78 Chapter 7 ______________________________________________________________________________ 7.49 We assume that the population standard deviation is known to be $5,000 and use Z values (which, for a sample size 100, will give approximately the same answers as a t distribution with an estimated standard deviation). For α = 0.05, the rejection cutoffs are at Z = 1.96 and Z = –1.96. These imply a sample mean below –980 or above 980: ⎡ ⎤ X−0 ⎡ ⎛ 5, 000 ⎞ ⎤ P [ Z > 1.96 ] = P ⎢ > 1.96 ⎥ = P ⎢ X > 1.96 ⎜ = P ⎡⎣ X > 980 ⎤⎦ ⎝ 100 ⎟⎠ ⎥⎦ ⎣ ⎣ 5, 000 / 100 ⎦ ⎡ ⎤ X−0 ⎡ ⎛ 5, 000 ⎞ ⎤ P [ Z < −1.96 ] = P ⎢ < −1.96 ⎥ = P ⎢ X < −1.96 ⎜ = P ⎡⎣ X < −980 ⎤⎦ ⎝ 100 ⎟⎠ ⎥⎦ ⎣ ⎣ 5, 000 / 100 ⎦ For α = 0.01, the rejection cutoffs are at Z = 2.575 and Z = –2.575. These imply a sample mean below –1,287.50 or above 1,287.5: ⎡ ⎤ X−0 ⎡ ⎛ 5, 000 ⎞ ⎤ P [ Z > 2.575 ] = P ⎢ > 2.575 ⎥ = P ⎢ X > 2.575 ⎜ = P ⎡⎣ X > 1287.5 ⎤⎦ ⎝ 100 ⎟⎠ ⎥⎦ ⎣ ⎣ 5, 000 / 100 ⎦ ⎡ ⎤ X−0 ⎡ ⎛ 5, 000 ⎞ ⎤ P [ Z < −2.575 ] = P ⎢ < −2.575 ⎥ = P ⎢ X < −2.575 ⎜ = P ⎡⎣ X < −1287.5 ⎤⎦ ⎝ 100 ⎟⎠ ⎥⎦ ⎣ ⎣ 5, 000 / 100 ⎦ a. If the population mean is $500, the probability of rejecting the null hypothesis is higher if α = 0.05 than if α = 0.01. If α = 0.05, the probability of rejecting the null hypothesis is 0.1685 + 0.0015 = 0.17: ⎡ X − 500 980 − 500 ⎤ P ⎡⎣ X > 980 ⎤⎦ = P ⎢ > ⎥ = P [ Z > 0.96 ] = 0.1685 ⎣ 5, 000 / 100 5, 000 / 100 ⎦ ⎡ X − 500 −980 − 500 ⎤ P ⎡⎣ X < −980 ⎤⎦ = P ⎢ > ⎥ = P [ Z < −2.96 ] = 0.0015 ⎣ 5, 000 / 100 5, 000 / 100 ⎦ If α = 0.01, the probability of rejecting the null hypothesis is 0.0576 + 0.0002 = 0.0578: ⎡ X − 500 1, 287.5 − 500 ⎤ P ⎡⎣ X > 1, 287.5 ⎤⎦ = P ⎢ > ⎥ = P [ Z > 1.575 ] = 0.0576 ⎣ 5, 000 / 100 5, 000 / 100 ⎦ ⎡ X − 500 −1, 287.5 − 500 ⎤ P ⎡⎣ X < −1, 287.5 ⎤⎦ = P ⎢ > ⎥ = P [ Z < −3.575 ] = 0.0002 5, 000 / 100 ⎦ ⎣ 5, 000 / 100 b. The probability of not rejecting the null hypothesis is equal to 1 minus the probability of rejecting the null hypothesis. Thus the probability of not rejecting the null hypothesis is higher if α = 0.01 (1 – 0.0578 = 0.9422) than if α = 0.05 (1 – 0.17 = 0.83). 7.50 a. The null hypothesis is that hiring and promotions are independent of race. b. There might not be enough observations in the disaggregated categories. c. The statistical significance might be one way in some categories and the opposite way in the other categories, so that the differences cancel out when the data are aggregated.

Essential Statistics, Regression, and Econometrics 1st edition Gary Smith Solution manual