Chapter 15. Correlation and Regression

Transcription

1 Correlation and Regression 15.1 a. Scatter plot b. The major ais has slope s / s and goes through the point (, ). Here = 70, s = 20, = 80, s = 10, and n = 7 so an equation for the line along the major ais is = 80 + (10/20) ( 70) or = c. Student Sum ID z z z z

2 15.3 obs ( ) 2 ( ) 2 z z z z Total Here n = 3, = 6/3 = 2, s = 2, = 6/3 = 2, s = 1, and r = 1/2 The correlation coefficient is obs ( ) 2 ( ) 2 z z z z Total So the 5 number summar is = 150/6 = 25, s = 500 / 5 = 10, = 24/6 = 4, s = 20 /5 = 2, and r = z z n 1 = 4.75 = = From Eercise 15.1, the point of averages is (, ) = (70,80) and the slope of the regression line is r s = 2.75 s = so the regression line is ˆ = or ˆ =

3 15.11 a. From Eercise 15.3 the point of averages is (, ) = (2, 2) and the slope of the regression line is rs /s = (0.5)(0.5) = Use the point-slope formula to find the regression line ˆ = ( 2) ˆ = + 6 or 4. b. SSE = 1.50 as seen from the following table: OBS ˆ e = ˆ e Totals = SSE s a. An equation for the regression line is ˆ = + r ( ). s Using values from Eercise 15.5, we have ˆ = ( 25) 10 or ŷ = ( 25) or ŷ = b. SSE = 1.95 as seen from the last entr in the following table. OBS ˆ ( ˆ ) Totals = SSE s e = SSE n 2 = = a. Since is unknown, estimate b = 18. b. Since = 69 =, use the regression line and estimate b = 18. c. = 73 is (73 )/s = 1.6 standard units. Estimate to be 1.6r = 1.28 standard units. So predict the forearm length to be s = inches. 87

4 d. = 64 is (64 )/s = 2 standard units. Now estimate the forearm length to be 2r = 1.6 standard units. Predict the forearm length to be 1.6s = inches a. 19 inches is (19 18)/1 = 1 standard unit. The area under the standard normal curve to the right of 1 is (use Table 3). So 15.87% of the men have forearm length over 19 inches. b. Use the result from Eercise 15.15c. Men who are 73 inches tall have an average forearm length of inches with standard deviation s = 0.60 inches. So 19 inches is ( )/0.60 = standard units. The area under the standard normal curve to the right of is So 67.96% of the men who are 73 inches tall have forearm lengths over 19 inches Use the appropriate regression line to compute. The average height of men whose wives are inches tall is = inches. The standard deviation of these heights is s = inches. So estimate the height of the man to be inches. Assuming their heights follow a normal curve, about 95% of these men will be within 1.96 standard deviations of average. This gives a prediction interval of ± or ± 4.96 inches. So a 95% prediction interval for the heights of these men is inches to inches The husbands IQ is ( )/15 = 1 standard unit. The average IQ of the wives is 1 r = 0.5 standard units. This is = IQ units. It is not equal to 130 because there are two regression lines, one for predicting IQ s of wives from the IQ of the husbands, and the other for predicting IQ s of husbands from the IQ of their wife a. A 95% confidence interval for β is b ± t 5,.975 or 0.04 ± So, 95%C.I.: < β < Here b = r s =.3 2 s 15 = 0.04 t. 975, 5 = , n 2 = =

5 b. Since the 95% confidence for β includes both positive and negative values, the hpothesis that β (or ) is positive is not supported at the 5% level. Test H o : β = 0 versus the alternative hpothesis Ha: β > 0. The P-value is Pr(b 0.04) = Pr(T ) = Pr(T ) Since the P-value is not small, there is insufficient evidence to conclude β > a. From Eercise 15.2, n = 6, r = 2/5, and s /s = 0.5 so b = r s s = 1 5 and n 2 = Also t 4,.95 = (Using Table 4). A 90% confidence interval for has limits b ± t 4,.95 or ± so 90% C.I.: 0.29 < β < b. The two-sided 90% confidence interval includes negative values for β, so there is insufficient evidence to conclude > 0 (or β > 0) at the 5% significance level. To test H o : β = 0 versus the alternative hpothesis H a : β > 0, the P-value is Pr(b 1 5 ) = Pr(T ) = Pr(T ) a. The regression line goes through (, ) = (75,85) with slope rs /s = An equation for the regression line is ŷ = b. If = 63, predict to be on the regression line. Thus, predict to be ŷ = c. A 95% confidence interval for β has bounds b ± t 6,.075 with b = 0.60, n 2 = and t 6,0.975 = (using Table 4). So the limits are β = 0.6± and the interval is 0.59 < β < d. Since the 95% confidence interval for β includes both negative and positive values, the alternative hpothesis β > 0 (or > 0) is not accepted at the 2.5% significance level. The P-value for H o : β = 0 versus H a : β > 0 is Pr(b >.6 H o ) = Pr(T 6 > ) = Pr(T >1.2343) The P-value is large, so the null hpothesis = 0 (equivalent to β = 0) is retained at the 5% significance level (because P-value > ). 89

6 15.35 Specimen Humerus Femur () () ( ) 2 ( ) 2 z z z z Totals Average s = s = a. r = z = = b. The major ais goes through the point (, ) = (310, 431.4) and has slope s /s = Thus, = ( 310). c. The regression line goes through the point (, ) = (310, 431.4) and has slope rs /s = Thus ŷ = ( 310). d. A 95% confidence interval for is b ± t 3,.975 = ± Here b = , n 2 = and t 3,0.975 = So the 95% confidence interval is 0.62 < <