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Stat+131+-+HW+3+-+solutions - 2.36(a r = 0.98 goes with the...

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Unformatted text preview: 2.36. (a) r = 0.98 goes with the Dividend Growth fund, which is most similar to the stocks represented by the Sci-P index. r = 0.81 goes with the Small Cap Stock fund; small U.S. companies should be somewhat similar to large U.S. companies. Finally, r = 0.35 goes with Emerging Markets, as these stocks would be the most different from those in the S&P index. (b) Positive correlations do not indicate that stocks went up. Rather, they indicate that when the S&P index rose, the other funds often did, too—and when the S&P index fell, the other funds were likely to fall. 2.44. See the solution to Exercise 2.22 for the scatterplot. r = —0.172—it is close to zero, because the relationship is a curve rather than a line; correlation measures the degree of iinear association. 2.48. Explanations and sketches will vary, but should note that correlation measures the strength of the association, not the slope of the line. The hypothetical Funds A and B mentioned in the report, for example, might be related by a linear formula with slope 2 (or 1/2). 2.50. (3) Because gender has a categorical (nominal) scale, we cannot compute the correlation between sex and anything. (There is a strong association between gender and income. Some writers and speakers use “correlation” as a synonym for “association.” It is much better to retain the more specific meaning.) (b) A correlation r = 1.09 is impossible because —1 5 r 5 1 always. ((2) Correlation has no units, so r = 0.23 bushei is incorrect. 2.64. (3) Because the slope is 0.0086 (in units of “proportion of perch eaten per perch count”), an increase of 10 in the perch count increases the proportion eaten by 0.086 (on the average). 0)) When the perch count is 0, the equation tells us that 12% (0.12) of those perch will be eaten. Of course, 12% of 0 is 0, so one could argue that this is in some sense correct, but computing the proportion eaten would require dividing by zero. 2.72. The means and standard deviations are :3 = 95 min, 3—: = 12.6611 cm, 5; = 53.3854 min, and s}, = 8.4967 cm; the correlation is r = 0.9958. For predicting length from time, the slope and intercept are in = r5343I '= 0.158 cmfmin and a1 = y— — b1)? i —2.39 cm, giving the equation 53 = —2.39 + 0.158;: (as in Exercise 2.69). For predicting time from length, the slope and intercept are [72 = rat/s), '= 6.26 min/'cm and a2 :3? — 62y— é 15.79 min, giving the equation :3 = 15.79 + 6.263). 2.78. (a) The slope is b = rsy/s, = (0.6)(8),f(30) = 0.16, and the intercept is a = )7 — bi = 30.2. (b) Julie’s predicted score is :33 = 78.2. (c) r2 = 0.36; only 36% of the variability in y is accounted for by the regression, so the estimate 3)” = 78.2 could be quite different from the real score. 2.96. (a) Apart from the outlier—circled for part (b)—the scatterplot shows a moderate linear negative association. ([1) With the outlier,r = —0.338’}'; without it,r* = —0.’}'866. (c) The two regression formulas are p = —492.6 — 33.?9): (the solid line, with all points), and y = —13’}'1.6 —75.52x (the dashed line, with the outlier omitted). The omitted point is also influential, as it has a noticeable impact on the line. Silicon (mglg) e§§§§§§ —22 —21.5 —21 —20.5 —20 —19.5 Isotope (96) 2.98. (a) Scatterplot below, left. The relationship seems linear. (b) The regression line is y = 151"}' + 0.0803): (y is stride rate, x is speed). (c) The residuals (reported by Minitab, then rounded to 3 decimal places) are 0.011, —0.001, —0.001, —0.011, —0.009, 0.003, 0.009. These add to 0.001. Results will vary with rounding, and also with the number of decimal places used in the regression equation. ((1) Residuals are positive for low and high speeds, negative for moderate speeds; this suggests that a curve (like a parabola) may be a better fit. No observations are particularly influential; the line would change very little if we omitted any point. 1min 'ro Stride rate (stepsleecond) m 1516171819202122 Speed (feeti'second) 0.01 0.005 0 -0.005 -0.01 -0.015 1516171819202122 Speed (feeti'second) Residual 2.100. The correlation is r = 0999. With individual runners, the correlation would be smaller (closer to 0), since using data from individual runners would increase the “scatter” on the scatterplot, thus decreasing the strength of the relationship. ...
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