# Question

There is another way of looking at Exercise 9.3-6, namely, as a two-factor analysis-of-variance problem with the levels of gender being female and male, the levels of age being less than 50 and at least 50, and the measurement for each subject being their cholesterol level. The data would then be set up as follows:

(a) Test HAB: γij = 0, i = 1, 2; j = 1, 2 (no interaction).

(b) Test HA: α1 = α2 = 0 (no row effect).

(c) Test HB: β1 = β2 = 0 (no column effect). Use a 5% significance level for each test.

(a) Test HAB: γij = 0, i = 1, 2; j = 1, 2 (no interaction).

(b) Test HA: α1 = α2 = 0 (no row effect).

(c) Test HB: β1 = β2 = 0 (no column effect). Use a 5% significance level for each test.

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