In this case, all the observations with zero as the predictor variable are placed in one group, and all of the observations with a one as the predictor variable are placed in a second group. A two-sample t-test is used to determine the probability that the two groups have the same mean.

17.2.1 Univariate Case

In this case, all the observations with zero as the predictor variable are placed in one group, and all of the observations with a one as the predictor variable are placed in a second group. A two-sample Student’s t-test is used to determine the probability that the two groups have the same mean.

More specifically, suppose you have n items that are split into two groups of sizes n₀ and n₁, and the respective sums of their continuous responses are s₀ and s₁. Further, let S be the sum of the squared responses, S = ∑ _i=1ⁿy_i², and let

We can then calculate the t statistic:

where the p-value is given by the tails of a two sided Student’s t distribution with n₀ + n₁ - 2 degrees of freedom: p = 2 × StudentT.

17.2.2 Multivariate Case

In this case, we have n observations, each with a k-dimensional response in an n × k matrix, Y. All the observations with zero as the predictor variable are placed in one group, g₀, of size n₀, and all of the observations with a one as the predictor variable are placed in a second group, g₁, of size n₁. Note that n=n₀+n₁. We then calculate a Hotelling T² statistic to compare the multivariate continuous responses in the two daughter nodes. Let the k-dimensional vector M contain the means of responses 1...k, M_j = . Let us define the k-dimensional vector S, where S_j = ∑ _i=1ⁿ. Define the k × k matrix, A as follows: A_i,j = . Define the k × 1 b as follows: b_j = -. We then solve the matrix equation Ax = b, and we calculate the T² statistic as follows:

and the p value is computed using the F distribution with k and n-k-1 degrees of freedom: p = Ftest(T²,k,n - k - 1).