Multiple Regression and Model Building 379 11.2 a. 0ˆβ = 506.346, 1ˆβ = −941.900, 2ˆβ = -429.060 b. ˆy = 506.346 − 941.900x1 − 429.060x2 c. SSE = 151,016, MSE = 8883, s = 94.251 We expect about 95% of the y-values to fall within ±2s or ±2(94.251) or ±188.502 units of the fitted regression equation. d. H0: β1 = 0 Ha: β1 ≠ 0 The test statistic is t = 11ˆˆ0941.900275.08sββ−−= = −3.42 The rejection region requires α/2 = .05/2 = .025 in each tail of the t distribution with df = n − (k + 1) = 20 - (2 + 1) = 17. From Table VI, Appendix B, t.025 = 2.110. The rejection region is t < −2.110 or t > 2.110. Since the observed value of the test statistic falls in the rejection region (t = −3.42 < −2.110), H0 is rejected. There is sufficient evidence to indicate β1 ≠ 0 at α = .05. e. For confidence coefficient .95, α = .05 and α/2 = .025. From Table VI, Appendix B, with df = n − (k + 1) = 20 − (2 + 1) = 17, t.025 = 2.110. The 95% confidence interval is: 2ˆβ ± t.0252ˆsβ ⇒ −429.060 ± 2.110(379.83) ⇒ −429.060 ± 801.441 ⇒ (−1230.501, 372.381) f. R2 = R-Sq = 45.9% . 45.9% of the total sample variation of the y values is explained by the model containing x1 and x2. R2a = R-Sq(adj) = 39.6%. 39.6% of the total sample variation of the y values is explained by the model containing x1 and x2, adjusted for the sample size and the number of parameters in the model. Multiple Regression and Model Building Chapter 11380 Chapter 11 g. To determine if at least one of the independent variables is significant in prediction y, we test: H0: β1 = β2 = 0 Ha: At least one βi ≠ 0 From the printout, the test statistic is F = 7.22 Since no α level was given, we will choose α = .05. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k = 2 and ν2 = n – (k + 1) = 20 – (2 + 1) = 17. From Table IX, Appendix B, F.05 = 3.59. The rejection region is F > 3.59. Since the observed value of the test statistic falls in the rejection region ( F = 7.22 > 3.59), H0 is rejected. There is sufficient evidence to indicate at least one of the variables, x1 or x2, is significant in predicting y at α = .05. h. The observed significance level of the test is p-value = 0.005. Since the p-value is so small, we will reject H0 for most reasonable values of α. There is sufficient evidence to indicate at least one of the variables, x1 or x2, is significant in predicting y at α greater than 0.005. 11.4 a. We are given 1ˆβ = 3.1, 1ˆsβ = 2.3, and n = 25. H0: β1 = 0 Ha: β1 > 0 The test statistic is t = 11ˆˆ03.12.3sββ−= = 1.35 The rejection region requires α = .05 in the upper tail of the t distribution with df = n − (k + 1) = 25 − (2 + 1) = 22. From Table VI, Appendix B, t.05 = 1.717. The rejection region is t > 1.717. Since the observed value of the test statistic does not fall in the rejection region (t = 1.35 >/ 1.717), H0 is not rejected. There is insufficient evidence to indicate β1 > 0 at α = .05. b. We are given 2ˆβ = .92, 2ˆsβ = .27, and n = 25. H0: β2 = 0 Ha: β2 ≠ 0 The test statistic is t = 22ˆˆ0.92.27sββ−= = 3.41Multiple Regression and Model Building 381 The rejection region requires α/2 = .05/2 = .025 in each tail of the t distribution with df = n − (k + 1) = 25 − (2 + 1) = 22. From Table VI, Appendix B, t.025 = 2.074. The rejection region is t < −2.074 or t > 2.074. Since the observed value of the test statistic falls in the rejection region (t = 3.41 > 2.074), reject H0. There is sufficient evidence to indicate β2 ≠ 0 at α = .05. c. For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table VI, Appendix B, with df = n − (k + 1) = 25 − (2 + 1) = 22, t.05 = 1.717. The confidence interval is: 1ˆβ ± t.051ˆsβ ⇒ 3.1 ± 1.717(2.3) ⇒ 3.1 ± 3.949 ⇒ (−.849, 7.049) We are 90% confident that β1 falls between −.849 and 7.049. d. For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From Table VI, Appendix B, with df = n − (k + 1) = 25 − (2 + 1) = 22, t.005 = 2.819. The confidence interval is: 2ˆβ ± t.0052ˆsβ ⇒ .92 ± 2.819(.27) ⇒ .92 ± .761 ⇒ (.159, 1.681) We are 99% confident that β2 falls between .159 and 1.681. 11.6 a. For x2 = 1 and x3 = 3, E(y) = 1 + 2x1 + 1 − 3(3) E(y) = 2x1 − 7 The graph is :382 Chapter 11 b. For x2 = −1 and x3 = 1 E(y) = 1 + 2x1 + (−1) − 3(1) E(y) = 2x1 − 3 The graph is: c. They are parallel, each with a slope of 2. They have different y-intercepts. d. The relationship will be parallel lines. 11.8 No. There may be other independent variables that are important that have not been included in the model, while there may also be some variables included in the model which are not important. The only conclusion is that at least one of the independent variables is a good predictor of y. 11.10 a. The first order model is: 01122334455E(y) ββ x β x β x β x β x=++ + + + b. R2 = .58. 58% of the total sample variation of the levels of trust is explained by the model containing the 5 independent variables. c. 22.58 516.57(1 .58) [66 (5 1)](1 ) [ ( 1)]RkFRnk== =−−+−−+ d. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = k = 5 and ν2 = n – (k + 1) = 66 – (5 + 1) = 60. From Table VIII, Appendix B, F.10 = 1.90. The rejection region is F > 1.96. Since the observed value of the test statistic falls in the rejection region (F = 16.57 > 1.96), H0 is rejected. There is sufficient evidence to indicate that at least one of the 5 independent variables is useful in the prediction of level of trust at α = .10. 11.12 a. The least squares prediction equation is: 123 4 56789ˆ3.70 .34 .49 .72 1.14 1.51 .26 .14 .10 .10 .yxxxxxxxxx=+ + + + + + − − −Multiple Regression and Model Building 383 b. 0ˆ3.70β= . This is estimate of the y-intercept. It has no …