chapter 1

Chapter 13: Nonlinear and Multiple Regression25. The point estimate of1ˆ1e β= e.17772≈ 1.194β is βˆ1= . 17772, so the estimate of the odds ratio is. That is , when the amount of experience increases by one year (i.e. aone unit increase in x), we estimate that the odds ratio increase by about 1.194. The z valueof 2.70 and its corresponding p-value of .007 imply that the null hypothesis H β 00:1=can be rejected at any of the usual significance levels (e.g., .10, .05, .025, .01). Therefore,there is clear evidence that β1is not zero, which means that experience does appear to affectthe likelihood of successfully performing the task. This is consistent with the confidenceinterval ( 1.05, 1.36) for the odds ratio given in the printout, since this interval does notcontain the value 1. A graph of πˆ appears below.p(x)0.90.80.70.60.50.40.30.20.10.00 10 20 30experienceSection 13.326.a. There is a slight curve to this scatter plot. It could be consistent with a quadraticregression.b. We desire R 2 , which we find in the output: R 2 = 93.8%c. β = β 0H vs H : at least one β ≠ 00:1 2=a408i. The test statistic isMSRf = = 22.51, and the corresponding p-value is .016. Since the p-value < .05,MSEwe reject H o and conclude that the model is useful.d. We want a 99% confidence interval, but the output gives us a 95% confidence interval of(452.71, 529.48), which can be rewritten as 491 .10 ± 38. 38 ; t 3. 182 , so38.38sy⋅ ˆ 14= = 12.06 ; Now, t. 005,3= 5. 8413.182491 .10 5.84112.06 = 491.10 ± 70.45 = 420.65,561.55e. β 00:2=. 025,3=, so the 99% C.I. is( ) ( )± .H vs : β ≠ 20H . The test statistic is t = -3.81, with a corresponding p-avalue of .032, which is < .05, so we reject H o . the quadratic term appears to be useful inthis model.