critical value is based on confidence intervals.

4. Calculate Your Z-Test Statistic

Compute the Z-test Statistic using the sample mean, μ_1 , the population mean, μ_0 , the number of data points in the sample, n and the population’s standard deviation, σ :

If the test statistic is greater (or lower depending on the test we are conducting) than the critical value, then the alternate hypothesis is true because the sample’s mean is statistically significant enough from the population mean.

Another way to think about this is if the sample mean is so far away from the population mean, the alternate hypothesis has to be true or the sample is a complete anomaly.

More on Data Science: Basic Probability Theory and Statistics Terms to Know

Z-Test Example

Let’s go through an example to fully understand the one-sample mean Z-test.

A school says that its pupils are, on average, smarter than other schools. It takes a sample of 50 students whose average IQ measures to be 110. The population, or the rest of the schools, has an average IQ of 100 and standard deviation of 20. Is the school’s claim correct?

The null and alternate hypotheses are:

Where we are saying that our sample, the school, has a higher mean IQ than the population mean.

Now, this is what’s called a right-sided, one-tailed test as our sample mean is greater than the population’s mean. So, choosing a critical value of 5 percent, which equals a Z-score of 1.96 , we can only reject the null hypothesis if our Z-test statistic is greater than 1.96.

If the school claimed its students’ IQs were an average of 90, then we would use a left-tailed test, as shown in the figure above. We would then only reject the null hypothesis if our Z-test statistic is less than -1.96.

Computing our Z-test statistic, we see:

Therefore, we have sufficient evidence to reject the null hypothesis, and the school’s claim is right.

Hope you enjoyed this article on Z-tests. In this post, we only addressed the most simple case, the one-sample mean test. However, there are other types of tests, but they all follow the same process just with some small nuances.  

Recent Data Science Articles

Microbe Notes

Z-Test: Formula, Examples, Uses, Z-Test vs T-Test

Table of Contents

Interesting Science Videos

Z-test Definition

z-test is a statistical tool used for the comparison or determination of the significance of several statistical measures, particularly the mean in a sample from a normally distributed population or between two independent samples.

• Like t-tests, z tests are also based on normal probability distribution.
• Z-test is the most commonly used statistical tool in research methodology, with it being used for studies where the sample size is large (n>30).
• In the case of the z-test, the variance is usually known.
• Z-test is more convenient than t-test as the critical value at each significance level in the confidence interval is the sample for all sample sizes.
• A z-score is a number indicating how many standard deviations above or below the mean of the population is.

Z-test formula

For the normal population with one sample:

where x̄    is the mean of the sample, and µ is the assumed mean, σ is the standard deviation, and n is the number of observations.

z-test for the difference in mean: 

where x̄ 1 and x̄ 2 are the means of two samples, σ is the standard deviation of the samples, and n1 and n2 are the numbers of observations of two samples.

One sample z-test (one-tailed z-test)

• One sample z-test is used to determine whether a particular population parameter, which is mostly mean, significantly different from an assumed value.
• It helps to estimate the relationship between the mean of the sample and the assumed mean.
• In this case, the standard normal distribution is used to calculate the critical value of the test.
• If the z-value of the sample being tested falls into the criteria for the one-sided tets, the alternative hypothesis will be accepted instead of the null hypothesis.
• A one-tailed test would be used when the study has to test whether the population parameter being tested is either lower than or higher than some hypothesized value.
• A one-sample z-test assumes that data are a random sample collected from a normally distributed population that all have the same mean and same variance.
• This hypothesis implies that the data is continuous, and the distribution is symmetric.
• Based on the alternative hypothesis set for a study, a one-sided z-test can be either a left-sided z-test or a right-sided z-test. 
• For instance, if our H 0 : µ 0 = µ and H a : µ < µ 0 , such a test would be a one-sided test or more precisely, a left-tailed test and there is one rejection area only on the left tail of the distribution.
• However, if H 0 : µ = µ 0 and H a : µ > µ 0 , this is also a one-tailed test (right tail), and the rejection region is present on the right tail of the curve.

Two sample z-test (two-tailed z-test)

• In the case of two sample z-test, two normally distributed independent samples are required.
• A two-tailed z-test is performed to determine the relationship between the population parameters of the two samples.
• In the case of the two-tailed z-test, the alternative hypothesis is accepted as long as the population parameter is not equal to the assumed value.
• The two-tailed test is appropriate when we have H 0 : µ = µ 0 and H a : µ ≠ µ 0 which may mean µ > µ 0 or µ < µ 0
• Thus, in a two-tailed test, there are two rejection regions, one on each tail of the curve.

Z-test examples

If a sample of 400 male workers has a mean height of 67.47 inches, is it reasonable to regard the sample as a sample from a large population with a mean height of 67.39 inches and a standard deviation of 1.30 inches at a 5% level of significance?

Taking the null hypothesis that the mean height of the population is equal to 67.39 inches, we can write:                           

H 0 : µ = 67 . 39 “

H a : µ ≠ 67 . 39 “

x̄ = 67 . 47 “, σ = 1 . 30 “, n = 400

Assuming the population to be normal, we can work out the test statistic z as under:

z-test applications

• Z-test is performed in studies where the sample size is larger, and the variance is known.
• It is also used to determine if there is a significant difference between the mean of two independent samples.
• The z-test can also be used to compare the population proportion to an assumed proportion or to determine the difference between the population proportion of two samples.

Z-test vs T-test (8 major differences)

References and Sources

• C.R. Kothari (1990) Research Methodology. Vishwa Prakasan. India.
• https://ncss-wpengine.netdna-ssl.com/wp-content/themes/ncss/pdf/Procedures/PASS/One-Sample_Z-Tests.pdf
• https://www.wallstreetmojo.com/z-test-vs-t-test/
• https://sites.google.com/site/fundamentalstatistics/chapter-13
• 3% – https://www.investopedia.com/terms/z/z-test.asp
• 2% – https://www.coursehero.com/file/61052903/Questions-statisticswpdf/
• 2% – https://towardsdatascience.com/everything-you-need-to-know-about-hypothesis-testing-part-i-4de9abebbc8a
• 2% – https://ncss-wpengine.netdna-ssl.com/wp-content/themes/ncss/pdf/Procedures/PASS/One-Sample_Z-Tests.pdf
• 1% – https://www.slideshare.net/MuhammadAnas96/ztest-with-examples
• 1% – https://www.mathandstatistics.com/learn-stats/hypothesis-testing/two-tailed-z-test-hypothesis-test-by-hand
• 1% – https://www.infrrr.com/proportions/difference-in-proportions-hypothesis-test-calculator
• 1% – https://keydifferences.com/difference-between-t-test-and-z-test.html
• 1% – https://en.wikipedia.org/wiki/Z-test
• 1% – http://www.sci.utah.edu/~arpaiva/classes/UT_ece3530/hypothesis_testing.pdf
• <1% – https://www.researchgate.net/post/Can-a-null-hypothesis-be-stated-as-a-difference
• <1% – https://www.isixsigma.com/tools-templates/hypothesis-testing/making-sense-two-sample-t-test/
• <1% – https://www.investopedia.com/terms/t/two-tailed-test.asp
• <1% – https://www.academia.edu/24313503/BIOSTATISTICS_AND_RESEARCH_METHODS_IN_PHARMACY_Pharmacy_C479_4_quarter_credits_A_Course_for_Distance_Learning_Prepared

About Author

Anupama Sapkota

2 thoughts on “Z-Test: Formula, Examples, Uses, Z-Test vs T-Test”

The formula for Z test provided for testing the single mean is wrong. The correct formula is wrong. Please check and correct it. It should be Z = (𝑥̅−𝜇)/𝜎/√n

Hi Ramnath, Sorry for the mistake. Thank you so much for the correction. We have updated the page with correct formula.

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Z Test: Definition & Two Proportion Z-Test

What is a z test.

For example, if someone said they had found a new drug that cures cancer, you would want to be sure it was probably true. A hypothesis test will tell you if it’s probably true, or probably not true. A Z test, is used when your data is approximately normally distributed (i.e. the data has the shape of a bell curve when you graph it).

When you can run a Z Test.

Several different types of tests are used in statistics (i.e. f test , chi square test , t test ). You would use a Z test if:

• Your sample size is greater than 30 . Otherwise, use a t test .
• Data points should be independent from each other. In other words, one data point isn’t related or doesn’t affect another data point.
• Your data should be normally distributed . However, for large sample sizes (over 30) this doesn’t always matter.
• Your data should be randomly selected from a population, where each item has an equal chance of being selected.
• Sample sizes should be equal if at all possible.

How do I run a Z Test?

Running a Z test on your data requires five steps:

• State the null hypothesis and alternate hypothesis .
• Choose an alpha level .
• Find the critical value of z in a z table .
• Calculate the z test statistic (see below).
• Compare the test statistic to the critical z value and decide if you should support or reject the null hypothesis .

You could perform all these steps by hand. For example, you could find a critical value by hand , or calculate a z value by hand . For a step by step example, watch the following video: Watch the video for an example:

Can’t see the video? Click here to watch it on YouTube. You could also use technology, for example:

• Two sample z test in Excel .
• Find a critical z value on the TI 83 .
• Find a critical value on the TI 89 (left-tail) .

Two Proportion Z-Test

Watch the video to see a two proportion z-test:

Can’t see the video? Click here to watch it on YouTube.

A Two Proportion Z-Test (or Z-interval) allows you to calculate the true difference in proportions of two independent groups to a given confidence interval .

There are a few familiar conditions that need to be met for the Two Proportion Z-Interval to be valid.

• The groups must be independent. Subjects can be in one group or the other, but not both – like teens and adults.
• The data must be selected randomly and independently from a homogenous population. A survey is a common example.
• The population should be at least ten times bigger than the sample size. If the population is teenagers for example, there should be at least ten times as many total teenagers as the number of teenagers being surveyed.
• The null hypothesis (H 0 ) for the test is that the proportions are the same.
• The alternate hypothesis (H 1 ) is that the proportions are not the same.

Example question: let’s say you’re testing two flu drugs A and B. Drug A works on 41 people out of a sample of 195. Drug B works on 351 people in a sample of 605. Are the two drugs comparable? Use a 5% alpha level .

Step 1: Find the two proportions:

• P 1 = 41/195 = 0.21 (that’s 21%)
• P 2 = 351/605 = 0.58 (that’s 58%).

Set these numbers aside for a moment.

Step 2: Find the overall sample proportion . The numerator will be the total number of “positive” results for the two samples and the denominator is the total number of people in the two samples.

• p = (41 + 351) / (195 + 605) = 0.49.

Set this number aside for a moment.

Solving the formula, we get: Z = 8.99

We need to find out if the z-score falls into the “ rejection region .”

Step 5: Compare the calculated z-score from Step 3 with the table z-score from Step 4. If the calculated z-score is larger, you can reject the null hypothesis.

8.99 > 1.96, so we can reject the null hypothesis .

Example 2:  Suppose that in a survey of 700 women and 700 men, 35% of women and 30% of men indicated that they support a particular presidential candidate. Let’s say we wanted to find the true difference in proportions of these two groups to a 95% confidence interval .

At first glance the survey indicates that women support the candidate more than men by about 5% . However, for this statistical inference to be valid we need to construct a range of values to a given confidence interval.

To do this, we use the formula for Two Proportion Z-Interval:

Plugging in values we find the true difference in proportions to be

Based on the results of the survey, we are 95% confident that the difference in proportions of women and men that support the presidential candidate is between about 0 % and 10% .

Check out our YouTube channel for more stats help and tips!

Hypothesis Testing

Hypothesis Tests (or Significance Tests) are statistical tests to see if a difference we observe is due to chance.

There are many different types of hypothesis tests for different scenarios, but they all have the same basic ideas. Below are the general steps to performing a hypothesis test:

• Formulate your Null and Alternative Hypotheses.
• Ho- Null Hypothesis : The null hypothesis is the hypothesis of no effect. It's the dull, boring hypothesis that says that nothing interesting is going on. If we are trying to test if a difference we observe is due to chance, the null says it is!
• Ha- Alternative Hypothesis : The alternative hypothesis is the opposite of the null. It's what you are trying to test. If we are trying to test if a difference we observe is due to chance, the alternative says it is not!

Think about what you would expect to get if you randomly sampled from the population, assuming the null is true. Compare your observed data and expected data and calculate the test statistic .

Calculate the probability of getting the data you got or something even more extreme if the null were true. This is called the p-value .

• Make your conclusion and interpret it in the context of the problem. If p is very low, we say that the data support rejecting the null hypothesis.

