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.

Age Gender 50 262 193 224 201 161 178 265 213 202 183 185 197 162 271 192 189 209 227 236 142 Female 192 253 248 278 232 267 289 Male