Test Statistic | DF | ||||
---|---|---|---|---|---|
Criterion | F | Num | Denom | P | |
Wilks' | 0.89178 | 1.621 | 4 | 110 | 0.174 |
Lawley-Hotelling | 0.11972 | 1.616 | 4 | 108 | 0.175 |
Pillai's | 0.10967 | 1.625 | 4 | 112 | 0.173 |
Roy's | 0.10400 |
Test Statistic | DF | ||||
---|---|---|---|---|---|
Criterion | F | Num | Denom | P | |
Wilks' | 0.85826 | 2.184 | 4 | 110 | 0.075 |
Lawley-Hotelling | 0.16439 | 2.219 | 4 | 108 | 0.072 |
Pillai's | 0.14239 | 2.146 | 4 | 112 | 0.080 |
Roy's | 0.15966 |
The p-values for the production method are statistically significant at the 0.10 significance level. The p-values for the manufacturing plant are not significant at the 0.10 significance level for any of the tests. The p-values for the interaction between plant and method are statistically significant at the 0.10 significance level. Because the interaction is statistically significant, the effect of the method depends on the plant.
Use the eigen analysis to assess how the response means differ between the levels of the different model terms. You should focus on the eigenvectors that correspond to high eigenvalues. To display the eigen analysis, go to Stat > ANOVA > General MANOVA > Results and select Eigen analysis under Display of Results .
Eigenvalue | 0.5848 | 0.00000 |
---|---|---|
Proportion | 1.0000 | 0.00000 |
Cumulative | 1.0000 | 1.00000 |
Eigenvector | 1 | 2 |
---|---|---|
Usability Rating | 0.144062 | -0.07870 |
Quality Rating | -0.003968 | 0.13976 |
In these results, the first eigenvalue for method (0.5848) is greater than the second eigenvalue (0.00000). Therefore, you should put higher importance on the first eigenvector. The first eigenvector for method is 0.144062, -0.003968. The highest absolute value within this vector is for the usability rating. This suggests that the means for usability have the largest difference between the factor levels for method. This information is helpful for assessing the means table.
Use the Means table to understand the statistically significant differences between the factor levels in your data. The mean of each group provides an estimate of each population mean. Look for differences between group means for terms that are statistically significant.
For main effects, the table displays the groups within each factor and their means. For interaction effects, the table displays all possible combinations of the groups. If an interaction term is statistically significant, do not interpret the main effects without considering the interaction effects.
To display the means, go to Stat > ANOVA > General MANOVA > Results , select Univariate analysis of variance , and enter the terms in Display least squares means corresponding to the terms .
Usability Rating | Quality Rating | |||
---|---|---|---|---|
Mean | SE Mean | Mean | SE Mean | |
Method | ||||
Method 1 | 4.819 | 0.165 | 5.242 | 0.193 |
Method 2 | 6.212 | 0.179 | 6.026 | 0.211 |
Plant | ||||
Plant A | 5.708 | 0.192 | 5.833 | 0.226 |
Plant B | 5.493 | 0.232 | 5.914 | 0.273 |
Plant C | 5.345 | 0.206 | 5.155 | 0.242 |
Method*Plant | ||||
Method 1 Plant A | 4.667 | 0.272 | 5.417 | 0.319 |
Method 1 Plant B | 4.700 | 0.298 | 5.400 | 0.350 |
Method 1 Plant C | 5.091 | 0.284 | 4.909 | 0.334 |
Method 2 Plant A | 6.750 | 0.272 | 6.250 | 0.319 |
Method 2 Plant B | 6.286 | 0.356 | 6.429 | 0.418 |
Method 2 Plant C | 5.600 | 0.298 | 5.400 | 0.350 |
In these results, the Means table shows how the mean usability and quality ratings varies by method, plant, and the method*plant interaction. Method and the interaction term are statistically significant at the 0.10 level. The table shows that method 1 and method 2 are associated with mean usability ratings of 4.819 and 6.212 respectively. The difference between these means is larger than the difference between the corresponding means for quality rating. This confirms the interpretation of the eigen analysis.