How low is “very low”?

The convention is to reject the null when P < 5% (P < 0.05) and call the result “significant”. There’s no particular justification for this value but it’s commonly used.

The P-value cut-off is called the significance level and is often represented by the Greek letter alpha (α).

The One Sample Z Test: One-sided Hypothesis

The first type of hypothesis test we are going to look at is the one-sample z-test. You can do a z-test for means or for proportions. This is the most simple type of hypothesis test and it uses z-scores and the normal curve. Let’s look at one below!

Hypothesis Test Example : Suppose a large university claims that the average ACT score of their incoming freshman class is 30, but we think the University may be inflating their average. To test the University’s claim we take a simple random sample of 50 students and find their average to be only 28.3 with an SD of 4. Perform a hypothesis test to test the claim. Here are the 4 steps:

• This can be written in symbols as well: Ho: μ = 30
• μ is the symbol for the population mean
• This can be written in symbols as well: Ha: μ < 30
• Our test statistic for the one sample z test is z! We can calculate z using our z-score formula for random variables since we are dealing with a sample of 50 students.

• In our case, the expected value (EV) is 30 since we are assuming our null hypothesis is true (until proven otherwise).
• Since we are dealing with means, our SE is found using the following formula:

Our z-score is -3. See the calculation below:

• Calculate the probability of getting the data you got or something even more extreme if the null were true. This is called the p-value . In this case, our p-value is going to be the area to the left of z = -3. We can use Python to calculate this by using norm.cdf(-3).
• We get that the p-value is 0.0013.
• This is the probability that we would get a sample average of 28.3 given that the null hypothesis was true (the true average was 30).

• Our p-value is less than 5% so we reject our Null Hypothesis. In other words, there is evidence of the Alternative Hypothesis (that the University is inflating their average).

The One Sample Z Test: Two-sided Hypothesis

Hypothesis Test Example : Now we're going to test the above claim but with a different alternative hypothesis. The large university still claims that the average ACT score of their incoming freshman class is 30, but now we think the University may be inflating or deflating their average. To test the University’s claim we take a simple random sample of 50 students and find their average to be only 28.3 with an SD of 4. Perform a hypothesis test to test the claim with our new alternative hypothesis. Here are the 4 steps:

• This can be written in symbols as well: Ho: μ != 30

Step 2 is the same as the one-sided example, so our z score is still -3.

Calculate the probability of getting the data you got or something even more extreme if the null were true. This is called the p-value . In this case, our p-value is going to be the area to the left or right of z = -3. We can use Python to calculate this by using 2*norm.cdf(-3).

• We get that the p-value is 0.0027.
• Our p-value is less than 5% so we reject our Null Hypothesis. In other words, there is evidence of the Alternative Hypothesis (that the University is inflating or deflating their average).

Example Walk-Throughs with Worksheets

Video 1: one sample z-test examples.

• Download Blank Worksheet (PDF)

Video 2: Two Sample z-test Examples

Video 3: z-tests in Python

Video 4: One Sample t-test Examples

Video 5: t-tests in Python

Hypothesis Testing

Hypothesis testing is a tool for making statistical inferences about the population data. It is an analysis tool that tests assumptions and determines how likely something is within a given standard of accuracy. Hypothesis testing provides a way to verify whether the results of an experiment are valid.

A null hypothesis and an alternative hypothesis are set up before performing the hypothesis testing. This helps to arrive at a conclusion regarding the sample obtained from the population. In this article, we will learn more about hypothesis testing, its types, steps to perform the testing, and associated examples.

What is Hypothesis Testing in Statistics?

Hypothesis testing uses sample data from the population to draw useful conclusions regarding the population probability distribution . It tests an assumption made about the data using different types of hypothesis testing methodologies. The hypothesis testing results in either rejecting or not rejecting the null hypothesis.

Hypothesis Testing Definition

Hypothesis testing can be defined as a statistical tool that is used to identify if the results of an experiment are meaningful or not. It involves setting up a null hypothesis and an alternative hypothesis. These two hypotheses will always be mutually exclusive. This means that if the null hypothesis is true then the alternative hypothesis is false and vice versa. An example of hypothesis testing is setting up a test to check if a new medicine works on a disease in a more efficient manner.

Null Hypothesis

The null hypothesis is a concise mathematical statement that is used to indicate that there is no difference between two possibilities. In other words, there is no difference between certain characteristics of data. This hypothesis assumes that the outcomes of an experiment are based on chance alone. It is denoted as \(H_{0}\). Hypothesis testing is used to conclude if the null hypothesis can be rejected or not. Suppose an experiment is conducted to check if girls are shorter than boys at the age of 5. The null hypothesis will say that they are the same height.

Alternative Hypothesis

The alternative hypothesis is an alternative to the null hypothesis. It is used to show that the observations of an experiment are due to some real effect. It indicates that there is a statistical significance between two possible outcomes and can be denoted as \(H_{1}\) or \(H_{a}\). For the above-mentioned example, the alternative hypothesis would be that girls are shorter than boys at the age of 5.

Hypothesis Testing P Value

In hypothesis testing, the p value is used to indicate whether the results obtained after conducting a test are statistically significant or not. It also indicates the probability of making an error in rejecting or not rejecting the null hypothesis.This value is always a number between 0 and 1. The p value is compared to an alpha level, \(\alpha\) or significance level. The alpha level can be defined as the acceptable risk of incorrectly rejecting the null hypothesis. The alpha level is usually chosen between 1% to 5%.

Hypothesis Testing Critical region

All sets of values that lead to rejecting the null hypothesis lie in the critical region. Furthermore, the value that separates the critical region from the non-critical region is known as the critical value.

Hypothesis Testing Formula

Depending upon the type of data available and the size, different types of hypothesis testing are used to determine whether the null hypothesis can be rejected or not. The hypothesis testing formula for some important test statistics are given below:

• z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). \(\overline{x}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation and n is the size of the sample.
• t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\). s is the sample standard deviation.
• \(\chi ^{2} = \sum \frac{(O_{i}-E_{i})^{2}}{E_{i}}\). \(O_{i}\) is the observed value and \(E_{i}\) is the expected value.

We will learn more about these test statistics in the upcoming section.

Types of Hypothesis Testing

Selecting the correct test for performing hypothesis testing can be confusing. These tests are used to determine a test statistic on the basis of which the null hypothesis can either be rejected or not rejected. Some of the important tests used for hypothesis testing are given below.

Hypothesis Testing Z Test

A z test is a way of hypothesis testing that is used for a large sample size (n ≥ 30). It is used to determine whether there is a difference between the population mean and the sample mean when the population standard deviation is known. It can also be used to compare the mean of two samples. It is used to compute the z test statistic. The formulas are given as follows:

• One sample: z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
• Two samples: z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing t Test

The t test is another method of hypothesis testing that is used for a small sample size (n < 30). It is also used to compare the sample mean and population mean. However, the population standard deviation is not known. Instead, the sample standard deviation is known. The mean of two samples can also be compared using the t test.

• One sample: t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\).
• Two samples: t = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\).

Hypothesis Testing Chi Square

The Chi square test is a hypothesis testing method that is used to check whether the variables in a population are independent or not. It is used when the test statistic is chi-squared distributed.

One Tailed Hypothesis Testing

One tailed hypothesis testing is done when the rejection region is only in one direction. It can also be known as directional hypothesis testing because the effects can be tested in one direction only. This type of testing is further classified into the right tailed test and left tailed test.

Right Tailed Hypothesis Testing

The right tail test is also known as the upper tail test. This test is used to check whether the population parameter is greater than some value. The null and alternative hypotheses for this test are given as follows:

\(H_{0}\): The population parameter is ≤ some value

\(H_{1}\): The population parameter is > some value.

If the test statistic has a greater value than the critical value then the null hypothesis is rejected

Left Tailed Hypothesis Testing

The left tail test is also known as the lower tail test. It is used to check whether the population parameter is less than some value. The hypotheses for this hypothesis testing can be written as follows:

\(H_{0}\): The population parameter is ≥ some value

\(H_{1}\): The population parameter is < some value.

The null hypothesis is rejected if the test statistic has a value lesser than the critical value.

Two Tailed Hypothesis Testing

In this hypothesis testing method, the critical region lies on both sides of the sampling distribution. It is also known as a non - directional hypothesis testing method. The two-tailed test is used when it needs to be determined if the population parameter is assumed to be different than some value. The hypotheses can be set up as follows:

\(H_{0}\): the population parameter = some value

\(H_{1}\): the population parameter ≠ some value

The null hypothesis is rejected if the test statistic has a value that is not equal to the critical value.

Hypothesis Testing Steps

Hypothesis testing can be easily performed in five simple steps. The most important step is to correctly set up the hypotheses and identify the right method for hypothesis testing. The basic steps to perform hypothesis testing are as follows:

• Step 1: Set up the null hypothesis by correctly identifying whether it is the left-tailed, right-tailed, or two-tailed hypothesis testing.
• Step 2: Set up the alternative hypothesis.
• Step 3: Choose the correct significance level, \(\alpha\), and find the critical value.
• Step 4: Calculate the correct test statistic (z, t or \(\chi\)) and p-value.
• Step 5: Compare the test statistic with the critical value or compare the p-value with \(\alpha\) to arrive at a conclusion. In other words, decide if the null hypothesis is to be rejected or not.

Hypothesis Testing Example

The best way to solve a problem on hypothesis testing is by applying the 5 steps mentioned in the previous section. Suppose a researcher claims that the mean average weight of men is greater than 100kgs with a standard deviation of 15kgs. 30 men are chosen with an average weight of 112.5 Kgs. Using hypothesis testing, check if there is enough evidence to support the researcher's claim. The confidence interval is given as 95%.

Step 1: This is an example of a right-tailed test. Set up the null hypothesis as \(H_{0}\): \(\mu\) = 100.

Step 2: The alternative hypothesis is given by \(H_{1}\): \(\mu\) > 100.

Step 3: As this is a one-tailed test, \(\alpha\) = 100% - 95% = 5%. This can be used to determine the critical value.

1 - \(\alpha\) = 1 - 0.05 = 0.95

0.95 gives the required area under the curve. Now using a normal distribution table, the area 0.95 is at z = 1.645. A similar process can be followed for a t-test. The only additional requirement is to calculate the degrees of freedom given by n - 1.

Step 4: Calculate the z test statistic. This is because the sample size is 30. Furthermore, the sample and population means are known along with the standard deviation.

z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).

\(\mu\) = 100, \(\overline{x}\) = 112.5, n = 30, \(\sigma\) = 15

z = \(\frac{112.5-100}{\frac{15}{\sqrt{30}}}\) = 4.56

Step 5: Conclusion. As 4.56 > 1.645 thus, the null hypothesis can be rejected.

Hypothesis Testing and Confidence Intervals

Confidence intervals form an important part of hypothesis testing. This is because the alpha level can be determined from a given confidence interval. Suppose a confidence interval is given as 95%. Subtract the confidence interval from 100%. This gives 100 - 95 = 5% or 0.05. This is the alpha value of a one-tailed hypothesis testing. To obtain the alpha value for a two-tailed hypothesis testing, divide this value by 2. This gives 0.05 / 2 = 0.025.

Related Articles:

• Probability and Statistics
• Data Handling

Important Notes on Hypothesis Testing

• Hypothesis testing is a technique that is used to verify whether the results of an experiment are statistically significant.
• It involves the setting up of a null hypothesis and an alternate hypothesis.
• There are three types of tests that can be conducted under hypothesis testing - z test, t test, and chi square test.
• Hypothesis testing can be classified as right tail, left tail, and two tail tests.

