![]() Where S is the pooled sample covariance matrix of X and Y, namely the null hypothesis H 0: μ X = μ Y.ĭefinition 1: The Two sample Hotelling’s T-square test statistic is We now look at a multivariate version of the problem, namely to test whether the population means of the k × 1 random vectors and Y are equal, i.e. Also note that by Property 1 of F Distribution, an equivalent test can be made using the test statistic t 2 and noting that The null hypothesis is rejected if | t| > t crit. It turns out that the t-test is pretty robust for violations of the normality assumption provided each population is relatively symmetric about its mean. Regarding the normality assumption, if n x and n y are sufficiently large, the Central Limit Theorem holds, and we can proceed as if the populations were normal. The sample for x and y are random with each element in the sample taken independently. ![]() ![]() The variances of the two populations are equal (homogeneity of variances).The populations of x and y have a normal distribution.The populations of x and y have unique means and there are no distinct sub-populations with different means.Provided the following assumptions are met It then follows that the t-statistic defined above has a t distribution with n x + n y – 2 degrees of freedom, i.e. Where s is the pooled standard deviation defined by Assuming that sample for x has size, mean and standard deviation respectively n x, x̄ and s x, and that the sample for y has size, mean and standard deviation respectively n y, ȳ and s y, we define the following statistic ![]() To test this hypothesis we create a random sample for each variable. In the univariate case, we have two independent random variables and want to determine whether the population means of the two random variables are equal, i.e. ![]()
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