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# Solution Manaul Of Of Business Statistics in Practice, 3rd Canadian Edition By Bruce

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## Solution Manaul Of Of Business Statistics in Practice, 3rd Canadian Edition By Bruce

CHAPTER 11—

Correlation Coefficient and Simple Linear Regression Analysis

11.1 [LO 1] Covariance is the strength of the linear relationship between x and y. If there is no covariance between x and y, then the linear relationship between these two variables looks like a circle or a zero. 11.2 [LO 2] a. r = 0.02; no linear relationship; might look like a circle or square or curvilinear. b. r = -0.85; is a strong negative; might look like a negative slope. c. r = 0.73; is a strong positive; might look like a positive slope. 11.3 [LO 2] sxy = 10.2281 / 23 = .4447 r = .4447 / (.7053)(.6515) = .9678 very strong positive relationship 11.4 [LO 3] r = .97 very strong positive relationship between age and amount donated (older people donate more money) r2 = .94 approximately 94% of the variance between age and average amount donated overlap 11.5 [S 11.2]  is the actual population (unknown) value of the correlation between two variables. 11.6 [S 11.2] Calculate t = and obtain its associated p-value. If p-value <  then you reject H0 and conclude that there is a strong linear relationship between the two variables. 11.7 [S 11.2] t = .23(14.07)/.97 = 3.34, tcritical at alpha = .001 is 3.09, therefore x and y are strongly related. 11.8 [S 11.2] t = .23(2.83)/.97 = 0.67, the linear relationship is no longer significant with only 10 people. 11.9 [LO 4] When there appears to be a linear relationship between y and x 11.10 [LO 4] the observed value of the dependent variable the mean value of the y when the value of the independent variable is x  = error term 11.11 [LO 4] : the change in the mean value of the dependent variable that is associated with a one-unit increase in the value of the independent variable. : the mean value of the dependent variable when the value of the independent variable is zero. 11.12 [LO 4] When data is observed in time sequence, the data is called time series data. Cross-sectional data is observed at a single point in time. 11.13 [LO 4] The straight line appearance on this data plot suggests that the simple linear regression model with a positive slope might be appropriate. 11.14 [LO 4] a. It is the mean of the service times required when the number of copiers is 4. b. It is the mean of the service times required when the number of copiers is 6. c. The slope parameter equals the change in the mean service time that is associated with each additional copier serviced. d. The intercept is the mean service time when there are no copiers. It fails to make practical sense because it requires service time when no copiers exist. e. All factors other than the number of copiers serviced. 11.15 [LO 4] The plot looks reasonably linear. 11.16 [LO 4] a. Mean demand when price difference is .10. b. Mean demand when price difference is –.05. c. Change in mean demand per dollar increase in price difference d. Mean demand when price difference = 0; yes e. Factors other than price difference; answers will vary. 11.17 [LO 4] a. b. Yes, the plot looks linear, positive slope 11.18 [LO 4] a. Mean labour cost when batch size = 60 b. Mean labour cost when batch size = 30 c. Change in mean labour cost per unit increase in batch size d. Mean labour cost when batch size = 0; No e. Factors will include buyer preferences, reputation of the manufacturer, etc. 11.19 [LO 4] a. b. Yes, the relationship looks to be linear with a positive slope. 11.20 [LO 4] a. Mean sales when 20 coupons are issued b. Mean sales when 18 coupons are issued c. Change in mean sales per one unit (coupon) issued d. Mean sales when no coupons have been issued. Yes e. Factors other than the number of coupons issued; answers will vary. 11.21 [LO 5] (1) Mean of error terms = 0 (2) Constant variance (3) Normality (4) Independence 11.22 [LO 6] Mean square error is the point estimate of the error variance. Standard error is the point estimate of the standard deviation of the error. 11.23 [LO 6] The quality or “goodness” of the fit of the least squares line to the observed data. 11.24 [LO 6] The “best” line that can be fitted to the observed data. The slope and the intercept of the least squares line. 11.25 [LO 6] Evaluate for the given value of x. 11.26 [LO 6] Because we do not know how y and x are related outside the experimental region. 