Examples on Hypothesis Testing

• Example 1: The average weight of a dumbbell in a gym is 90lbs. However, a physical trainer believes that the average weight might be higher. A random sample of 5 dumbbells with an average weight of 110lbs and a standard deviation of 18lbs. Using hypothesis testing check if the physical trainer's claim can be supported for a 95% confidence level. Solution: As the sample size is lesser than 30, the t-test is used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) > 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 5, s = 18. \(\alpha\) = 0.05 Using the t-distribution table, the critical value is 2.132 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = 2.484 As 2.484 > 2.132, the null hypothesis is rejected. Answer: The average weight of the dumbbells may be greater than 90lbs
• Example 2: The average score on a test is 80 with a standard deviation of 10. With a new teaching curriculum introduced it is believed that this score will change. On random testing, the score of 38 students, the mean was found to be 88. With a 0.05 significance level, is there any evidence to support this claim? Solution: This is an example of two-tail hypothesis testing. The z test will be used. \(H_{0}\): \(\mu\) = 80, \(H_{1}\): \(\mu\) ≠ 80 \(\overline{x}\) = 88, \(\mu\) = 80, n = 36, \(\sigma\) = 10. \(\alpha\) = 0.05 / 2 = 0.025 The critical value using the normal distribution table is 1.96 z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) z = \(\frac{88-80}{\frac{10}{\sqrt{36}}}\) = 4.8 As 4.8 > 1.96, the null hypothesis is rejected. Answer: There is a difference in the scores after the new curriculum was introduced.
• Example 3: The average score of a class is 90. However, a teacher believes that the average score might be lower. The scores of 6 students were randomly measured. The mean was 82 with a standard deviation of 18. With a 0.05 significance level use hypothesis testing to check if this claim is true. Solution: The t test will be used. \(H_{0}\): \(\mu\) = 90, \(H_{1}\): \(\mu\) < 90 \(\overline{x}\) = 110, \(\mu\) = 90, n = 6, s = 18 The critical value from the t table is -2.015 t = \(\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}\) t = \(\frac{82-90}{\frac{18}{\sqrt{6}}}\) t = -1.088 As -1.088 > -2.015, we fail to reject the null hypothesis. Answer: There is not enough evidence to support the claim.

go to slide go to slide go to slide

Book a Free Trial Class

FAQs on Hypothesis Testing

What is hypothesis testing.

Hypothesis testing in statistics is a tool that is used to make inferences about the population data. It is also used to check if the results of an experiment are valid.

What is the z Test in Hypothesis Testing?

The z test in hypothesis testing is used to find the z test statistic for normally distributed data . The z test is used when the standard deviation of the population is known and the sample size is greater than or equal to 30.

What is the t Test in Hypothesis Testing?

The t test in hypothesis testing is used when the data follows a student t distribution . It is used when the sample size is less than 30 and standard deviation of the population is not known.

What is the formula for z test in Hypothesis Testing?

The formula for a one sample z test in hypothesis testing is z = \(\frac{\overline{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\) and for two samples is z = \(\frac{(\overline{x_{1}}-\overline{x_{2}})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}\).

What is the p Value in Hypothesis Testing?

The p value helps to determine if the test results are statistically significant or not. In hypothesis testing, the null hypothesis can either be rejected or not rejected based on the comparison between the p value and the alpha level.

What is One Tail Hypothesis Testing?

When the rejection region is only on one side of the distribution curve then it is known as one tail hypothesis testing. The right tail test and the left tail test are two types of directional hypothesis testing.

What is the Alpha Level in Two Tail Hypothesis Testing?

To get the alpha level in a two tail hypothesis testing divide \(\alpha\) by 2. This is done as there are two rejection regions in the curve.

Approximate Hypothesis Tests: the z Test and the t Test

This chapter presents two common tests of the hypothesis that a population mean equals a particular value and of the hypothesis that two population means are equal: the z test and the t test. These tests are approximate : They are based on approximations to the probability distribution of the test statistic when the null hypothesis is true, so their significance levels are not exactly what they claim to be. If the sample size is reasonably large and the population from which the sample is drawn has a nearly normal distribution —a notion defined in this chapter—the nominal significance levels of the tests are close to their actual significance levels. If these conditions are not met, the significance levels of the approximate tests can differ substantially from their nominal values. The z test is based on the normal approximation ; the t test is based on Student's t curve, which approximates some probability histograms better than the normal curve does. The chapter also presents the deep connection between hypothesis tests and confidence intervals, and shows how to compute approximate confidence intervals for the population mean of nearly normal populations using Student's t -curve.

where \(\phi\) is the pooled sample percentage of the two samples. The estimate of \(SE(\phi^{t-c})\) under the null hypothesis is

\[ se = s^*\times(1/n_t + 1/n_c)^{1/2}, \]

where \(n_t\) and \(n_c\) are the sizes of the two samples. If the null hypothesis is true, the Z statistic,

\[ Z=\phi^{t-c}/se, \]

is the original test statistic \(\phi^{t-c}\) in approximately standard units , and Z has a probability histogram that is approximated well by the normal curve , which allowed us to select the rejection region for the approximate test.

This strategy—transforming a test statistic approximately to standard units under the assumption that the null hypothesisis true, and then using the normal approximation to determine the rejection region for the test—works to construct approximate hypothesis tests in many other situations, too. The resulting hypothesis test is called a z test. Suppose that we are testing a null hypothesis using a test statistic \(X\) , and the following conditions hold:

• We have a probability model for how the observations arise, assuming the null hypothesis is true. Typically, the model is that under the null hypothesis, the data are like random draws with or without replacement from a box of numbered tickets.
• Under the null hypothesis, the test statistic \(X\) , converted to standard units, has a probability histogram that can be approximated well by the normal curve.
• Under the null hypothesis, we can find the expected value of the test statistic, \(E(X)\) .
• Under the null hypothesis, either we can find the SE of the test statistic, \(SE(X)\) , or we can estimate \(SE(X)\) accurately enough to ignore the error of the estimate of the SE. Let se denote either the exact SE of \(X\) under the null hypothesis, or the estimated value of \(SE(X)\) under the null hypothesis.

Then, under the null hypothesis, the probability histogram of the Z statistic

\[ Z = (X-E(X))/se \]

is approximated well by the normal curve, and we can use the normal approximation to select the rejection region for the test using \(Z\) as the test statistic. If the null hypothesis is true,

\[ P(Z < z_a) \approx a \]

\[ P(Z > z_{1-a} ) \approx a, \]

\[ P(|Z| > z_{1-a/2} ) \approx a. \]

These three approximations yield three different z tests of the hypothesis that \(\mu = \mu_0\) at approximate significance level \(a\) :

• Reject the null hypothesis whenever \(Z (left-tail z test)
• Reject the null hypothesis whenever \(Z > z_{1-a}\) (right-tail z test)
• Reject the null hypothesis whenever \(|Z|> z_{1-a/2}\) (two-tail z test)

The word "tail" refers to the tails of the normal curve: In a left-tail test, the probability of a Type I error is approximately the area of the left tail of the normal curve, from minus infinity to \(z_a\) . In a right-tail test, the probability of a Type I error is approximately the area of the right tail of the normal curve, from \(z_{1-a}\) to infinity. In a two-tail test, the probability of a Type I error is approximately the sum of the areas of both tails of the normal curve, the left tail from minus infinity to \(z_{a/2}\) and the right tail from \(z_{1-a/2}\) to infinity. All three of these tests are called z tests. The observed value of Z is called the z score .

Which of these three tests, if any, should one use? The answer depends on the probability distribution of Z when the alternative hypothesis is true. As a rule of thumb, if, under the alternative hypothesis, \(E(Z) , use the left-tail test. If, under the alternative hypothesis, \(E(Z) > 0\) , use the right-tail test. If, under the alternative hypothesis, it is possible that \(E(Z) and it is possible that \(E(Z) > 0\) , use the two-tail test. If, under the alternative hypothesis, \(E(Z) = 0\) , consult a statistician. Generally (but not always), this rule of thumb selects the test with the most power for a given significance level.

P values for z tests

Each of the three z tests gives us a family of procedures for testing the null hypothesis at any (approximate) significance level \(a\) between 0 and 100%—we just use the appropriate quantile of the normal curve. This makes it particularly easy to find the P value for a z test. Recall that the P value is the smallest significance level for which we would reject the null hypothesis, among a family of tests of the null hypothesis at different significance levels.

Suppose the z score (the observed value of \(Z\) ) is \(x\) . In a left-tail test, the P value is the area under the normal curve to the left of \(x\) : Had we chosen the significance level \(a\) so that \(z_a=x\) , we would have rejected the null hypothesis, but we would not have rejected it for any smaller value of \(a\) , because for all smaller values of \(a\) , \(z_a . Similarly, for a right-tail z test, the P value is the area under the normal curve to the right of \(x\) : If \(x=z_{1-a}\) we would reject the null hypothesis at approximate significance level \(a\) , but not at smaller significance levels. For a two-tail z test, the P value is the sum of the area under the normal curve to the left of \(-|x|\) and the area under the normal curve to the right of \(|x|\) .

Finding P values and specifying the rejection region for the z test involves the probability distribution of \(Z\) under the assumption that the null hypothesis is true. Rarely is the alternative hypothesis sufficiently detailed to specify the probability distribution of \(Z\) completely, but often the alternative does help us choose intelligently among left-tail, right-tail, and two-tail z tests. This is perhaps the most important issue in deciding which hypothesis to take as the null hypothesis and which as the alternative: We calculate the significance level under the null hypothesis, and that calculation must be tractable.

However, to construct a z test, we need to know the expected value and SE of the test statistic under the null hypothesis. Usually it is easy to determine the expected value, but often the SE must be estimated from the data. Later in this chapter we shall see what to do if the SE cannot be estimated accurately, but the shape of the distribution of the numbers in the population is known. The next section develops z tests for the population percentage and mean, and for the difference between two population means.

Examples of z tests

The central limit theorem assures us that the probability histogram of the sample mean of random draws with replacement from a box of tickets—transformed to standard units—can be approximated increasingly well by a normal curve as the number of draws increases. In the previous section, we learned that the probability histogram of a sum or difference of independent sample means of draws with replacement also can be approximated increasingly well by a normal curve as the two sample sizes increase. We shall use these facts to derive z tests for population means and percentages and differences of population means and percentages.

z Test for a Population Percentage

Suppose we have a population of \(N\) units of which \(G\) are labeled "1" and the rest are labeled "0." Let \(p = G/N\) be the population percentage. Consider testing the null hypothesis that \(p = p_0\) against the alternative hypothesis that \(p \ne p_0\) , using a random sample of \(n\) units drawn with replacement. (We could assume instead that \(N >> n\) and allow the draws to be without replacement.)

Under the null hypothesis, the sample percentage

\[ \phi = \frac{\mbox{# tickets labeled "1" in the sample}}{n} \]

has expected value \(E(\phi) = p_0\) and standard error

\[ SE(\phi) = \sqrt{\frac{p_0 \times (1 - p_0)}{n}}. \]

Let \(Z\) be \(\phi\) transformed to standard units :

\[ Z = (\phi - p_0)/SE(\phi). \]

Provided \(n\) is large and \(p_0\) is not too close to zero or 100% (say \(n \times p > 30\) and \(n \times (1-p) > 30)\) , the probability histogram of \(Z\) will be approximated reasonably well by the normal curve, and we can use it as the Z statistic in a z test. For example, if we reject the null hypothesis when \(|Z| > 1.96\) , the significance level of the test will be about 95%.

z Test for a Population Mean

The approach in the previous subsection applies, mutatis mutandis , to testing the hypothesis that the population mean equals a given value, even when the population contains numbers other than just 0 and 1. However, in contrast to the hypothesis that the population percentage equals a given value, the null hypothesis that a more general population mean equals a given value does not specify the SD of the population, which poses difficulties that are surmountable (by approximation and estimation) if the sample size is large enough. (There are also nonparametric methods that can be used.)