However, because the Method*Plant interaction term is also statistically significant, do not interpret the main effects without considering the interaction effects. For example, the table for the interaction term shows that with method 1, plant C is associated with the highest usability rating and the lowest quality rating. However, with method 2, plant A is associated with the highest usability rating and a quality rating that is nearly equal to the highest quality rating.
When you perform General MANOVA , you can choose to calculate the univariate statistics to examine the individual responses. The univariate results can provide a more intuitive understanding of the relationships in your data. However, the univariate results can differ from the multivariate results.
To display the univariate results, go to Stat > ANOVA > General MANOVA > Results and select Univariate analysis of variance under Display of Results .
Source | DF | Seq SS | Adj SS | Adj MS | F | P |
---|---|---|---|---|---|---|
Method | 1 | 31.264 | 29.074 | 29.0738 | 32.72 | 0.000 |
Plant | 2 | 1.366 | 1.499 | 0.7495 | 0.84 | 0.436 |
Method*Plant | 2 | 7.099 | 7.099 | 3.5494 | 3.99 | 0.024 |
Error | 56 | 49.754 | 49.754 | 0.8885 | ||
Total | 61 | 89.484 |
Source | DF | Seq SS | Adj SS | Adj MS | F | P |
---|---|---|---|---|---|---|
Method | 1 | 8.8587 | 9.2196 | 9.2196 | 7.53 | 0.008 |
Plant | 2 | 6.7632 | 7.0572 | 3.5286 | 2.88 | 0.064 |
Method*Plant | 2 | 0.7074 | 0.7074 | 0.3537 | 0.29 | 0.750 |
Error | 56 | 68.5900 | 68.5900 | 1.2248 | ||
Total | 61 | 84.9194 |
In these results, the p-value for the main effect of method and the method*plant interaction effect are statistically significant at the 0.10 level in the model for usability rating. The main effects of both method and plant are statistically significant in the model for quality rating. You can conclude that changes in these variables are associated with changes in the response variables.
Use the residual plots to help you determine whether the model is adequate and meets the assumptions of the analysis. If the assumptions are not met, the model may not fit the data well and you should use caution when you interpret the results.
When you perform General MANOVA , Minitab displays residual plots for all response variables that are in your model. You must determine whether the residual plots for all response variables indicate that the model meets the assumptions.
For more information on how to handle patterns in the residual plots, go to Residual plots for General MANOVA and click the name of the residual plot in the list at the top of the page.
Use the residuals versus fits plot to verify the assumption that the residuals are randomly distributed and have constant variance. Ideally, the points should fall randomly on both sides of 0, with no recognizable patterns in the points.
Pattern | What the pattern may indicate |
---|---|
Fanning or uneven spreading of residuals across fitted values | Nonconstant variance |
Curvilinear | A missing higher-order term |
A point that is far away from zero | An outlier |
A point that is far away from the other points in the x-direction | An influential point |
Use the normal probability plot of the residuals to verify the assumption that the residuals are normally distributed. The normal probability plot of the residuals should approximately follow a straight line.
Pattern | What the pattern may indicate |
---|---|
Not a straight line | Nonnormality |
A point that is far away from the line | An outlier |
Changing slope | An unidentified variable |
You are now leaving support.minitab.com.
Click Continue to proceed to:
Statistical Methods and Data Analytics
This page shows an example of multivariate analysis of variance (manova) in SPSS with footnotes explaining the output. The data used in this example are from the following experiment.
A researcher randomly assigns 33 subjects to one of three groups. The first group receives technical dietary information interactively from an on-line website. Group 2 receives the same information from a nurse practitioner, while group 3 receives the information from a video tape made by the same nurse practitioner. Each subject then made three ratings: difficulty, usefulness, and importance of the information in the presentation. The researcher looks at three different ratings of the presentation (difficulty, usefulness and importance) to determine if there is a difference in the modes of presentation. In particular, the researcher is interested in whether the interactive website is superior because that is the most cost-effective way of delivering the information. In the dataset, the ratings are presented in the variables useful , difficulty and importance . The variable group indicates the group to which a subject was assigned.
