Solutions to the Chapter 5 practice problems
5.1 (a) The 95% confidence interval for [pic] is [pic] that is[pic]
(b) Calculate the t-statistic:
[pic]The p-value for the test [pic] vs. [pic] is
The p-value is less than 0.01, so we can reject the null hypothesis at the 5% significance level, and also at the 1% significancelevel.
(c) The t-statistic is
The p-value for the test [pic] vs. [pic] is
The p-value is larger than 0.10, so we cannot reject the null hypothesis at the 10%, 5%or 1% significance level. Because [pic] is not rejected at the 5% level, this value is contained in the 95% confidence interval.
(d) The 99% confidence interval for β0 is [pic] that is, [pic]5.5 (a) The estimated gain from being in a small class is 13.9 points. This is equal to approximately 1/5 of the standard deviation in test scores, a moderate increase.
(b) The t-statistic is[pic] which has a p-value of 0.00. Thus the null hypothesis is rejected at the 5% (and 1%) level.
(c) 13.9 ± 2.58 ( 2.5 ’ 13.9 ± 6.45.
5. 6 (a) The question asks whether the variability in testscores in large classes is the same as the variability in small classes. It is hard to say. On the one hand, teachers in small classes might able to spend more time bringing all of the students along,reducing the poor performance of particularly unprepared students. On the other hand, most of the variability in test scores might be beyond the control of the teacher.
(b) The formula in 5.3 isvalid for heteroskesdasticity or homoskedasticity; thus inferences are valid in either case.
5.7 (a) The t-statistic is [pic] with a p-value of 0.03; since the p-value is less than 0.05, thenull hypothesis is rejected at the 5% level.
(b) 3.2 ± 1.96 ( 1.5 ’ 3.2 ± 2.94
(c) Yes. If Y and X are independent, then β1 ’ 0; but this null hypothesis was rejected at the 5% level in part...