Consider testing the null hypothesis that the population mean \(\mu\) is equal to a specific null value \(\mu_0\) , against the alternative hypothesis that \(\mu , on the basis of a random sample with replacement of size \(n\) . Recall that the sample mean \(M\) of \(n\) random draws with or without replacement from a box of numbered tickets is an unbiased estimator of the population mean \(\mu\) : If

\[ M = \frac{\mbox{sum of sample values}}{n}, \]

\[ E(M) = \mu = \frac{\mbox{sum of population values}}{N}, \]

where \(N\) is the size of the population. The population mean determines the expected value of the sample mean. The SE of the sample mean of a random sample with replacement is

\[ \frac{SD(\mbox{box})}{\sqrt{n}}, \]

where SD(box) is the SD of the list of all the numbers in the box, and \(n\) is the sample size. As a special case, the sample percentage \phi of \(n\) independent random draws from a 0-1 box is an unbiased estimator of the population percentage p , with SE equal to

\[ \sqrt{\frac{p\times(1-p)}{n}}. \]

In testing the null hypothesis that a population percentage \(p\) equals \(p_0\) , the null hypothesis specifies not only the expected value of the sample percentage \phi, it automatically specifies the SE of the sample percentage as well, because the SD of the values in a 0-1 box is determined by the population percentage \(p\) :

\[ SD(box) = \sqrt{p\times(1-p)}. \]

The null hypothesis thus gives us all the information we need to standardize the sample percentage under the null hypothesis. In contrast, the SD of the values in a box of tickets labeled with arbitrary numbers bears no particular relation to the mean of the values, so the null hypothesis that the population mean \(\mu\) of a box of tickets labeled with arbitrary numbers equals a specific value \(\mu_0\) determines the expected value of the sample mean, but not the standard error of the sample mean. To standardize the sample mean to construct a z test for the value of a population mean, we need to estimate the SE of the sample mean under the null hypothesis. When the sample size is large, the sample standard deviation s> is likely to be close to the SD of the population, and

\[ se=\frac{s}{\sqrt{n}} \]

is likely to be an accurate estimate of \(SE(M)\) . The central limit theorem tells us that when the sample size \(n\) is large, the probability histogram of the sample mean, converted to standard units, is approximated well by the normal curve. Under the null hypothesis,

\[ E(M) = \mu_0, \]

and thus when \(n\) is large

\[ Z = \frac{M-\mu_0}{s/\sqrt{n}} \]

has expected value zero, and its probability histogram is approximated well by the normal curve, so we can use \(Z\) as the Z statistic in a z test. If the alternative hypothesis is true, the expected value of \(Z\) could be either greater than zero or less than zero, so it is appropriate to use a two-tail z test. If the alternative hypothesis is \(\mu > \mu_0\) , then under the alternative hypothesis, the expected value of \(Z\) is greater than zero, and it is appropriate to use a right-tail z test. If the alternative hypothesis is \(\mu , then under the alternative hypothesis, the expected value of \(Z\) is less than zero, and it is appropriate to use a left-tail z test.

z Test for a Difference of Population Means

Consider the problem of testing the hypothesis that two population means are equal, using random samples from the two populations. Different sampling designs lead to different hypothesis testing procedures. In this section, we consider two kinds of random samples from the two populations: paired samples and independent samples , and construct z tests appropriate for each.

Paired Samples

Consider a population of \(N\) individuals, each of whom is labeled with two numbers. For example, the \(N\) individuals might be a group of doctors, and the two numbers that label each doctor might be the annual payments to the doctor by an HMO under the terms of the current contract and under the terms of a proposed revision of the contract. Let the two numbers associated with individual \(i\) be \(c_i\) and \(t_i\) . (Think of \(c\) as control and \(t\) as treatment . In this example, control is the current contract, and treatment is the proposed contract.) Let \(\mu_c\) be the population mean of the \(N\) values

\[ \{c_1, c_2, \ldots, c_N \}, \]

and let \(\mu_t\) be the population mean of the \(N\) values

\[ \{t_1, t_2, \ldots, t_N\}. \]

Suppose we want to test the null hypothesis that

\[ \mu = \mu_t - \mu_c = \mu_0 \]

against the alternative hypothesis that \(\mu . With \(\mu_0=\$0\) , this null hypothesis is that the average annual payment to doctors under the proposed revision would be the same as the average payment under the current contract, and the alternative is that on average doctors would be paid less under the new contract than under the current contract. With \(\mu_0=-\$5,000\) , this null hypothesis is that the proposed contract would save the HMO an average of $5,000 per doctor, compared with the current contract; the alternative is that under the proposed contract, the HMO would save even more than that. With \(\mu_0=\$1,000\) , this null hypothesis is that doctors would be paid an average of $1,000 more per year under the new contract than under the old one; the alternative hypothesis is that on average doctors would be paid less than an additional $1,000 per year under the new contract—perhaps even less than they are paid under the current contract. For the remainder of this example, we shall take \(\mu_0=\$1,000\) .

The data on which we shall base the test are observations of both \(c_i\) and \(t_i\) for a sample of \(n\) individuals chosen at random with replacement from the population of \(N\) individuals (or a simple random sample of size \(n ): We select \(n\) doctors at random from the \(N\) doctors under contract to the HMO, record the current annual payments to them, and calculate what the payments to them would be under the terms of the new contract. This is called a paired sample , because the samples from the population of control values and from the population of treatment values come in pairs: one value for control and one for treatment for each individual in the sample. Testing the hypothesis that the difference between two population means is equal to \(\mu_0\) using a paired sample is just the problem of testing the hypothesis that the population mean \(\mu\) of the set of differences

\[ d_i = t_i - c_i, \;\; i= 1, 2, \ldots, N, \]

is equal to \(\mu_0\) . Denote the \(n\) (random) observed values of \(c_i\) and \(t_i\) by \(\{C_1, C_2, \ldots, C_n\}\) and \(\{T_1, T_2, \ldots, T_n \}\) , respectively. The sample mean \(M\) of the differences between the observed values of \(t_i\) and \(c_i\) is the difference of the two sample means:

\[ M = \frac{(T_1-C_1)+(T_2-C_2) + \cdots + (T_n-C_n)}{n} = \frac{T_1+T_2+ \cdots + T_n}{n} - \frac{C_1+C_2+ \cdots + C_n}{n} \]

\[ = (\mbox{sample mean of observed values of } t_i) - (\mbox{sample mean of observed values of } c_i). \]

\(M\) is an unbiased estimator of \(\mu\) , and if n is large, the normal approximation to its probability histogram will be accurate. The SE of \(M\) is the population standard deviation of the \(N\) values \(\{d_1, d_2, \ldots, d_N\}\) , which we shall denote \(SD_d\) , divided by the square root of the sample size, \(n^{1/2}\) . Let \(sd\) denote the sample standard deviation of the \(n\) observed differences \((T_i - C_i), \;\; i=1, 2, \ldots, n\) :

\[ sd = \sqrt{\frac{(T_1-C_1-M)^2 + (T_2-C_2-M)^2 + \cdots + (T_n-C_n-M)^2}{n-1}} \]

(recall that \(M\) is the sample mean of the observed differences). If the sample size \(n\) is large, sd is very likely to be close to SD( d ), and so, under the null hypothesis,

\[ Z = \frac{M-\mu_0}{sd/n^{1/2}} \]

has expected value zero, and when \(n\) is large the probability histogram of \(Z\) can be approximated well by the normal curve. Thus we can use \(Z\) as the Z statistic in a z test of the null hypothesis that \(\mu=\mu_0\) . Under the alternative hypothesis that \(\mu (doctors on the average are paid less than an additional $1,000 per year under the new contract), the expected value of \(Z\) is less than zero, so we should use a left-tail z test. Under the alternative hypothesis \(\mu\ne\mu_0\) (on average, the difference in average annual payments to doctors is not an increase of $1,000, but some other number instead), the expected value of \(Z\) could be positive or negative, so we would use a two-tail z test. Under the alternative hypothesis that \(\mu>\mu_0\) (on average, under the new contract, doctors are paid more than an additional $1,000 per year), the expected value of \(Z\) would be greater than zero, so we should use a right-tail z test.

Independent Samples

Consider two separate populations of numbers, with population means \(\mu_t\) and \(\mu_c\) , respectively. Let \(\mu=\mu_t-\mu_c\) be the difference between the two population means. We would like to test the null hypothesis that \(\mu=\mu_0\) against the alternative hypothesis that \(\mu>0\) . For example, let \(\mu_t\) be the average annual payment by an HMO to doctors in the Los Angeles area, and let \(\mu_c\) be the average annual payment by the same HMO to doctors in the San Francisco area. Then the null hypothesis with \(\mu_0=0\) is that the HMO pays doctors in the two regions the same amount annually, on average; the alternative hypothesis is that the average annual payment by the HMO to doctors differs between the two areas. Suppose we draw a random sample of size \(n_t\) with replacement from the first population, and independently draw a random sample of size \(n_c\) with replacement from the second population. Let \(M_t\) and \(M_c\) be the sample means of the two samples, respectively, and let

\[ M = M_t - M_c \]

be the difference between the two sample means. Because the expected value of \(M_t\) is \(\mu_t\) and the expected value of \(M_c\) is \(\mu_c\) , the expected value of \(M\) is

\[ E(M) = E(M_t - M_c) = E(M_t) - E(M_c) = \mu_t - \mu_c = \mu. \]

Because the two random samples are independent , \(M_t\) and \(-M_c\) are independent random variables, and the SE of their sum is

\[ SE(M) = (SE^2(M_t) + SE^2(M_c))^{1/2}. \]

Let \(s_t\) and \(s_c\) be the sample standard deviations of the two samples, respectively. If \(n_t\) and \(n_c\) are both very large, the two sample standard deviations are likely to be close to the standard deviations of the corresponding populations, and so \(s_t/n_t^{1/2}\) is likely to be close to \(SE(M_t)\) , and \(s_c/n_c^{1/2}\) is likely to be close to \(SE(M_c)\) . Therefore, the pooled estimate of the standard error

\[ se_\mbox{diff} = ( (s_t/n_t^{1/2})^2 + (s_c/n_c^{1/2})^2)^{1/2} = \sqrt{ s_t^2/n_t + s_c^2/n_c} \]

is likely to be close to \(SE(M)\) . Under the null hypothesis, the statistic

\[ Z = \frac{M - \mu_0}{se_\mbox{diff}} = \frac{M_1 - M_2 - \mu_0}{\sqrt{ s_t^2/n_t + s_c^2/n_c}} \]

has expected value zero and its probability histogram is approximated well by the normal curve, so we can use it as the Z statistic in a z test.

Under the alternative hypothesis

\[ \mu = \mu_t - \mu_c > \mu_0, \]

the expected value of \(Z\) is greater than zero, so it is appropriate to use a right-tail z test.

If the alternative hypothesis were \(\mu \ne \mu_0\) , under the alternative the expected value of \(Z\) could be greater than zero or less than zero, so it would be appropriate to use a two-tail z test. If the alternative hypothesis were \(\mu , under the alternative the expected value of \(Z\) would be less than zero, so it would be appropriate to use a left-tail z test.

The following exercises check that you can compute the z test for a population mean or a difference of population means. The exercises are dynamic: the data will tend to change when you reload the page.

For the nominal significance level of the z test for a population mean to be approximately correct, the sample size typically must be large. When the sample size is small, two factors limit the accuracy of the z test: the normal approximation to the probability distribution of the sample mean can be poor, and the sample standard deviation can be an inaccurate estimate of the population standard deviation, so se is not an accurate estimate of the SE of the test statistic Z . For nearly normal populations , defined in the next subsection, the probability distribution of the sample mean is nearly normal even when the sample size is small, and the uncertainty of the sample standard deviation as an estimate of the population standard deviation can be accounted for by using a curve that is broader than the normal curve to approximate the probability distribution of the (approximately) standardized test statistic. The broader curve is Student's t curve . Student's t curve depends on the sample size: The smaller the sample size, the more spread out the curve.

Nearly Normally Distributed Populations

A list of numbers is nearly normally distributed if the fraction of values in any range is close to the area under the normal curve for the corresponding range of standard units—that is, if the list has mean \(\mu\) and standard deviation SD, and for every pair of values \(a < b\) ,

\[ \mbox{ the fraction of numbers in the list between } a \mbox{ and } b \approx \mbox{the area under the normal curve between } (a - \mu)/SD \mbox{ and } (b - \mu)/SD. \]

A list is nearly normally distributed if the normal curve is a good approximation to the histogram of the list transformed to standard units. The histogram of a list that is approximately normally distributed is (nearly) symmetric about some point, and is (nearly) bell-shaped.

No finite population can be exactly normally distributed, because the area under the normal curve between every two distinct values is strictly positive—no matter how large or small the values nor how close together they are. No population that contains only a finite number of distinct values can be exactly normally distributed, for the same reason. In particular, populations that contain only zeros and ones are not approximately normally distributed, so results for the sample mean of samples drawn from nearly normally distributed populations need not apply to the sample percentage of samples drawn from 0-1 boxes. Such results will be more accurate for the sample percentage when the population percentage is close to 50% than when the population percentage is close to 0% or 100%, because then the histogram of population values is more nearly symmetric.

Suppose a population is nearly normally distributed. Then a histogram of the population is approximately symmetric about the mean of the population. The fraction of numbers in the population within ±1 SD of the mean of the population is about 68%, the fraction of numbers within ±2 SD of the mean of the population is about 95%, and the fraction of numbers in the population within ±3 SD of the mean of the population is about 99.7%.

The following exercises check that you understand what it means for a list to be nearly normally distributed. The exercises are dynamic: the data tend to change when you reload the page.

Student's t -curve

Student's t curve is similar to the normal curve, but broader. It is positive, has a single maximum, and is symmetric about zero. The total area under Student's t curve is 100%. Student's t curve approximates some probability histograms more accurately than the normal curve does. There are actually infinitely many Student t curves, one for each positive integer value of the degrees of freedom. As the degrees of freedom increases, the difference between Student's t curve and the normal curve decreases.

