Appendix F: Solutions to Problems

CHAPTER 2

• 2.1 Slope: .444 g/cm; standard error of slope: .0643 g/cm; intercept: −6.01 g; standard error of intercept: 2.39 g; correlation coefficient: .925; standard error of the estimate: .964 g; F: 47.7.

• 2.2 Yes. A simple regression yields r = .96 (t = 7.8 with 5 degrees of freedom; P < .001) and s_y|x = 3.9.

• 2.3 Yes. Regress female breast cancer rate B on male lung cancer rate L to obtain = 400 + .282 L, P = .02.

• 2.4 The regression equation is Ĉ = 1.36 log CFU − .04 log CFU/% M, with R² = .649, s_y|x = .87 log CFU, and F = 96.1 with 1 and 52 degrees of freedom. Both the slope (t = −9.8; P < .001; 95 percent confidence interval, −.049 to −.032) and intercept (t = 5.21; P < .001; 95 percent confidence interval, .84 to 1.88) are highly significant. This linear model provides a significant explanation of the observed variability of the dependent variable.

• 2.5 A. Based on the mean data, the regression equation is = 1.67 mmHg/(mL · min · kg) − .276 (mmHg · h)/(mL · min · ng)Δr, with r = .957, s_y|x =.97 mmHg/(mL · min · kg), and a t statistic for the slope of −4.63 with 2 degrees of freedom (P = .044). Thus, the change in vascular resistance seems to be significantly related to the change in renin production by the kidney. B. Using the raw data points, we find the similar regression equation = .183 mmHg/(mL · min · kg) −.162 (mmHg · h)/(mL · min · ng)Δr, with r = .632, s_y|x = 2.15 mmHg/(mL · min · kg), and the t statistic for the slope is −4.32 with 28 degrees of freedom (P < .001). Thus, we still conclude, based on the t statistic, that the change in vascular resistance is significantly related to the change in renin production by the kidney. C. Whereas the regression equations are similar, the equation computed using the group mean values had a much higher correlation than the one based on the raw data (.957 versus .632). Likewise, the estimate of the variability about the regression plane s_y|x is much smaller when computed using the mean values [.97 vs. 2.15 mmHg/(mL · min · kg)]. The regression equation computed from the group means seems to provide a much better prediction of the data than the one computed from the raw data. The difference in the results of the two regression analyses lies in the difference between the mean values and the raw data. By computing the means of each group, then analyzing these four data points with regression, we have effectively thrown away most of the variability in the observations. Thus, s_y|x is artificially reduced and is no longer an unbiased estimate of ...