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Stat 13 http://www.stat.ucla.edu/~dinov/courses_students.html Chapter 12 Problems/Solutions All Problems are from: Myra L. Samuels and Jeffrey A. Witmer, Statistics for the Life Sciences, 3rd edition, Prentice-Hall (2003) 12.5. SOCR (http://www.socr.ucla.edu) output is given below. (a) cob-wt = 316 - 0.721 plant-density (b) scatterplot (not shown) shows a strong negative linear association between cob-weight (gm grain/cob) and plant density (# plants / pot). (c) as plant density increases by 1 plant per plot, cob weight decreases by 0.72 gm of grain per cob, on average. (d) sY = sqrt(11831.8/19) = 25 gm and sY/X = sqrt(1337.3/18) = 8.6 gm (e) Predictions of cob weight based on the regression model tend to be off by 8.6 gm on average. Equivalently, the data points deviate above or below the regression line by 8.6 gm on average. Regression Analysis The regression equation is cob-wt = 316 - 0.721 plant-density Predictor Coef StDev T P Constant 316.376 8.000 39.55 0.000 plant-de -0.72063 0.06063 -11.89 0.000 S = 8.619 R-Sq = 88.7% R-Sq(adj) = 88.1% Analysis of Variance Source DF SS MS F P Regression 1 10495 10495 141.26 0.000 Residual Error 18 1337 74 Total 19 11832 12.6 (a) The slope and intercept of the regression line are b_1 = -927.75/1303 = -.7120; b_0 = y-bar - b_1x-bar = 23.64 - (-.7120)(11.5) = 31.83 The fitted regression line is y-hat = 31.83 - .7120X (c) s_Y|X = sqrt[SS(resid)/df] = sqrt[16.7812/(12-2)] = 1.3 12.8a: b1 = 161.40/50667 = .0003186; b0 = .210 – (.0003186)(433.3) = .0720.The fitted regression line is Y = .072 - .0003186X. 12.8c: As altitude of origin goes up by 1 m, respiration rate goes up by .0003186 mul/hr-mg, on average. 12.8d: s_{Y|X} = sqrt[.013986/10] = .0374 12.12 The intercept of the regression line b0 is based on all 12 data points, not just on the two point for which X = 0. If there is a linear relationship between X and Y (a scatter plot of the data strongly suggest that there is), then the best estimate of the average for Y at any given value of X is given by the regression line, taking into account all of the data. In contrast, the average (33.3 + 31.0)/2 = 32.15 ignores most of the data. 12.14a: (See Exercise 12.5 for b0 and b1. (i) plugging in plant-density = 100 plants gives a predicted cob-wt of 316 - 0.721(100) = 244.34 gm (ii) plugging in plant-density = 120 plants gives a predicted cob-wt of 316 - 0.721(120) =229.93 gm 12.14b: (i) (224.34)(100) = 24434 gm = 24.43 kg (ii) (229.928)(12) = 27591 gm = 27.6 kg 12.15 Using the fitted regression line found in Exercise 12.6 above, we substitute X = 15. This yields y-hat = 31.83 - (.7120)(15) = 21.1. Thus, we estimate that the mean fungus growth would be 21.1 mm at a laetisaric acid concentration of 15 microg/ml. According to the linear model, the standard deviation of fungus growth does not depend on X. Our estimate of this standard deviation from the regression line is the Residual Standard Deviation sigma_{Y|X} = sqrt[SS(resid)/(n-2)] = sqrt[16.7812/10] = 1.3 mm. Thus we estimate that the standard deviation of fungus growth would be 1.3 mm at a laetisaric acid concentration of 15 microg/ml. For X = 15, we have y-hat = 21.1 +/- 1.3 mm. 12.19a: b1 = 81.90/2800 = 0.02925 ng/min (the rate of incorporation) b0 = 0.83 – (0.02925)(30) = -0.05s_{y|x} = sqrt[SS(resid)/(n-2)] = sqrt[0.035225/5] = 0.0839 To construct a 95% confident interval, we consult the z-table (Table 4) with df = n-2 = 7-2 =5; the multiplier is t_{4,0.025} = 2.571. The confidence interval is b1 +/- t_{4,0.025}SEb1 = 0.02925 +/- (2.571)(0.00159) or 0.0252 < beta1 < 0.033 ng/min 12.19ba: We are 95% confident that the rate at which leucine is incorporated into protein in the population of all Xenopus oocytes is between 0.0252 ng/min and 0.0333 ng/min 12.21a: SEb1 = 8.6/sqrt(20209) = 0.0605, so 95% CI for b1 is -0.7206 +/- (2.101)(0.0605) or -0.7206 +/- 0.1271 or (-0.848 , -0.593) 12.21b: We are 95% confident that as plant density increases by 1 plant per plot, average cob weight decreases by between 0.848 gm and 0.593 gm of grain per cob. 12.22a: From Exercise 12.6, s_Y|X = sqrt[SS(resid)/df] = sqrt[16.7812/(12-2)] = 1.3 The standard error of the slope is SE_b1 = s_Y|X / sqrt[sum(x - x-bar)^2] = 1.3/sqrt[1303] = 0.36 12.22b: H0: Leatisaric acid has no effect on fungus growth (beta_1 = 0) HA: Laetisaric acid inhibits fungus growth (beta_1 < 0) t_s = -.7120/0.36 = -19.8. With df = 10, the t-table (Table 4) gives t_.0005 = 4.587. Thus the P-value < .0005, so we reject H0. There is sufficient evidence (P-value < .0005) to conclude that laetisaric acid inhibits fungus growth. 12.27a: r = 82.8977/sqrt[(28465.7)(.363708)] = .8147 12.27b: s_Y = sqrt[(.363708/(13-1)] = .1741 gm s_Y|X = sqrt[SS(resid)/df] = sqrt[.1223/(13-2)] = .1054 gm .1054/.1741 = .605; sqrt[1 - .8147^2] = .580 12.27c: b_1 = 82.8977/28465.7 = .002912; b_0 = 2.174 - (.002912)(443.8) = .882 The fitted regession line is y-hat = .882 - .002912X 12.28a: r = -14563.1/sqrt[ (20209)*(11831.8) ] = -0.942 12.28b: from Exercise 12.5 (d), sY = 25 gm and sY/X = 8.6 gm, so sY/X / sY = 0.344 further, sqrt(1 - r2) = 0.3356, which is nearly equal to 0.344, so the approximate relationship is indeed verified. 12.28c: b1 = -14563.1/20209.0 = -.7206; b0 = 224.1 – (-.7206)(128.05) = 316.4 The fitted regression line is Y = 316.4 - .7206X.12.30: Let X = age and let Y = blood pressure. The Residual Standard Deviation is s_{Y|X} = sqrt[1 - r^2](s_Y)sqrt[(n-1)/(n-2)] = sqrt[1 - .43^2](19.5)sqrt[2668/2667] = 17.6 mm Hg. s_{Y|X} = sqrt[(y - y-hat)^2/(n-2)] is a measure of the variability about the regression line y-hat = b1x + b0. But s_Y = sqrt[(y - y-bar)^2/(n-1)] is a measure of the variability about the mean y-bar. 12.41a: with (iii), 12.41b: with (ii), and 12.41c: with (i). 12.45a: The slope and intercept of the regression line are b1 = -.342/.1512 = -2.262 b0 = 1.117 – (-2.262)(.12) = 1.39 The fitted regression line is Y = 1.39 – 2.262X. 12.45c: s_{Y|X} =


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