Consider a population of \(N\) units labeled with numbers. Let \(\mu\) denote the population mean of the \(N\) numbers, and let SD denote the population standard deviation of the \(N\) numbers. Let \(M\) denote the sample mean of a random sample of size \(n\) drawn with replacement from a population, and let s> denote the sample standard deviation of the sample. The expected value of \(M\) is \(\mu\) , and the SE of \(M\) is \(SD/n^{1/2}\) . Let

\[ Z = (M - \mu)/(SD/n^{1/2}). \]

Then the expected value of \(Z\) is zero, the SE of \(Z\) is 1, and if \(n\) is large enough, the normal curve is a good approximation to the probability histogram of \(Z\) . The closer to normal the distribution of values in the population is, the smaller \(n\) needs to be for the normal curve to be a good approximation to the distribution of \(Z\) . Consider the statistic

\[ T = \frac{M - \mu}{s/n^{1/2}}, \]

which replaces SD by its estimated value (the sample standard deviation \(s\) ). If \(n\) is large enough, \(s\) is very likely to be close to SD, so \(T\) will be close to \(Z\) ; the normal curve will be a good approximation to the probability histogram of \(T\) ; and we can use \(T\) as the Z statistic in a z test of hypotheses about \(\mu\) .

For many populations, when the sample size is small—say less than 25, but the accuracy depends on the population—the normal curve is not a good approximation to the probability histogram of \(T\) . For nearly normally distributed populations, when the sample size is intermediate—say 25–100, but again this depends on the population—the normal curve is a good approximation to the probability histogram of \(Z\) , but not to the probability histogram of \(T\) , because of the variability of the sample standard deviation s> from sample to sample, which tends to broaden the probability distribution of \(T\) (i.e., to make \(SE(T)>1\) ).

When you first load this page, the degrees of freedom will be set to 25, and the region from -1.96 to 1.96 will be hilighted. The area under the normal curve between ±1.96 is 95%, but for Student's t curve with 25 degrees of freedom, the area is about 93.9%: Student's t curve with d.f.=25 is broader than the normal curve. Increase the degrees of freedom to 200; you will see that the Student t curve gets slightly narrower, and the area under the curve between ±1.96 is about 94.9%.

We define quantiles of Student t curves in the same way we defined quantiles of the normal curve: For any number a between 0 and 100%, the a quantile of Student's t curve with \(d.f.=d\) , \(t_{d,a}\) , is the unique value such that the area under the Student t curve with d degrees of freedom from minus infinity to \(t_{d,a}\) is equal to \(a\) . For example, \(t_{d,0.5} = 0\) for all values of \(d\) . Generally, the value of \(t_{d,a}\) depends on the degrees of freedom \(d\) . The probability calculator allows you to find quantiles of Student's t curve.

t test for the Mean of a Nearly Normally Distributed Population

We can use Student's t curve to construct approximate tests of hypotheses about the population mean \(\mu\) when the population standard deviation is unknown, for intermediate values of the sample size \(n\) . The approach is directly analogous to the z test, but instead of using a quantile of the normal curve, we use the corresponding quantile of Student's t curve (with the appropriate number of degrees of freedom). However, for the test to be accurate when \(n\) is small or intermediate, the distribution of values in the population must be nearly normal for the test to have approximately its nominal level. This is a somewhat bizarre restriction: It may require a very large sample to detect that the population is not nearly normal—but if the sample is very large, we can use the z test instead of the t test, so we don't need to rely as much on the assumption. It is my opinion that the t test is over-taught and overused—because its assumptions are not verifiable in the situations where it is potentially useful.

Consider testing the null hypothesis that \(\mu=\mu_0\) using the sample mean \(M\) and sample standard deviation s> of a random sample of size \(n\) drawn with replacement from a population that is known to have a nearly normal distribution. Define

\[ T = \frac{M - \mu_0}{s/n^{1/2}}. \]

Under the null hypothesis, if \(n\) is not too small, Student's t curve with \(n-1\) degrees of freedom will be an accurate approximation to the probability histogram of \(T\) , so

\[ P(T < t_{n-1,a}), \]

\[ P(T > t_{n-1,1-a}), \]

\[ P(|T| > t_{n-1,1-a/2}) \]

all are approximately equal to \(a\) . As we saw earlier in this chapter for the Z statistic, these three approximations give three tests of the null hypothesis \(\mu=\mu_0\) at approximate significance level \(a\) —a left-tail t test, a right-tail t test, and a two-tail t test:

• Reject the null hypothesis if \(T (left-tail)
• Reject the null hypothesis if \(T > t_{n-1,1-a}\) (right-tail)
• Reject the null hypothesis if \(|T| > t_{n-1,1-a/2}\) (two-tail)

To decide which t test to use, we can apply the same rule of thumb we used for the z test:

• Use a left-tail t test if, under the alternative hypothesis, the expected value of \(T\) is less than zero.
• Use a right-tail t test if, under the alternative hypothesis, the expected value of \(T\) is greater than zero.
• Use a two-tail t test if, under the alternative hypothesis, the expected value of \(T\) is not zero, but could be less than or greater than zero.
• Consult a statistician for a more appropriate test if, under the alternative hypothesis, the expected value of \(T\) is zero.

P-values for t tests are computed in much the same way as P-values for z tests. Let t be the observed value of \(T\) (the t score). In a left-tail t test, the P-value is the area under Student's t curve with \(n-1\) degrees of freedom, from minus infinity to \(t\) . In a right-tail t test, the P-value is the area under Student's t curve with \(n-1\) degrees of freedom, from \(t\) to infinity. In a two-tail t test, the P-value is the total area under Student's t curve with \(n-1\) degrees of freedom between minus infinity and \(-|t|\) and between \(|t|\) and infinity.

There are versions of the t test for comparing two means, as well. Just like for the z test, the method depends on how the samples from the two populations are drawn. For example, if the two samples are paired (if we are sampling individuals labeled with two numbers and for each individual in the sample, we observe both numbers), we may base the t test on the sample mean of the paired differences and the sample standard deviation of the paired differences. Let \(\mu_1\) and \(\mu_2\) be the means of the two populations, and let

\[ \mu = \mu_1 - \mu_2. \]

The \(T\) statistic to test the null hypothesis that \(\mu=\mu_0\) is

\[ T = \frac{(\mbox{sample mean of differences}) - \mu_0 }{(\mbox{sample standard deviation of differences})/n^{1/2}}, \]

and the appropriate curve to use to find the rejection region for the test is Student's t curve with \(n-1\) degrees of freedom, where \(n\) is the number of individuals (differences) in the sample.

Two-sample t tests for a difference of means using independent samples depend on additional assumptions, such as equality of the two population standard deviations; we shall not present such tests here. The following exercises check your ability to compute t tests. The exercises are dynamic: the data tend to change when you reload the page.

Hypothesis Tests and Confidence Intervals

There is a deep connection between hypothesis tests about parameters, and confidence intervals for parameters. If we have a procedure for constructing a level \(100\% \times (1-a)\) confidence interval for a parameter \(\mu\) , then the following rule is a two-sided significance level \(a\) test of the null hypothesis that \(\mu = \mu_0\) :

reject the null hypothesis if the confidence interval does not contain \(\mu_0\).

Similarly, suppose we have an hypothesis-testing procedure that lets us test the null hypothesis that \(\mu=\mu_0\) for any value of \(\mu_0\) , at significance level \(a\) . Define

\(A\) = (all values of \(\mu_0\) for which we would not reject the null hypothesis that \(\mu = \mu_0\)).

Then \(A\) is a \(100\% \times (1-a)\) confidence set for \(\mu\) :

\[ P( A \mbox{ contains the true value of } \mu ) = 100\% \times (1-a). \]

(A confidence set is a generalization of the idea of a confidence interval: a \(1-a\) confidence set for the parameter \(\mu\) is a random set that has probability \(1-a\) of containing \(\mu\) . As is the case with confidence intervals, the probability makes sense only before collecting the data.) The set \(A\) might or might not be an interval, depending on the nature of the test. If one starts with a two-tail z test or two-tail t test, one ends up with a confidence interval rather than a more general confidence set.

Confidence Intervals Using Student's t curve

The t test lets us test the hypothesis that the population mean \(\mu\) is equal to \(\mu_0\) at approximate significance level a using a random sample with replacement of size n from a population with a nearly normal distribution. If the sample size n is small, the actual significance level is likely to differ considerably from the nominal significance level. Consider a two-sided t test of the hypothesis \(\mu=\mu_0\) at significance level \(a\) . If the sample mean is \(M\) and the sample standard deviation is \(s\) , we would not reject the null hypothesis at significance level \(a\) if

\[ \frac{|M-\mu_0|}{s/n^{1/2}} \le t_{n-1,1-a/2}. \]

We rearrange this inequality:

\[ -t_{n-1,1-a/2} \le \frac{M-\mu_0}{s/n^{1/2}} \le t_{n-1,1-a/2} \]

\[ -t_{n-1,1-a/2} \times s/n^{1/2} \le M - \mu_0 \le t_{n-1,1-a/2} \times s/n^{1/2} \]

\[ -M - t_{n-1,1-a/2} \times s/n^{1/2} \le - \mu_0 \le -M + t_{n-1,1-a/2} \times s/n^{1/2} \]

\[ M + t_{n-1,1-a/2} \times s/n^{1/2} \le \mu_0 \le M - t_{n-1,1-a/2} \times s/n^{1/2} \]

That is, we would not reject the hypothesis \(\mu = \mu_0\) provided \(\mu_0\) is in the interval

\[ [M - t_{n-1,1-a/2} \times s/n^{1/2}, M + t_{n-1,1-a/2} \times s/n^{1/2}]. \]

Therefore, that interval is a \(100\%-a\) confidence interval for \(\mu\) :

\[ P([M - t_{n-1,1-a/2} \times s/n^{1/2}, M + t_{n-1,1-a/2} \times s/n^{1/2}] \mbox{ contains } \mu) \approx 1-a. \]

The following exercise checks that you can use Student's t curve to construct a confidence interval for a population mean. The exercise is dynamic: the data tend to change when you reload the page.

In hypothesis testing, a Z statistic is a random variable whose probability histogram is approximated well by the normal curve if the null hypothesis is correct: If the null hypothesis is true, the expected value of a Z statistic is zero, the SE of a Z statistic is approximately 1, and the probability that a Z statistic is between \(a\) and \(b\) is approximately the area under the normal curve between \(a\) and \(b\) . Suppose that the random variable \(Z\) is a Z statistic. If, under the alternative hypothesis, \(E(Z) , the appropriate z test to test the null hypothesis at approximate significance level \(a\) is the left-tailed z test: Reject the null hypothesis if \(Z , where \(z_a\) is the \(a\) quantile of the normal curve. If, under the alternative hypothesis, \(E(Z)>0\) , the appropriate z test to test the null hypothesis at approximate significance level \(a\) is the right-tailed z test: Reject the null hypothesis if \(Z>z_{1-a}\) . If, under the alternative hypothesis, \(E(Z)\ne 0 \) but could be greater than 0 or less than 0, the appropriate z test to test the null hypothesis at approximate significance level \(a\) is the two-tailed z test: reject the null hypothesis if \(|Z|>z_{1-a/2}\) . If, under the alternative hypothesis, \(E(Z)=0\) , a z test probably is not appropriate—consult a statistician. The exact significance levels of these tests differ from \(a\) by an amount that depends on how closely the normal curve approximates the probability histogram of \(Z\) .

Z statistics often are constructed from other statistics by transforming approximately to standard units, which requires knowing the expected value and SE of the original statistic on the assumption that the null hypothesis is true. Let \(X\) be a test statistic; let \(E(X)\) be the expected value of \(X\) if the null hypothesis is true, and let \(se\) be approximately equal to the SE of \(X\) if the null hypothesis is true. If \(X\) is a sample sum of a large random sample with replacement, a sample mean of a large random sample with replacement, or a sum or difference of independent sample means of large samples with replacement,

\[ Z = \frac{X-E(X)}{se} \]

is a Z statistic.

Consider testing the null hypothesis that a population percentage \(p\) is equal to the value \(p_0\) on the basis of the sample percentage \phi of a random sample of size \(n\) with replacement. Under the null hypothesis, \(E(\phi)=p_0\) and

\[ SE(\phi) = \sqrt{\frac{p_0\times(1-p_0)}{n}}, \]

and if \(n\) is sufficiently large (say \(n \times p > 30\) and \(n \times (1-p)>30\) , but this depends on the desired accuracy), the normal approximation to

\[ Z = \frac{\phi-p_0}{\sqrt{(p_0 \times (1-p_0))/n}} \]

will be reasonably accurate, so \(Z\) can be used as the Z statistic in a z test of the null hypothesis \(p=p_0\) .

