Example 81 Pottery Data (MANOVA) Section
Before carrying out a MANOVA, first check the model assumptions:
 The data from group i has common mean vector \(\boldsymbol{\mu}_{i}\)
 The data from all groups have common variancecovariance matrix \(\Sigma\).
 Independence: The subjects are independently sampled.
 Normality: The data are multivariate normally distributed.
Assumptions

Assumption 1: The data from group i has common mean vector \(\boldsymbol{\mu}_{i}\)
This assumption says that there are no subpopulations with different mean vectors. Here, this assumption might be violated if pottery collected from the same site had inconsistencies.

Assumption 3: Independence: The subjects are independently sampled. This assumption is satisfied if the assayed pottery are obtained by randomly sampling the pottery collected from each site. This assumption would be violated if, for example, pottery samples were collected in clusters. In other applications, this assumption may be violated if the data were collected over time or space.

Assumption 4: Normality: The data are multivariate normally distributed.
 For large samples, the Central Limit Theorem says that the sample mean vectors are approximately multivariate normally distributed, even if the individual observations are not.
 For the pottery data, however, we have a total of only N = 26 observations, including only two samples from Caldicot. With small N, we cannot rely on the Central Limit Theorem.
Diagnostic procedures are based on the residuals, computed by taking the differences between the individual observations and the group means for each variable:
\(\hat{\epsilon}_{ijk} = Y_{ijk}\bar{Y}_{i.k}\)
Thus, for each subject (or pottery sample in this case), residuals are defined for each of the p variables. Then, to assess normality, we apply the following graphical procedures:
 Plot the histograms of the residuals for each variable. Look for a symmetric distribution.
 Plot a matrix of scatter plots. Look for elliptical distributions and outliers.
 Plot threedimensional scatter plots. Look for elliptical distributions and outliers.
If the histograms are not symmetric or the scatter plots are not elliptical, this would be evidence that the data are not sampled from a multivariate normal distribution in violation of Assumption 4. In this case, a normalizing transformation should be considered.
Download the text file containing the data here: pottery.txt
Using SAS
The SAS program below will help us check this assumption.
Download the SAS Program here: potterya.sas
View the video explanation of the SAS code.Using Minitab
Minitab procedures are not shown separately.
These can be handled using procedures already known.
 Histograms suggest that, except for sodium, the distributions are relatively symmetric. However, the histogram for sodium suggests that there are two outliers in the data. Both of these outliers are in Llanadyrn.
 Two outliers can also be identified from the matrix of scatter plots.
 Removal of the two outliers results in a more symmetric distribution for sodium.
The results of MANOVA can be sensitive to the presence of outliers. One approach to assessing this would be to analyze the data twice, once with the outliers and once without them. The results may then be compared for consistency. The following analyses use all of the data, including the two outliers.
Assumption 2: The data from all groups have common variancecovariance matrix \(\Sigma\).
This assumption can be checked using Bartlett's test for homogeneity of variancecovariance matrices. To obtain Bartlett's test, let \(\Sigma_{i}\) denote the population variancecovariance matrix for group i . Consider testing:
\(H_0\colon \Sigma_1 = \Sigma_2 = \dots = \Sigma_g\)
against
\(H_0\colon \Sigma_i \ne \Sigma_j\) for at least one \(i \ne j\)
Under the alternative hypothesis, at least two of the variancecovariance matrices differ on at least one of their elements. Let:
\(\mathbf{S}_i = \dfrac{1}{n_i1}\sum\limits_{j=1}^{n_i}\mathbf{(Y_{ij}\bar{y}_{i.})(Y_{ij}\bar{y}_{i.})'}\)
denote the sample variancecovariance matrix for group i . Compute the pooled variancecovariance matrix
\(\mathbf{S}_p = \dfrac{\sum_{i=1}^{g}(n_i1)\mathbf{S}_i}{\sum_{i=1}^{g}(n_i1)}= \dfrac{\mathbf{E}}{Ng}\)
Bartlett's test is based on the following test statistic:
\(L' = c\left\{(Ng)\log \mathbf{S}_p  \sum_{i=1}^{g}(n_i1)\log\mathbf{S}_i\right\}\)
where the correction factor is
\(c = 1\dfrac{2p^2+3p1}{6(p+1)(g1)}\left\{\sum_\limits{i=1}^{g}\dfrac{1}{n_i1}\dfrac{1}{Ng}\right\}\)
The version of Bartlett's test considered in the lesson of the twosample Hotelling's Tsquare is a special case where g = 2. Under the null hypothesis of homogeneous variancecovariance matrices, L' is approximately chisquare distributed with
\(\dfrac{1}{2}p(p+1)(g1)\)
degrees of freedom. Reject \(H_0\) at level \(\alpha\) if
\(L' > \chi^2_{\frac{1}{2}p(p+1)(g1),\alpha}\)
Example 82: Pottery Data Section
Using SAS
Here we will use the Pottery SAS program.
Download the SAS Program here: pottery2.sas
View the video explanation of the SAS code.Using Minitab
Minitab procedures are not shown separately.
These can be handled using procedures already known.
Analysis
We find no statistically significant evidence against the null hypothesis that the variancecovariance matrices are homogeneous (L' = 27.58; d.f. = 45; p = 0.98).
Notes
 If we were to reject the null hypothesis of homogeneity of variancecovariance matrices, then we would conclude that assumption 2 is violated.
 MANOVA is not robust to violations of the assumption of homogeneous variancecovariance matrices.
 If the variancecovariance matrices are determined to be unequal then the solution is to find a variancestabilizing transformation.
 Note that the assumptions of homogeneous variancecovariance matrices and multivariate normality are often violated together.
 Therefore, a normalizing transformation may also be a variancestabilizing transformation.