When the variances of the two populations are unequal (as indicated by notably unequal sample variances), we use a modified version of the t-test. In particular we use the following t-statistic
We now test the null hypothesis H0: μx = μy using the fact that t ~ T(m) where m is defined as
(see Two Sample t Test with Unequal Variances). As we have seen several times now, this is equivalent to
where t2 can be expressed as:
where z̄ = x̄ – ȳ and µz = µx – µy.
We now look at a multivariate version of the problem, namely to test whether the population means of the k × 1 random vectors X and Y are equal, i.e. the null hypothesis H0: μX = μY, under the assumption that the covariance matrices are not necessarily equal.
Definition 1: The modified two sample Hotelling’s T-square test statistic is
Observation: Note the similarity between the expression for T2 and the expression for t2 given above. Also note that if nX = nY, then this definition of T2 is equivalent to that in Definition 1 of Hotelling’s T-square for Independent Samples.
Theorem 1: For nX and nY sufficiently large, T2 ~ χ2(k).
Observation: For small nX and nY, T2 is not sufficiently accurate and a better estimate is achieved using the following theorem
Theorem 2: Under the null hypothesis,
where n = nX + nY – 1 and m is defined as follows:
If F > Fcrit then we reject the null hypothesis.
Example 1: Repeat Example 1 of Hotelling’s T-square for Independent Samples using the data in Figure 1.
Figure 1 – Data for Example 1
Once again, we employ Box’s test, obtaining the results shown in Figure 2.
Figure 2 – Box’s Test for Example 1
This time we see that p-value < α = .05, and so we conclude that there is evidence that the covariance matrices are unequal (or that the data is not multivariate normally distributed), although we have somewhat forced the issue since usually a significance level of α = .001 instead of α = .05 is used for Box’s Test.
As a result we will use the T2 test with unequal covariance matrices. This analysis is shown in Figure 3.
Figure 3 – Analysis for Example 1
We conclude there is a significant difference between the drug and the placebo in treating the symptoms.
The simultaneous 1 – α confidence interval for μi is given by the expression
For n sufficiently large, we could use the following expression instead
Once again we can use Bonferroni confidence intervals instead.