We are interested in how the variability in the three ratings can be explained by a subject’s group. Group is a categorical variable with three possible values: 1, 2 or 3. Because we have multiple dependent variables that cannot be combined, we will choose to use manova. Our null hypothesis in this analysis is that a subject’s group has no effect on any of the three different ratings, and we can test this hypothesis on the dataset, manova.sav .
We can start by examining the three outcome variables.
Next, we can enter our manova command. In SPSS, manova can be conducted through the generalized linear model function, GLM. In the manova command, we first list the outcome variables, then indicate any categorical factors after “by” and any covariates after “with”. Here, group is a categorical factor. We must also indicate the lowest and highest values found in group . We are also asking SPSS to print the eigenvalues generated. These will be useful in seeing how the test statistics are calculated.
a. Case summary – This provides counts of the observations to be included in the manova and the counts of observations to be dropped due to missing data or data that falls out-of-range. For example, a record where the value for group is 4, after we have specified that the maximum value for group is 3, would be considered out-of-range.
b. Effect – This indicates the predictor variable in question. In our model, we are looking at the effect of group .
c. Value – This is the test statistic for the given effect and multivariate statistic listed in the prior column. For each predictor variable, SPSS calculates four test statistics. All of these test statistics are calculated using the eigenvalues of the model (see superscript m). See superscripts h, i, j and k for explanations of each of the tests.
d. Approx. F – This is the approximate F statistic for the given effect and test statistic.
e. Hypoth. DF – This is the number of degrees of freedom in the model.
f. Error DF – This is the number of degrees of freedom associated with the model errors. There are instances in manova when the degrees of freedom may be a non-integer.
g. Sig. of F – This is the p-value associated with the F statistic and the hypothesis and error degrees of freedom of a given effect and test statistic. The null hypothesis that a given predictor has no effect on either of the outcomes is evaluated with regard to this p-value. For a given alpha level, if the p-value is less than alpha, the null hypothesis is rejected. If not, then we fail to reject the null hypothesis. In this example, we reject the null hypothesis that group has no effect on the three different ratings at alpha level .05 because the p-values are all less than .05.
h. Pillais – This is Pillai’s Trace, one of the four multivariate criteria test statistics used in manova. We can calculate Pillai’s trace using the generated eigenvalues (see superscript m). Divide each eigenvalue by (1 + the eigenvalue), then sum these ratios. So in this example, you would first calculate 0.89198790/(1+0.89198790) = 0.471455394, 0.00524207/(1+0.00524207) = 0.005214734, and 0/(1+0)=0. When these are added, we arrive at Pillai’s trace: (0.471455394 + 0.005214734 + 0) = .47667.
i. Hotellings – This is Lawley-Hotelling’s Trace. It is very similar to Pillai’s Trace. It is the sum of the eigenvalues (see superscript m) and is a direct generalization of the F statistic in ANOVA. We can calculate the Hotelling-Lawley Trace by summing the characteristic roots listed in the output: 0.8919879 + 0.00524207 + 0 = 0.89723.
j. Wilks – This is Wilk’s Lambda. This can be interpreted as the proportion of the variance in the outcomes that is not explained by an effect. To calculate Wilks’ Lambda, for each eigenvalue, calculate 1/(1 + the eigenvalue), then find the product of these ratios. So in this example, you would first calculate 1/(1+0.8919879) = 0.5285446, 1/(1+0.00524207) = 0.9947853, and 1/(1+0)=1. Then multiply 0.5285446 * 0.9947853 * 1 = 0.52579.