Consider testing the null hypothesis that a population mean \(\mu\) is equal to the value \(\mu_0\) , on the basis of the sample mean \(M\) of a random sample of size \(n\) with replacement. Let \(s\) denote the sample standard deviation. Under the null hypothesis, \(E(M)=\mu_0\) , and if \(n\) is large,

\[ SE(M)=SD/n^{1/2} \approx s/n^{1/2}, \]

and the normal approximation to

\[ Z = \frac{M-\mu_0}{s/n^{1/2}} \]

will be reasonably accurate, so \(Z\) can be used as the Z statistic in a z test of the null hypothesis \(\mu=\mu_0\) .

Consider a population of \(N\) individuals, each labeled with two numbers. The \(i\) th individual is labeled with the numbers \(c_i\) and \(t_i\) , \(i=1, 2, \ldots, N\) . Let \(\mu_c\) be the population mean of the \(N\) values \(\{c_1, \ldots, c_N\}\) and let \(\mu_t\) be the population mean of the \(N\) values \(\{t_1, \ldots, t_N \}\) . Let \(\mu=\mu_t-\mu_c\) be the difference between the two population means. Consider testing the null hypothesis that \(\mu=\mu_0\) on the basis of a paired random sample of size \(n\) with replacement from the population: that is, a random sample of size \(n\) is drawn with replacement from the population, and for each individual \(i\) in the sample, \(c_i\) and \(t_i\) are observed. This is equivalent to testing the hypothesis that the population mean of the \(N\) values \(\{(t_1-c_1), \ldots, (t_N-c_N)\}\) is equal to \(\mu_0\) , on the basis of the random sample of size \(n\) drawn with replacement from those \(N\) values. Let \(M_t\) be the sample mean of the \(n\) observed values of \(t_i\) and let \(M_c\) be the sample mean of the \(n\) observed values of \(c_i\) . Let \(sd\) denote the sample standard deviation of the \(n\) observed differences \(\{(t_i-c_i)\}\) . Under the null hypothesis, the expected value of \(M_t-M_c\) is \(\mu_0\) , and if \(n\) is large,

\[ SE(M_t-M_c) \approx sd/n^{1/2}, \]

and the normal approximation to the probability histogram of

\[ Z = \frac{M_t-M_c-\mu_0}{sd/n^{1/2}} \]

will be reasonably accurate, so \(Z\) can be used as the Z statistic in a z test of the null hypothesis that \(\mu_t-\mu_c=\mu_0\) .

Consider testing the hypothesis that the difference ( \(\mu_t-\mu_c\) ) between two population means, \(\mu_c\) and \(\mu_t\) , is equal to \(\mu_0\) , on the basis of the difference ( \(M_t-M_c\) ) between the sample mean \(M_c\) of a random sample of size \(n_c\) with replacement from the first population and the sample mean \(M_t\) of an independent random sample of size \(n_t\) with replacement from the second population. Let \(s_c\) denote the sample standard deviation of the sample of size \(n_c\) from the first population and let \(s_t\) denote the sample standard deviation of the sample of size \(n_t\) from the second population. If the null hypothesis is true,

\[ E(M_t-M_c)=\mu_0, \]

and if \(n_c\) and \(n_t\) are both large,

\[ SE(M_t-M_c) \approx \sqrt{s_t^2/n_t + s_c^2/n_c} \]

\[ Z = \frac{M_t-M_c-\mu_0}{\sqrt{s_t^2/n_t + s_c^2/n_c}} \]

A list of numbers is nearly normally distributed if the fraction of numbers between any pair of values, \(a , is approximately equal to the area under the normal curve between \((a-\mu)/SD\) and \((b-\mu)/SD\) , where \(\mu\) is the mean of the list and SD is the standard deviation of the list.

Student's t curve with \(d\) degrees of freedom is symmetric about 0, has a single bump centered at 0, and is broader and flatter than the normal curve. The total area under Student's t curve is 1, no matter what \(d\) is; as \(d\) increases, Student's t curve gets narrower, its peak gets higher, and it becomes closer and closer to the normal curve.

Let \(M\) be the sample mean of a random sample of size \(n\) with replacement from a population with mean \(\mu\) and a nearly normal distribution, and let \(s\) be the sample standard deviation of the random sample. For moderate values of \(n\) ( \(n or so), Student's t curve approximates the probability histogram of \((M-\mu)/(s/n^{1/2})\) better than the normal curve does, which can lead to an approximate hypothesis test about \(\mu\) that is more accurate than the z test.

Consider testing the null hypothesis that the mean \(\mu\) of a population with a nearly normal distribution is equal to \(\mu_0\) from a random sample of size \(n\) with replacement. Let

\[ T=\frac{M-\mu_0}{s/n^{1/2}}, \]

where \(M\) is the sample mean and \(s\) is the sample standard deviation. The tests that reject the null hypothesis if \(T (left-tail t test), if \(T>t_{n-1,1-a}\) (right-tail t test), or if \(|T|>t_{n-1,1-a/2}\) (two-tail t test) all have approximate significance level \(a\) . How close the nominal significance level \(a\) is to the true significance level depends on the distribution of the numbers in the population, the sample size \(n\) , and \(a\) . The same rule of thumb for selecting whether to use a left, right, or two-tailed z test (or not to use a z test at all) works to select whether to use a left, right, or two-tailed t test: If, under the alternative hypothesis, \(E(T) , use a left-tail test. If, under the alternative hypothesis, \(E(T) > 0 \) , use a right-tail test. If, under the alternative hypothesis, \(E(T)\) could be less than zero or greater than zero, use a two-tail test. If, under the alternative hypothesis, \(E(T) = 0 \) , consult an expert. Because the t test differs from the z test only when the sample size is small, and from a small sample it is not possible to tell whether the population has a nearly normal distribution, the t test should be used with caution.

A \(1-a\) confidence set for a parameter \(\mu\) is like a \(1-a\) confidence interval for a parameter \(\mu\) : It is a random set of values that has probability \(1-a\) of containing the true value of \(\mu\) . The difference is that the set need not be an interval.

There is a deep duality between hypothesis tests about a parameter \(\mu\) and confidence sets for \(\mu\) . Given a procedure for constructing a \(1-a\) confidence set for \(\mu\) , the rule reject the null hypothesis that \(\mu=\mu_0\) if the confidence set does not contain \(\mu\) is a significance level \(a\) test of the null hypothesis that \(\mu=\mu_0\) . Conversely, given a family of significance level \(a\) hypothesis tests that allow one to test the hypothesis that \(\mu=\mu_0\) for any value of \(\mu_0\) , the set of all values \(\mu_0\) for which the test does not reject the null hypothesis that \(\mu=\mu_0\) is a \(1-a\) confidence set for \(\mu\) .

• alternative hypothesis
• central limit theorem
• confidence interval
• confidence set
• expected value
• independent
• independent random variable
• mutatis mutandis
• nearly normal distribution
• normal approximation
• normal curve
• null hypothesis
• pooled bootstrap estimate of the population SD
• pooled bootstrap estimate of the SE
• population mean
• population percentage
• population standard deviation
• probability
• probability distribution
• probability histogram
• random sample
• random variable
• rejection region
• sample mean
• sample percentage
• sample size
• sample standard deviation
• significance level
• simple random sample
• standard deviation (SD)
• standard error (SE)
• standard unit
• Student's t curve
• test statistic
• two-tailed test
• Type I error
• Z statistic

10 Chapter 10: Hypothesis Testing with Z

Setting up the hypotheses.

When setting up the hypotheses with z, the parameter is associated with a sample mean (in the previous chapter examples the parameters for the null used 0). Using z is an occasion in which the null hypothesis is a value other than 0. For example, if we are working with mothers in the U.S. whose children are at risk of low birth weight, we can use 7.47 pounds, the average birth weight in the US, as our null value and test for differences against that. For now, we will focus on testing a value of a single mean against what we expect from the population.

Using birthweight as an example, our null hypothesis takes the form: H 0 : μ = 7.47 Notice that we are testing the value for μ, the population parameter, NOT the sample statistic ̅X (or M). We are referring to the data right now in raw form (we have not standardized it using z yet). Again, using inferential statistics, we are interested in understanding the population, drawing from our sample observations. For the research question, we have a mean value from the sample to use, we have specific data is – it is observed and used as a comparison for a set point.

As mentioned earlier, the alternative hypothesis is simply the reverse of the null hypothesis, and there are three options, depending on where we expect the difference to lie. We will set the criteria for rejecting the null hypothesis based on the directionality (greater than, less than, or not equal to) of the alternative.

If we expect our obtained sample mean to be above or below the null hypothesis value (knowing which direction), we set a directional hypothesis. O ur alternative hypothesis takes the form based on the research question itself. In our example with birthweight, this could be presented as H A : μ > 7.47 or H A : μ < 7.47. 

Note that we should only use a directional hypothesis if we have a good reason, based on prior observations or research, to suspect a particular direction. When we do not know the direction, such as when we are entering a new area of research, we use a non-directional alternative hypothesis. In our birthweight example, this could be set as H A : μ ≠ 7.47

In working with data for this course we will need to set a critical value of the test statistic for alpha (α) for use of test statistic tables in the back of the book. This is determining the critical rejection region that has a set critical value based on α.

Determining Critical Value from α

We set alpha (α) before collecting data in order to determine whether or not we should reject the null hypothesis. We set this value beforehand to avoid biasing ourselves by viewing our results and then determining what criteria we should use.

When a research hypothesis predicts an effect but does not predict a direction for the effect, it is called a non-directional hypothesis . To test the significance of a non-directional hypothesis, we have to consider the possibility that the sample could be extreme at either tail of the comparison distribution. We call this a two-tailed test .

Figure 1. showing a 2-tail test for non-directional hypothesis for z for area C is the critical rejection region.

When a research hypothesis predicts a direction for the effect, it is called a directional hypothesis . To test the significance of a directional hypothesis, we have to consider the possibility that the sample could be extreme at one-tail of the comparison distribution. We call this a one-tailed test .

Figure 2. showing a 1-tail test for a directional hypothesis (predicting an increase) for z for area C is the critical rejection region.

Determining Cutoff Scores with Two-Tailed Tests

Typically we specify an α level before analyzing the data. If the data analysis results in a probability value below the α level, then the null hypothesis is rejected; if it is not, then the null hypothesis is not rejected. In other words, if our data produce values that meet or exceed this threshold, then we have sufficient evidence to reject the null hypothesis ; if not, we fail to reject the null (we never “accept” the null). According to this perspective, if a result is significant, then it does not matter how significant it is. Moreover, if it is not significant, then it does not matter how close to being significant it is. Therefore, if the 0.05 level is being used, then probability values of 0.049 and 0.001 are treated identically. Similarly, probability values of 0.06 and 0.34 are treated identically. Note we will discuss ways to address effect size (which is related to this challenge of NHST).

When setting the probability value, there is a special complication in a two-tailed test. We have to divide the significance percentage between the two tails. For example, with a 5% significance level, we reject the null hypothesis only if the sample is so extreme that it is in either the top 2.5% or the bottom 2.5% of the comparison distribution. This keeps the overall level of significance at a total of 5%. A one-tailed test does have such an extreme value but with a one-tailed test only one side of the distribution is considered.

Figure 3. Critical value differences in one and two-tail tests. Photo Credit

Let’s re view th e set critical values for Z.

We discussed z-scores and probability in chapter 8.  If we revisit the z-score for 5% and 1%, we can identify the critical regions for the critical rejection areas from the unit standard normal table.

• A two-tailed test at the 5% level has a critical boundary Z score of +1.96 and -1.96
• A one-tailed test at the 5% level has a critical boundary Z score of +1.64 or -1.64
• A two-tailed test at the 1% level has a critical boundary Z score of +2.58 and -2.58
• A one-tailed test at the 1% level has a critical boundary Z score of +2.33 or -2.33.

Review: Critical values, p-values, and significance level

There are two criteria we use to assess whether our data meet the thresholds established by our chosen significance level, and they both have to do with our discussions of probability and distributions. Recall that probability refers to the likelihood of an event, given some situation or set of conditions. In hypothesis testing, that situation is the assumption that the null hypothesis value is the correct value, or that there is no effec t. The value laid out in H 0 is our condition under which we interpret our results. To reject this assumption, and thereby reject the null hypothesis, we need results that would be very unlikely if the null was true.

