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17.2 Normally Distributed Response Binomial Predictor

17.2 Normally Distributed Response Binomial Predictor

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=1nyi2, and let

          2   2
     S---sn00---s1n1
SD =  n0 + n1 - 2 .

We can then calculate the t statistic:

    (       )
      s1n1 - sn00
T = ∘-SD---SD-,
      n0-+ n1-

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(- ∣T ∣,n0 + n1 - 2).

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, Mj = ∑ni=1Yi,j
   n. Let us define the k-dimensional vector S, where Sj = i=1n(Yi,j - Mj). Define the k × k matrix, A as follows: Ai,j = ∑k  ∑l
--l=1-nm-=11SlSm. Define the k × 1 b as follows: bj = ∑ni=1Yi,j-∑i-∈g1Yi,j-
      n0 -∑i∈g1Yi,j-
  n1. We then solve the matrix equation Ax = b, and we calculate the T2 statistic as follows:

     ∣∣(n - 2) n0n1xT b∣∣n - k- 1
T 2 = ∣∣n-- 1---nn0n1xTb∣∣k-(n---2),
              n

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).