k. Roys – This is Roy’s Largest Root. Note there are two definitions of Roy’s Largest Root depending on whether you use SPSS MANOVA or SPSS GLM. In MANOVA, the calculation is 0.8919879/(1+0.8919879) = 0.4714544. In GLM, Roy’s Largest root is defined as the largest eigenvalue, which is 0.8919879. Based on this definition, Roy’s Largest Root can behave differently from the other three test statistics. In instances where the other three are not significant and Roy’s is significant, the effect should be considered not significant. You can reference the following article for a detailed description of this discrepancy:
Kuhfeld, W. F. (1986). A note on Roy’s largest root. Psychometrika , 51 (3), 479-481.
l. Note – This indicates that the F statistic for Wilk’s Lambda was calculated exactly. For the other test statistics, the F values are approximate (as indicated by the column heading).
m. Eigenvalues and Canonical Correlations – This section of output provides the eigenvalues from the product of the sum-of-squares matrix of the model and the sum-of-squares matrix of the errors. There is one eigenvalue for each of the three eigenvectors of the product of the model sum of squares matrix and the error sum of squares matrix, a 3×3 matrix. Because only two are listed here, we can assume the third eigenvalue is zero. These values can be used to calculate the four multivariate test statistics.
n. Univariate F-tests – The manova procedure provides both univariate and multivariate output. This section of output provides summarized output from a one-way anova for each of the outcomes in the manova. Each row corresponds to a different one-way anova, one for each dependent variable in the manova. While the manova tested a single hypothesis, each line in this output corresponds to a test of a different hypothesis. Generally, if your manova suggests that an effect is significant, you would expect at least one of these one-way anova tests to indicate that the effect is significant on a single outcome.
Your Name (required)
Your Email (must be a valid email for us to receive the report!)
Comment/Error Report (required)
How to cite this page
Content preview.
Arcu felis bibendum ut tristique et egestas quis:
8.3 - test statistics for manova.
SAS uses four different test statistics based on the MANOVA table:
\(\Lambda^* = \dfrac{|\mathbf{E}|}{|\mathbf{H+E}|}\)
Here, the determinant of the error sums of squares and cross-products matrix E is divided by the determinant of the total sum of squares and cross-products matrix T = H + E . If H is large relative to E , then | H + E | will be large relative to | E |. Thus, we will reject the null hypothesis if Wilks lambda is small (close to zero).
\(T^2_0 = trace(\mathbf{HE}^{-1})\)
Here, we are multiplying H by the inverse of E ; then we take the trace of the resulting matrix. If H is large relative to E , then the Hotelling-Lawley trace will take a large value. Thus, we will reject the null hypothesis if this test statistic is large.
\(V = trace(\mathbf{H(H+E)^{-1}})\)
Here, we are multiplying H by the inverse of the total sum of squares and cross products matrix T = H + E . If H is large relative to E , then the Pillai trace will take a large value. Thus, we will reject the null hypothesis if this test statistic is large.
Here, we multiply H by the inverse of E and then compute the largest eigenvalue of the resulting matrix. If H is large relative to E , then Roy's root will take a large value. Thus, we will reject the null hypothesis if this test statistic is large.
Recall: The trace of a p x p matrix
\(\mathbf{A} = \left(\begin{array}{cccc}a_{11} & a_{12} & \dots & a_{1p}\\ a_{21} & a_{22} & \dots & a_{2p} \\ \vdots & \vdots & & \vdots \\ a_{p1} & a_{p2} & \dots & a_{pp}\end{array}\right)\)
is equal to
\(trace(\mathbf{A}) = \sum_{i=1}^{p}a_{ii}\)
Statistical tables are not available for the above test statistics. However, each of the above test statistics has an F approximation: The following details the F approximations for Wilks lambda. Details for all four F approximations can be found on the SAS website .