Now recall that values of z which fall in the tails of the standard normal distribution represent unlikely values. That is, the proportion of the area under the curve as or more extreme than z is very small as we get into the tails of the distribution. Our significance level corresponds to the area under the tail that is exactly equal to α: if we use our normal criterion of α = .05, then 5% of the area under the curve becomes what we call the rejection region (also called the critical region) of the distribution. This is illustrated in Figure 4.

Figure 4: The rejection region for a one-tailed test

The shaded rejection region takes us 5% of the area under the curve. Any result which falls in that region is sufficient evidence to reject the null hypothesis.

The rejection region is bounded by a specific z-value, as is any area under the curve. In hypothesis testing, the value corresponding to a specific rejection region is called the critical value, z crit (“z-crit”) or z* (hence the other name “critical region”). Finding the critical value works exactly the same as finding the z-score corresponding to any area under the curve like we did in Unit 1. If we go to the normal table, we will find that the z-score corresponding to 5% of the area under the curve is equal to 1.645 (z = 1.64 corresponds to 0.0405 and z = 1.65 corresponds to 0.0495, so .05 is exactly in between them) if we go to the right and -1.645 if we go to the left. The direction must be determined by your alternative hypothesis, and drawing then shading the distribution is helpful for keeping directionality straight.

Suppose, however, that we want to do a non-directional test. We need to put the critical region in both tails, but we don’t want to increase the overall size of the rejection region (for reasons we will see later). To do this, we simply split it in half so that an equal proportion of the area under the curve falls in each tail’s rejection region. For α = .05, this means 2.5% of the area is in each tail, which, based on the z-table, corresponds to critical values of z* = ±1.96. This is shown in Figure 5.

Figure 5: Two-tailed rejection region

Thus, any z-score falling outside ±1.96 (greater than 1.96 in absolute value) falls in the rejection region. When we use z-scores in this way, the obtained value of z (sometimes called z-obtained) is something known as a test statistic, which is simply an inferential statistic used to test a null hypothesis.

Calculate the test statistic: Z

Now that we understand setting up the hypothesis and determining the outcome, let’s examine hypothesis testing with z!  The next step is to carry out the study and get the actual results for our sample. Central to hypothesis test is comparison of the population and sample means. To make our calculation and determine where the sample is in the hypothesized distribution we calculate the Z for the sample data.

Make a decision

To decide whether to reject the null hypothesis, we compare our sample’s Z score to the Z score that marks our critical boundary. If our sample Z score falls inside the rejection region of the comparison distribution (is greater than the z-score critical boundary) we reject the null hypothesis.

The formula for our z- statistic has not changed:

To formally test our hypothesis, we compare our obtained z-statistic to our critical z-value. If z obt > z crit , that means it falls in the rejection region (to see why, draw a line for z = 2.5 on Figure 1 or Figure 2) and so we reject H 0 . If z obt < z crit , we fail to reject. Remember that as z gets larger, the corresponding area under the curve beyond z gets smaller. Thus, the proportion, or p-value, will be smaller than the area for α, and if the area is smaller, the probability gets smaller. Specifically, the probability of obtaining that result, or a more extreme result, under the condition that the null hypothesis is true gets smaller.

Conversely, if we fail to reject, we know that the proportion will be larger than α because the z-statistic will not be as far into the tail. This is illustrated for a one- tailed test in Figure 6.

Figure 6. Relation between α, z obt , and p

When the null hypothesis is rejected, the effect is said to be statistically significant . Do not confuse statistical significance with practical significance. A small effect can be highly significant if the sample size is large enough.

Why does the word “significant” in the phrase “statistically significant” mean something so different from other uses of the word? Interestingly, this is because the meaning of “significant” in everyday language has changed. It turns out that when the procedures for hypothesis testing were developed, something was “significant” if it signified something. Thus, finding that an effect is statistically significant signifies that the effect is real and not due to chance. Over the years, the meaning of “significant” changed, leading to the potential misinterpretation.

Review: Steps of the Hypothesis Testing Process

The process of testing hypotheses follows a simple four-step procedure. This process will be what we use for the remained of the textbook and course, and though the hypothesis and statistics we use will change, this process will not.

Step 1: State the Hypotheses

Your hypotheses are the first thing you need to lay out. Otherwise, there is nothing to test! You have to state the null hypothesis (which is what we test) and the alternative hypothesis (which is what we expect). These should be stated mathematically as they were presented above AND in words, explaining in normal English what each one means in terms of the research question.

Step 2: Find the Critical Values

Next, we formally lay out the criteria we will use to test our hypotheses. There are two pieces of information that inform our critical values: α, which determines how much of the area under the curve composes our rejection region, and the directionality of the test, which determines where the region will be.

Step 3: Compute the Test Statistic

Once we have our hypotheses and the standards we use to test them, we can collect data and calculate our test statistic, in this case z . This step is where the vast majority of differences in future chapters will arise: different tests used for different data are calculated in different ways, but the way we use and interpret them remains the same.

Step 4: Make the Decision

Finally, once we have our obtained test statistic, we can compare it to our critical value and decide whether we should reject or fail to reject the null hypothesis. When we do this, we must interpret the decision in relation to our research question, stating what we concluded, what we based our conclusion on, and the specific statistics we obtained.

Example: Movie Popcorn

Let’s see how hypothesis testing works in action by working through an example. Say that a movie theater owner likes to keep a very close eye on how much popcorn goes into each bag sold, so he knows that the average bag has 8 cups of popcorn and that this varies a little bit, about half a cup. That is, the known population mean is μ = 8.00 and the known population standard deviation is σ =0.50. The owner wants to make sure that the newest employee is filling bags correctly, so over the course of a week he randomly assesses 25 bags filled by the employee to test for a difference (n = 25). He doesn’t want bags overfilled or under filled, so he looks for differences in both directions. This scenario has all of the information we need to begin our hypothesis testing procedure.

Our manager is looking for a difference in the mean cups of popcorn bags compared to the population mean of 8. We will need both a null and an alternative hypothesis written both mathematically and in words. We’ll always start with the null hypothesis:

H 0 : There is no difference in the cups of popcorn bags from this employee H 0 : μ = 8.00

Notice that we phrase the hypothesis in terms of the population parameter μ, which in this case would be the true average cups of bags filled by the new employee.

Our assumption of no difference, the null hypothesis, is that this mean is exactly

the same as the known population mean value we want it to match, 8.00. Now let’s do the alternative:

H A : There is a difference in the cups of popcorn bags from this employee H A : μ ≠ 8.00

In this case, we don’t know if the bags will be too full or not full enough, so we do a two-tailed alternative hypothesis that there is a difference.

Our critical values are based on two things: the directionality of the test and the level of significance. We decided in step 1 that a two-tailed test is the appropriate directionality. We were given no information about the level of significance, so we assume that α = 0.05 is what we will use. As stated earlier in the chapter, the critical values for a two-tailed z-test at α = 0.05 are z* = ±1.96. This will be the criteria we use to test our hypothesis. We can now draw out our distribution so we can visualize the rejection region and make sure it makes sense

Figure 7: Rejection region for z* = ±1.96

Step 3: Calculate the Test Statistic

Now we come to our formal calculations. Let’s say that the manager collects data and finds that the average cups of this employee’s popcorn bags is ̅X = 7.75 cups. We can now plug this value, along with the values presented in the original problem, into our equation for z:

So our test statistic is z = -2.50, which we can draw onto our rejection region distribution:

Figure 8: Test statistic location

Looking at Figure 5, we can see that our obtained z-statistic falls in the rejection region. We can also directly compare it to our critical value: in terms of absolute value, -2.50 > -1.96, so we reject the null hypothesis. We can now write our conclusion:

When we write our conclusion, we write out the words to communicate what it actually means, but we also include the average sample size we calculated (the exact location doesn’t matter, just somewhere that flows naturally and makes sense) and the z-statistic and p-value. We don’t know the exact p-value, but we do know that because we rejected the null, it must be less than α.

Effect Size

When we reject the null hypothesis, we are stating that the difference we found was statistically significant, but we have mentioned several times that this tells us nothing about practical significance. To get an idea of the actual size of what we found, we can compute a new statistic called an effect size. Effect sizes give us an idea of how large, important, or meaningful a statistically significant effect is.

For mean differences like we calculated here, our effect size is Cohen’s d :

Effect sizes are incredibly useful and provide important information and clarification that overcomes some of the weakness of hypothesis testing. Whenever you find a significant result, you should always calculate an effect size

Table 1. Interpretation of Cohen’s d

Example: Office Temperature

Let’s do another example to solidify our understanding. Let’s say that the office building you work in is supposed to be kept at 74 degree Fahrenheit but is allowed

to vary by 1 degree in either direction. You suspect that, as a cost saving measure, the temperature was secretly set higher. You set up a formal way to test your hypothesis.

You start by laying out the null hypothesis:

H 0 : There is no difference in the average building temperature H 0 : μ = 74

Next you state the alternative hypothesis. You have reason to suspect a specific direction of change, so you make a one-tailed test:

H A : The average building temperature is higher than claimed H A : μ > 74

Now that you have everything set up, you spend one week collecting temperature data:

You calculate the average of these scores to be 𝑋̅ = 76.6 degrees. You use this to calculate the test statistic, using μ = 74 (the supposed average temperature), σ = 1.00 (how much the temperature should vary), and n = 5 (how many data points you collected):

z = 76.60 − 74.00 = 2.60    = 5.78

          1.00/√5            0.45

This value falls so far into the tail that it cannot even be plotted on the distribution!

Figure 7: Obtained z-statistic

You compare your obtained z-statistic, z = 5.77, to the critical value, z* = 1.645, and find that z > z*. Therefore you reject the null hypothesis, concluding: Based on 5 observations, the average temperature (𝑋̅ = 76.6 degrees) is statistically significantly higher than it is supposed to be, z = 5.77, p < .05.

d = (76.60-74.00)/ 1= 2.60

The effect size you calculate is definitely large, meaning someone has some explaining to do!

Example: Different Significance Level

First, let’s take a look at an example phrased in generic terms, rather than in the context of a specific research question, to see the individual pieces one more time. This time, however, we will use a stricter significance level, α = 0.01, to test the hypothesis.

We will use 60 as an arbitrary null hypothesis value: H 0 : The average score does not differ from the population H 0 : μ = 50

We will assume a two-tailed test: H A : The average score does differ H A : μ ≠ 50

We have seen the critical values for z-tests at α = 0.05 levels of significance several times. To find the values for α = 0.01, we will go to the standard normal table and find the z-score cutting of 0.005 (0.01 divided by 2 for a two-tailed test) of the area in the tail, which is z crit * = ±2.575. Notice that this cutoff is much higher than it was for α = 0.05. This is because we need much less of the area in the tail, so we need to go very far out to find the cutoff. As a result, this will require a much larger effect or much larger sample size in order to reject the null hypothesis.

We can now calculate our test statistic.  The average of 10 scores is M = 60.40 with a µ = 60. We will use σ = 10 as our known population standard deviation. From this information, we calculate our z-statistic as:

Our obtained z-statistic, z = 0.13, is very small. It is much less than our critical value of 2.575. Thus, this time, we fail to reject the null hypothesis. Our conclusion would look something like:

Notice two things about the end of the conclusion. First, we wrote that p is greater than instead of p is less than, like we did in the previous two examples. This is because we failed to reject the null hypothesis. We don’t know exactly what the p- value is, but we know it must be larger than the α level we used to test our hypothesis. Second, we used 0.01 instead of the usual 0.05, because this time we tested at a different level. The number you compare to the p-value should always be the significance level you test at. Because we did not detect a statistically significant effect, we do not need to calculate an effect size. Note: some statisticians will suggest to always calculate effects size as a possibility of Type II error. Although insignificant, calculating d = (60.4-60)/10 = .04 which suggests no effect (and not a possibility of Type II error).

Review Considerations in Hypothesis Testing

Errors in hypothesis testing.

Keep in mind that rejecting the null hypothesis is not an all-or-nothing decision. The Type I error rate is affected by the α level: the lower the α level the lower the Type I error rate. It might seem that α is the probability of a Type I error. However, this is not correct. Instead, α is the probability of a Type I error given that the null hypothesis is true. If the null hypothesis is false, then it is impossible to make a Type I error. The second type of error that can be made in significance testing is failing to reject a false null hypothesis. This kind of error is called a Type II error.

