Lesson_10_11_Review_Problem_Solutions_F12

The strength of the linear relationship

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Unformatted text preview: ; t α / 2,n − 2 = t.025,10 = 2.228 or t < −t α / 2,n − 2 = −t.025,10 = −2.228 s b1 = t= sε (n − 1)s 2 x = 1.347 (12 − 1)(749.7) = .0148 b 1 − β1 .0582 − 0 = = 3.93 (Excel: t = 3.93, p–value = .0028. There is enough evidence to s b1 .0148 infer a linear relationship between advertising and sales. 7 e b1 ± t α / 2,n −2 s b1 = .0582 ± 2.228(.0148) = .0582 ± .0330 LCL = .0252, UCL = .0912 f R 2 = s2 xy s2s2 xy = (43.66) 2 = .6067 (Excel: R 2 = .6066). 60.67% of the variation in sales is (749.7)(4.191) explained by the variation in advertising. g There is evidence of a linear relationship. For each additional dollar of advertising sales increase, on average by .0582. PROBLEM # 10.24 We know that the coefficient of determination is r2 = 1  ­ (SSE/SST). Therefore the coefficient of determination is 1  ­ (40.0/200.0) = 0.80. PROBLEM # 10.25 a. The monthly mortgage payment would likely be directly related to the market value of the house, the interest rate, the size of the house, or the monthly taxes and insurance, among other variables. b. The monthly mortgage payment would likely be inversely related to the age of the house, among other variables. c. The monthly mortgage payment would likely be unrelated to the amount of chocolate consumed by the owners, and a wide variety of other variables. 8 PROBLEM # 10.26 ­ A tire company has carried out tests in which rolling resistance (pounds) and inflation pressure (pounds per square inch, or psi) have been measured for psi values ranging from 20 to 45. The regression analysis is summarized in the following MiniTab printout: Regression Analysis The regression equation is ROLRESIS = 9.45 – 0.0811 PSI Predictor Constant PSI S= 0.8808 Coef 9.450  ­0.08113 StDev 1.228 0.03416 R ­Sq = 23.9% T 7.69  ­2.38 P 0.000 0.029 R ­Sq (adj) = 19.6% Analysis of Variance Source Regression Error Total DF 1 18 19 SS 4.3766 13.9657 18.3422 MS 4.3766 0.7759 F 5.64 P 0.029 a. The least squares regression line is: ROLRESIS = 9.450 - 0.08113 PSI. b. 23.9% of the variation in rolling resistance is explained by the regression line. (See R ­sq.) c. The slope of the line differs from zero at the 0.029 level of significance. Minitab used a two ­tail t ­test to reach this conclusion. d. The coefficient of correlation differs from zero at the 0.029 level of significance. (See the ANOVA table.) This is the same level found in part c. These levels will always be the same because the tests are equivalent. e. The 95% confidence interval for the slope of the population regression line is: b1 ± t s b1 =  ­0.08113 ± 2.101(0.03416) =  ­0.08113 ± 0.07177, or from  ­0.15290 to  ­0.00936. 9...
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