1. Wilks Lambda
\begin{align} \text{Starting with }&& \Lambda^* &= \dfrac{|\mathbf{E}|}{|\mathbf{H+E}|}\\ \text{Let, }&& a &= N-g - \dfrac{p-g+2}{2},\\ &&\text{} b &= \left\{\begin{array}{ll} \sqrt{\frac{p^2(g-1)^2-4}{p^2+(g-1)^2-5}}; &\text{if } p^2 + (g-1)^2-5 > 0\\ 1; & \text{if } p^2 + (g-1)^2-5 \le 0 \end{array}\right. \\ \text{and}&& c &= \dfrac{p(g-1)-2}{2} \\ \text{Then}&& F &= \left(\dfrac{1-\Lambda^{1/b}}{\Lambda^{1/b}}\right)\left(\dfrac{ab-c}{p(g-1)}\right) \overset{\cdot}{\sim} F_{p(g-1), ab-c} \\ \text{Under}&& H_{o} \end{align}
Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
If we fail to reject the null hypothesis of MANOVA at an alpha level, does that mean that all the underlying univariate ANOVA tests will not be significant as well at the same alpha level?
Not necessarily. But if you're planning to do each individual ANOVA anyway (presumably adjusting for multiple comparisons), why bother with the MANOVA?
If you fail to reject the null hypothesis of MANOVA at some alpha level, then you can report your result as " There is no statistically significant difference ..." And you do not need any follow up tests.
Sign up or log in, post as a guest.
Required, but never shown
By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy .
IMAGES
COMMENTS
Covariances etc. enter the assumptions and the computations of MANOVA, not the null hypothesis. $\endgroup$ - amoeba. Commented Jan 13, 2015 at 21:41 $\begingroup$ @amoeba, I didn't like For each response variable. To me it sounds like (or I read it as) "testing is done univarietly on each" (and then somehow combined). $\endgroup$
The Multivariate Analysis of Variance (MANOVA) is the multivariate analog of the Analysis of Variance (ANOVA) procedure used for univariate data. We will introduce the Multivariate Analysis of Variance with the Romano-British Pottery data example. ... Under the null hypothesis of homogeneous variance-covariance matrices, L' is approximately chi ...
Thus we reject the null hypothesis when Wilk's Lambda is close to zero. By Property 1, . Hotelling-Lawley Trace: = trace(HE-1) H is large compared to E when Hotelling-Lawley Trace is large. In this case, we reject the null hypothesis. Pillai-Bartlett Trace: V = trace(H(H+E)-1) If H is large compared to E then this statistic will be large ...
Formulation of Hypotheses: In a MANOVA, you have two types of hypotheses - null and alternative. The null hypothesis posits that the population means of the dependent variables are equal across different groups, while the alternative hypothesis suggests that at least one dependent variable's mean is different.
needed to reject the null hypothesis. This means that it is more di cult to reject H 0 (since we reject for small ) unless the null hypothesis is false for the new variables. I.e., adding new variables for which the populations are equal makes it harder to reject the null hypothesis. I When v H = 1;2 or p = 1;2, Wilks' is equivalent to an F ...
Detailed Walkthrough of the MANOVA Process. Define the Hypothesis: Clearly state the null and alternative hypotheses regarding the relationships between the independent and dependent variables. Choose the MANOVA Test: Select the appropriate MANOVA test based on the study design and hypothesis. Standard tests include Wilks' Lambda, Pillai's ...
The null hypothesis addresses whether there are any differences between groups of means. As in ANOVA, this is accomplished by partitioning variance. Therefore, as for ANOVA, the test is whether the variance in the MANOVA model exceeds the residual variance.
the null hypothesis is also a test of whether the vectors (columns) of means are equal across groups. A significant result indicates that one or more of the dependent variable means differ among groups. Although usually a set of univariate ANOVA comparisons will be consistent with the MANOVA, there are some
Here we could consider testing the null hypothesis that all of the treatment mean vectors are identical, \(H_0\colon \boldsymbol{\mu_1 = \mu_2 = \dots = \mu_g}\) ... This is the same null hypothesis that we tested in the One-way MANOVA. We would test this against the alternative hypothesis that there is a difference between at least one pair of ...
under the null hypothesis. The likelihood ratio statistic is a monotone function of T2(n 1 + n 2 2). This extends the two-sample t-test for multivariate observations. When we have several ... MANOVA is applied to this dataset of size 10 120. The p-value suggests that the di erence in e ects of drug compounds is statistically signi cant. A ...