Statistical Power

The statistical power of a research design is the probability of rejecting the null hypothesis given the sample size and expected relationship strength. Statistical power is the complement of the probability of committing a Type II error. Clearly, researchers should be interested in the power of their research designs if they want to avoid making Type II errors. In particular, they should make sure their research design has adequate power before collecting data. A common guideline is that a power of .80 is adequate. This means that there is an 80% chance of rejecting the null hypothesis for the expected relationship strength.

Given that statistical power depends primarily on relationship strength and sample size, there are essentially two steps you can take to increase statistical power: increase the strength of the relationship or increase the sample size. Increasing the strength of the relationship can sometimes be accomplished by using a stronger manipulation or by more carefully controlling extraneous variables to reduce the amount of noise in the data (e.g., by using a within-subjects design rather than a between-subjects design). The usual strategy, however, is to increase the sample size. For any expected relationship strength, there will always be some sample large enough to achieve adequate power.

Inferential statistics uses data from a sample of individuals to reach conclusions about the whole population. The degree to which our inferences are valid depends upon how we selected the sample (sampling technique) and the characteristics (parameters) of population data. Statistical analyses assume that sample(s) and population(s) meet certain conditions called statistical assumptions.

It is easy to check assumptions when using statistical software and it is important as a researcher to check for violations; if violations of statistical assumptions are not appropriately addressed then results may be interpreted incorrectly.

Learning Objectives

Having read the chapter, students should be able to:

• Conduct a hypothesis test using a z-score statistics, locating critical region, and make a statistical decision including.
• Explain the purpose of measuring effect size and power, and be able to compute Cohen’s d.

Exercises – Ch. 10

• List the main steps for hypothesis testing with the z-statistic. When and why do you calculate an effect size?
• z = 1.99, two-tailed test at α = 0.05
• z = 1.99, two-tailed test at α = 0.01
• z = 1.99, one-tailed test at α = 0.05
• You are part of a trivia team and have tracked your team’s performance since you started playing, so you know that your scores are normally distributed with μ = 78 and σ = 12. Recently, a new person joined the team, and you think the scores have gotten better. Use hypothesis testing to see if the average score has improved based on the following 8 weeks’ worth of score data: 82, 74, 62, 68, 79, 94, 90, 81, 80.
• A study examines self-esteem and depression in teenagers.  A sample of 25 teens with a low self-esteem are given the Beck Depression Inventory.  The average score for the group is 20.9.  For the general population, the average score is 18.3 with σ = 12.  Use a two-tail test with α = 0.05 to examine whether teenagers with low self-esteem show significant differences in depression.
• You get hired as a server at a local restaurant, and the manager tells you that servers’ tips are $42 on average but vary about $12 (μ = 42, σ = 12). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don’t know if you will make more or less in tips. After working 16 shifts, you find that your average nightly amount is $44.50 from tips. Test for a difference between this value and the population mean at the α = 0.05 level of significance.

Answers to Odd- Numbered Exercises – Ch. 10

1. List hypotheses. Determine critical region. Calculate z.  Compare z to critical region. Draw Conclusion.  We calculate an effect size when we find a statistically significant result to see if our result is practically meaningful or important

5. Step 1: H 0 : μ = 42 “My average tips does not differ from other servers”, H A : μ ≠ 42 “My average tips do differ from others”

Introduction to Statistics for Psychology Copyright © 2021 by Alisa Beyer is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

Share This Book

Learn Math and Stats with Dr. G

A shortcut is the longest distance between two points.

One-Tailed z-test Hypothesis Test By Hand

Example: Suppose it is up to you to determine if a certain state receives significantly more public school funding (per student) than the USA average. You know that the USA mean public school yearly funding is $6300 per student per year, with a standard deviation of $400. Next, suppose you collect a sample (n = 100) and determine that the sample mean for New Jersey (per student per year) is $8801. Use the z-test and the correct Ho and Ha to run a hypothesis test to determine if New Jersey receives significantly more funding for public school education (per student per year).

NOTE: This entire example works the same way if you have a dataset. Using the dataset, you would need to first calculate the sample mean. To run a z-test, it is generally expected that you have a larger sample size (30 or more) and that you have information about the population mean and standard deviation. If you do not have this information, it is sometimes best to use the t-test.

Step 1: Set up your hypothesis Hypothesis: The mean per student per year funding in New Jersey is significantly greater than the average per student per year funding over the entire USA.

Step 2: Create Ho and Ha NOTE: There are many ways to write out Ho.

Ho: mean per student per year funding for New Jersey = mean per student per year funding for the USA

This can also be written as the following. Ho: New Jersey mean – Population mean = 0

Ha: mean per student per year funding for New Jersey > mean per student per year funding for the USA

NOTICE1: The Ha in this example is ONE-TAILED because we are interested in seeing if New Jersey is significantly GREATER than the population mean. In a two-tailed test, the Ha contains a NOT EQUAL and the test will see if there is a significant difference (greater or smaller). NOTICE2: The Ho is the null hypothesis and so always contains the equal sign as it is the case for which there is no significant difference between the two groups.

Step 3: Calculate the z-test statistic Now, calculate the test statistic. In this example, we are using the z-test and are doing this by hand. However, there are many applications that run such tests. This Site has several examples under the Stats Apps link. z = (sample mean – population mean) / [population standard deviation/sqrt(n)] z = (8801 – 6300) / [400/sqrt(100)] z = 2501 / [400/10] z = 2501 / [40] z = 62.5

So, the z-test result, also called the test statistic is 62.5.

Step 4: Using the z-table , determine the rejection regions for you z-test.

To do this, you must first select an alpha value . The alpha value is the percentage chance that you will reject the null (choose to go with your Ha research hypothesis as you conclusion) when in fact the Ho really true (and your research Ha should not be selected). This is also called a Type I error (choosing Ha when Ho is actually correct). The smaller the alpha, the smaller the percentage of error, BUT the smaller the rejection regions and more difficult to reject Ho. Most research uses alpha at .05, which creates only a 5% chance of Type I error. However, in cases such as medical research, the alpha is set much smaller.

In our case, we will use alpha = .05

This is ONE-TAILED test, therefore the rejection region is any z-test value greater than the critical z value for a one-tailed test with alpha = .05. The critical value for one-tailed z-test at alpha = .05 is 1.645.

HOW TO Find Critical Values and Rejection Regions

Therefore, the rejection region is any value GREATER than 1.645.

Step 5: Create a conclusion Our z-test result is 62.5. This is very large! 62.5 is MUCH LARGER than 1.645 and so the result of the z test is INSIDE the rejection region.

(The rejection region is all portions of the curve to the right of 1.645. Why to the right? Because this is a one-tailed test for which we are looking at whether the sample is GREATER than the population).

Note: The p –value in this case would be the probability of getting a result of 62.5 randomly given that the rejection region starts at 1.645. The probability of getting 62.5 is nearly zero and so the p –value = 0. The actual calculation of the p value is mathematically complicated and is often done using a software like Excel, SPSS, SAS, R, etc.

This z-test has a significant result. Conclusion: The funding for New Jersey public schools (per student per year) is significantly GREATER than the average funding per student per year for the USA.

http://www.ascd.org/publications/educational-leadership/may02/vol59/num08/Unequal-School-Funding-in-the-United-States.aspx

Tutorial Playlist

Statistics tutorial, everything you need to know about the probability density function in statistics, the best guide to understand central limit theorem, an in-depth guide to measures of central tendency : mean, median and mode, the ultimate guide to understand conditional probability.

A Comprehensive Look at Percentile in Statistics

The Best Guide to Understand Bayes Theorem

Everything you need to know about the normal distribution, an in-depth explanation of cumulative distribution function, a complete guide to chi-square test, what is hypothesis testing in statistics types and examples, understanding the fundamentals of arithmetic and geometric progression, the definitive guide to understand spearman’s rank correlation, mean squared error: overview, examples, concepts and more, all you need to know about the empirical rule in statistics, the complete guide to skewness and kurtosis, a holistic look at bernoulli distribution.

All You Need to Know About Bias in Statistics

A Complete Guide to Get a Grasp of Time Series Analysis

The Key Differences Between Z-Test Vs. T-Test

The Complete Guide to Understand Pearson's Correlation

A complete guide on the types of statistical studies, everything you need to know about poisson distribution, your best guide to understand correlation vs. regression, the most comprehensive guide for beginners on what is correlation, hypothesis testing in statistics - types | examples.

Lesson 10 of 24 By Avijeet Biswal

Table of Contents

In today’s data-driven world, decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

The Ultimate Ticket to Top Data Science Job Roles

What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life - 

• A teacher assumes that 60% of his college's students come from lower-middle-class families.
• A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

• Here, x̅ is the sample mean,
• μ0 is the population mean,
• σ is the standard deviation,
• n is the sample size.

How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

Your Dream Career is Just Around The Corner!

Null Hypothesis and Alternative Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average. 

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

Become a Data Scientist with Hands-on Training!

Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

Steps in Hypothesis Testing

Hypothesis testing is a statistical method to determine if there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. Here’s a breakdown of the typical steps involved in hypothesis testing:

Formulate Hypotheses

• Null Hypothesis (H0): This hypothesis states that there is no effect or difference, and it is the hypothesis you attempt to reject with your test.
• Alternative Hypothesis (H1 or Ha): This hypothesis is what you might believe to be true or hope to prove true. It is usually considered the opposite of the null hypothesis.

Choose the Significance Level (α)

The significance level, often denoted by alpha (α), is the probability of rejecting the null hypothesis when it is true. Common choices for α are 0.05 (5%), 0.01 (1%), and 0.10 (10%).

Select the Appropriate Test

Choose a statistical test based on the type of data and the hypothesis. Common tests include t-tests, chi-square tests, ANOVA, and regression analysis. The selection depends on data type, distribution, sample size, and whether the hypothesis is one-tailed or two-tailed.

Collect Data

Gather the data that will be analyzed in the test. This data should be representative of the population to infer conclusions accurately.

Calculate the Test Statistic

Based on the collected data and the chosen test, calculate a test statistic that reflects how much the observed data deviates from the null hypothesis.

Determine the p-value

The p-value is the probability of observing test results at least as extreme as the results observed, assuming the null hypothesis is correct. It helps determine the strength of the evidence against the null hypothesis.

Make a Decision

Compare the p-value to the chosen significance level:

• If the p-value ≤ α: Reject the null hypothesis, suggesting sufficient evidence in the data supports the alternative hypothesis.
• If the p-value > α: Do not reject the null hypothesis, suggesting insufficient evidence to support the alternative hypothesis.

Report the Results

Present the findings from the hypothesis test, including the test statistic, p-value, and the conclusion about the hypotheses.

Perform Post-hoc Analysis (if necessary)

Depending on the results and the study design, further analysis may be needed to explore the data more deeply or to address multiple comparisons if several hypotheses were tested simultaneously.

Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

Chi-Square 

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

Become a Data Scientist through hands-on learning with hackathons, masterclasses, webinars, and Ask-Me-Anything! Start learning now!

Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

Become a Data Scientist With Real-World Experience

Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

• The null hypothesis is (H0 <= 90) or less change.
• A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true]. 

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

Our Data Scientist Master's Program covers core topics such as R, Python, Machine Learning, Tableau, Hadoop, and Spark. Get started on your journey today!

Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

• It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
• Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
• Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
• Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

Learn All The Tricks Of The BI Trade

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

If you are interested in statistics of data science and skills needed for such a career, you ought to explore the Post Graduate Program in Data Science.

If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

2. What is H0 and H1 in statistics?

In statistics, H0​ and H1​ represent the null and alternative hypotheses. The null hypothesis, H0​, is the default assumption that no effect or difference exists between groups or conditions. The alternative hypothesis, H1​, is the competing claim suggesting an effect or a difference. Statistical tests determine whether to reject the null hypothesis in favor of the alternative hypothesis based on the data.

3. What is a simple hypothesis with an example?

A simple hypothesis is a specific statement predicting a single relationship between two variables. It posits a direct and uncomplicated outcome. For example, a simple hypothesis might state, "Increased sunlight exposure increases the growth rate of sunflowers." Here, the hypothesis suggests a direct relationship between the amount of sunlight (independent variable) and the growth rate of sunflowers (dependent variable), with no additional variables considered.

4. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

• Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
• Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
• Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

Find our PL-300 Microsoft Power BI Certification Training Online Classroom training classes in top cities:

About the Author

Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

Recommended Resources

Free eBook: Top Programming Languages For A Data Scientist

Normality Test in Minitab: Minitab with Statistics

Machine Learning Career Guide: A Playbook to Becoming a Machine Learning Engineer

• PMP, PMI, PMBOK, CAPM, PgMP, PfMP, ACP, PBA, RMP, SP, and OPM3 are registered marks of the Project Management Institute, Inc.