8.11 - Forming a MANOVA table

8.11 - Forming a MANOVA table

The partitioning of the total sum of squares and cross products matrix may be summarized in the multivariate analysis of variance table as shown below:

MANOVA
Source d.f. SSP
Blocks b - 1 B
Treatments a - 1 H
Error (a - 1)(b - 1) E
Total ab - 1 T

SSP stands for the sum of squares and cross products discussed above.

To test the null hypothesis that the treatment mean vectors are equal, compute a Wilks Lambda using the following expression:

\(\Lambda^* = \dfrac{|\mathbf{E}|}{|\mathbf{H+E}|}\)

This is the determinant of the error sum of squares and cross-products matrix divided by the determinant of the sum of the treatment sum of squares and cross-products plus the error sum of squares and cross-products matrix.

Under the null hypothesis, this has an F-approximation. The approximation is quite involved and will not be reviewed here. Instead, let's take a look at our example where we will implement these concepts.

Example 8-11: Rice Data

Rice data can be downloaded here: rice.csv

The program below shows the analysis of the rice data.

Download the SAS Program here: rice.sas

 

Note: In the upper right-hand corner of the code block you will have the option of copying ( ) the code to your clipboard or downloading ( ) the file to your computer.

options ls=78;
title "Two-Way MANOVA: Rice Data";

data rice;
  infile "D:\Statistics\STAT 505\data\rice.csv" firstobs=2 delimiter=',';
  input block variety $ height tillers;
  run;

 /*
  * The class statement specifies block and variety as categorical variables.
  * The model statement specifies responses height and tillers with predictors
  * block and variety. lsmeans displays the least-squares means for variety.
  * The manova statement requests the test for response mean vectors across
  * levels of variety, and the printe and printh options display sums of 
  * squares and cross products for error and the hypothesis, respectively.
  */

proc glm data=rice;
  class block variety;
  model height tillers=block variety;
  lsmeans variety;
  manova h=variety / printe printh;
  run;

Performing a two-way MANOVA

To carry out the two-way MANOVA test in Minitab:

  1. Open the ‘rice’ data set in a new worksheet.
  2. For convenience, rename the columns: block, variety, height, and tillers, from left to right.
  3. Stat > ANOVA > General MANOVA
    1. Highlight and select height and tillers to move them to the Responses window.
    2. Highlight and select block and variety to move them to the Model window.
    3. Choose 'OK'. The MANOVA results for the tests for block and variety are separately displayed in the results area.

Analysis

  • We reject the null hypothesis that the variety mean vectors are identical \(( \Lambda = 0.342; F = 2.60 ; d f = 6,22 ; p = 0.0463 )\). At least two varieties differ in means for height and/or number of tillers.
  • Results of the ANOVAs on the individual variables:
    Variable F SAS p-value Bonferroni p-value
    Height 4.19 0.030 0.061
    Tillers 1.27 0.327 0.654

    Each test is carried out with 3 and 12 d.f. Because we have only 2 response variables, a 0.05 level test would be rejected if the p-value is less than 0.025 under a Bonferroni correction. Thus, if a strict \(α = 0.05\) level is adhered to, then neither variable shows a significant variety effect. However, if a 0.1 level test is considered, we see that there is weak evidence that the mean heights vary among the varieties (F = 4.19; d. f. = 3, 12).

  • The Mean Heights are presented in the following table:
    Variety Mean Standard Error
    A 58.4 1.62
    B 50.6 1.62
    C 55.2 1.62
    D 53.0 1.62
  • Variety A is the tallest, while variety B is the shortest. The standard error is obtained from:

    \(SE(\bar{y}_{i.k}) = \sqrt{\dfrac{MS_{error}}{b}} = \sqrt{\dfrac{13.125}{5}} = 1.62\)

  • Looking at the partial correlation (found below the error sum of squares and cross-products matrix in the output), we see that height is not significantly correlated with the number of tillers within varieties \(( r = - 0.278 ; p = 0.3572 )\).

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