Stat 407 Lab 14 MANOVA Fall 2001 SOLUTION
In this lab we examine the use S-Plus for multivariate analysis of variance (MANOVA) of the crabs data,
that we studied early on in the course.
1. Based on your memory of previous analyses of this data, guess the relative magnitude of Wilks Λ∗ . Would
you expect it to be close to 0 or close to 1 or in half-way between?
All individual responses expected here. (Nothing to grade.)
2. Generate histograms of the variables, conditioned on species. Comment on the multivariate normality of
the data, and the homogeneity or lack of it of the group variance-covariance.
Based on these histograms, these variables look reasonably consistent with normality. There is some skewness and some multimodality.
3. Compute the overall sample mean, and the means for each Sp.Sex group.
X̄ = [15.6 12.7 32.1 36.4 14.0]0
X̄1 = [14.8 11.7 32.0 36.8 13.4]0
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X̄2 = [13.3 12.1 28.1 32.6 11.8]0
X̄3 = [16.6 12.3 33.7 37.2 15.3]0
X̄4 = [17.6 14.8 34.6 39.0 15.6]0
4. Run the MANOVA analysis. Report the results. What hypothesis is this statistic testing? (You will
need to make the variable Sp.Sex a factor, by changing its variable type. Then choose the Statistics− >
Multivariate − > MANOVA. In the MANOVA control panel select FL, RW, CW,CL, BD as the response
variables, and S.Sex as the main effect. In the results tab, choose Wilks as the test statistic, and check the
boxes Short Output, ANOVA table, Estimated Effects. Save the fitted values into a new data set.)
Wilks Λ∗ = 0.024, which has an approximate F statistic = 101.8, with df 15, 530.4, and associated pvalue= 0.
5. What are the Estimated coefficients, that are computed in the MANOVA analysis?
The estimated coefficients should the the treatment effects. The intercept should correspond to the overall
mean, and the other coefficients to the first 3 group effects. The 4th treatments effect should be obtained
from the other 3 treatments. But these don’t seem to match with the numbers from S-Plus.
6. What are the fitted values that are computed in the MANOVA table?
The fitted values are the group means.
7. Run univariate ANOVA analyses comparing the means on each variable. Report the results. Why is
ANOVA inadequate for comparing the multivariate means? Which variables are the most important in the
mean difference based on the univariate ANOVAs?
Variable F Statistic p-value
FL
31.9
0
RW
40.7
0
CW
9.22
0.003
CL
5.21
0.024
BD
25.7
0
ANOVA doesn’t take the variance-covariance between variables into account in the mean testing.
8. Write down the numbers to compute the between (B) sums of squares matrix without actually computing
it.
B can be computed using the group means and the overall mean:
B = 50[14.8 − 15.6 11.7 − 12.7 32.0 − 32.1 36.8 − 36.4 13.4 − 14.0]0 [14.8 − 15.6 11.7 − 12.7 32.0 − 32.1 36.8 −
+ 50[13.3 − 15.6 12.1 − 12.7 28.1 − 32.1 32.6 − 36.4 11.8 − 14.0]0 [13.3 − 15.6 12.1 − 12.7 28.1 − 32.1 32.6 −
+ 50[16.6 − 15.6 12.3 − 12.7 33.7 − 32.1 37.2 − 36.4 15.3 − 14.0]0 [16.6 − 15.6 12.3 − 12.7 33.7 − 32.1 37.2 −
+ 50[17.6 − 15.6 14.8 − 12.7 34.6 − 32.1 39.0 − 36.4 15.6 − 14.0]0 [17.6 − 15.6 14.8 − 12.7 34.6 − 32.1 39.0 −
9. Explain how you would compute the within (W) sums of squares matrix, without actually calculating it.
Subtract the group means from the respective data value, then compute the cross-product of this vector.
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