MANOVA The E and H matrices can be used in di erent ways to test the null hypothesis. Wilks' Test Statistic is = jEj jE + Hj The null is rejected if < ;p;v H;v E where v H is the degrees of freedom for the hypothesis, k 1, and v E is degrees of freedom for error, k(n 1). Critical values are in Table A9. The test statistic can instead be converted
8.2 - The Multivariate Approach: One-way Multivariate Analysis of Variance (One-way MANOVA) Now we will consider the multivariate analog, the Multivariate Analysis of Variance, often abbreviated as MANOVA. ... This says that the null hypothesis is false if at least one pair of treatments is different on at least one variable.
In statistics, multivariate analysis of variance (MANOVA) is a procedure for comparing multivariate sample means. ... An algorithm for the distribution of the Roy's largest root under the null hypothesis was derived in [7] while the distribution under the alternative is studied in. [8]
Multivariate ANOVA (MANOVA) extends the capabilities of analysis of variance (ANOVA) by assessing multiple dependent variables simultaneously. ANOVA statistically tests the differences between three or more group means. For example, if you have three different teaching methods and you want to evaluate the average scores for these groups, you ...
As the p value is non-significant (p > 0.05) for Mardia's Skewness and Kurtosis test, we fail to reject the null hypothesis and conclude that data follows multivariate normality.. Here both Skewness and Kurtosis p value should be > 0.05 for concluding the multivariate normality.. Homogeneity of the variance-covariance matrices. We will use Box's M test to assess the homogeneity of the ...
MANOVA is the multivariate analogue to Hotelling's T2. The purpose of MANOVA is to test whether the vectorsof means for the two or more groups are sampled from the same sampling distribution. Just as Hotelling's T2will provide a measure of the likelihood of picking two random vectors of means out of the same hat, MANOVA gives a measure of the ...
The dialog box Post Hoc tests is used to conduct a separate comparison between factor levels, this is useful if the MANOVA includes factors have more than two factor levels. In our case we select two factors and each has only two factor levels (male/female and pass/fail). The MANOVA's F-test will test the null hypothesis that all means are ...
The null hypothesis is that there is no association between the term and the response. Usually, a significance level (denoted as α or alpha) of 0.05 works well. A significance level of 0.05 indicates a 5% risk of concluding that an association exists when there is no actual association. ... When you perform General MANOVA, Minitab displays ...
The Multivariate analysis of variance (MANOVA) procedure provides regression analysis and analysis of variance for multiple dependent variables by one or more factor variables or covariates. The factor variables divide the population into groups. Using this general linear model procedure, you can test null hypotheses about the effects of factor ...
The null hypothesis that a given predictor has no effect on either of the outcomes is evaluated with regard to this p-value. For a given alpha level, if the p-value is less than alpha, the null hypothesis is rejected. ... While the manova tested a single hypothesis, each line in this output corresponds to a test of a different hypothesis ...
8.2 - The Multivariate Approach: One-way Multivariate Analysis of Variance (One-way MANOVA) 8.3 - Test Statistics for MANOVA; 8.4 - Example: Pottery Data - Checking Model Assumptions; 8.5 - Example: MANOVA of Pottery Data; 8.6 - Orthogonal Contrasts; 8.7 - Constructing Orthogonal Contrasts; 8.8 - Hypothesis Tests; 8.9 - Randomized Block Design ...
The following examples show how to decide to reject or fail to reject the null hypothesis in both a one-way ANOVA and two-way ANOVA. Example 1: One-Way ANOVA. Suppose we want to know whether or not three different exam prep programs lead to different mean scores on a certain exam. To test this, we recruit 30 students to participate in a study ...
answered Apr 27, 2017 at 15:49. 369 2 8. Add a comment. If you fail to reject the null hypothesis of MANOVA at some alpha level, then you can report your result as " There is no statistically significant difference ..." And you do not need any follow up tests.