11.27 [LO 6] 11.28 [LO 6] 11.29 [LO 6] 11.30 [LO 6] 11.31 [LO 6] 11.32 [LO 6] a. number of errors decreases with age but the relationship is not very strong. b. s2 = 79.7857/25 = 3.1914 s = 1.7864 c. bo = 16.417 b1 = -.489 11.33 [LO 6] a. b0 = 11.4641 b1 = 24.6022 b0 – 0 copiers, 11.46 minutes of service. b1 – each additional copier adds 24.6022 minutes of service on average. No. The interpretation of b0 does not make practical sense since it indicates that 11.46 minutes of service would be required for a customer with no copiers. b. = 11.4641 + 24.6022(4) = 109.873, or 109.9 minutes 11.34 [LO 6] a. = 7.814 + 2.665 (.10) = 8.0805 b. 11.35 [LO 6] a. xi yi xi yi 5 71 25 355 62 663 3844 41106 35 381 1225 13335 12 138 144 1656 83 861 6889 71463 14 145 196 2030 46 493 2116 22678 52 548 2704 28496 23 251 529 5773 100 1024 10000 102400 41 435 1681 17835 75 772 5625 57900 b. is the estimated increase in mean labor cost (10.1463) for every 1 unit increase in the batch size. b0 is the estimated mean labor cost (18.4875) when batch size = 0; no. c. d. = 18.4880 + 10.1463(60) = 627.266 11.36 [LO 6] a. Results of regression in Excel are: SUMMARY OUTPUT Regression Statistics Multiple R 0.938804 R Square 0.881352 Adjusted R Square 0.866521 Standard Error 10.54132 Observations 10 ANOVA df SS MS F Significance F Regression 1 6603.44464 6603.445 59.42656 5.7E-05 Residual 8 888.955357 111.1194 Total 9 7492.4 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 47.80357 14.3500413 3.33125 0.010365 14.7123 80.89485 14.7123 80.89485 Coupons Issued 5.723214 0.74242009 7.708862 5.7E-05 4.011189 7.435239 4.011189 7.435239 b. is the estimated increase in mean sales for every coupon issued. b0 is the estimated mean sales price when number of coupons = 0. c. d. = 47.80 + 5.72 (20) = 162.20 That is, \$162.20 11.37 [S 11.6] a. Strong ( = .05) evidence that the regression relationship is significant. b. Very strong ( = .01) evidence that the regression relationship is significant. 11.38 [S 11.6] Explanations will vary. 11.39 [S 11.6] a. b0 = 11.4641 b1 = 24.6022 b. SSE = 191.7017 c. = .8045 t = 30.580 t = b1/ = 24.602 /.8045 = 30.580 d. t = 30.580; df = 9; t.025 = 2.262; Reject , strong evidence of a significant relationship between x and y. e. t = 30.580; t.005 = 3.250; Reject , very strong evidence of a significant relationship between x and y. f. p value = .000; Reject at all , extremely strong evidence of a significant relationship between x and y. g. [24.6022 2.262(.8045)] = [22.782, 26.422] h. = 3.4390 t = 3.334 t = b0 / = 11.464 / 3.439 = 3.334 i. p value = .0087; Reject at all  except .001 11.40 [S 11.6] a. b. SSE = 2.80 c. t = / = 2.665 /.2585 = 10.31 d. t = 10.31, df = 28; t.025 = 2.05; Reject , strong evidence of a significant relationship between x and y. e. t = 10.31, t.005 = 2.76; Reject , very strong evidence of a significant relationship between x and y. f. p-value = less than .001; reject at each value of  g. [2.665 ± 2.048(.2585)] = [2.136, 3.194] h. t = / = 7.814 / 0.0799 = 97.82 i. p-value less than .001; reject at each value of . 11.41 [S 11.6] a. b. SSE = 746.7624 c. t = / = 10.1463 /.0866 = 117.1344 d. t = 117.13, df = 10; t.025 = 2.23; Reject , strong evidence of a significant relationship between x and y. e. t = 117.13, t.005 = 3.17; Reject , very strong evidence of a significant relationship between x and y. f. p-value = .000; reject at each value of  g. [10.1463 ± 2.228(.0866)] = [9.953, 10.339] h. t = / = 18.488 / 4.677 = 3.95 i. p-value = .003; fail to reject at  = .001. Reject at all other values of  11.42 [S 11.6] a. = 47.80 = 5.72 b. SSE = 888.96 c. = .74 t = 7.71 t = / = 5.72 /.74 = 7.7 d. t = 7.71, df = 8; t.025 = 2.31; Reject , strong evidence of a significant relationship between x and y. e. t = 7.71, t.005 = 3.36; Reject , very strong evidence of a significant relationship between x and y. f. p value = .000 Reject at all  g. [4.01, 7.44] h. = 14.4 t = 3.33 t = / = 47.80 / 14.4 = 3.33 i. p value = .010 Reject at all  except .01 and .001 11.43 [S 11.6] 95% C.I. for β1 = -.87 to -.11 11.44 [S 11.7] The distance between xo and , the average of the previously observed values of x. 11.45 [S 11.7] A confidence interval is for the mean value of y. A prediction interval is for an individual value of y. 11.46 [S 11.7] The smaller the distance value, the shorter the lengths of the intervals. 11.47 [S 11.7] a. 109.873, [106.721, 113.025] b. 109.873, [98.967, 120.779] c. We have x = 4, distance value So confidence interval is: this compares (within rounding) to the computer generated output. For the prediction interval with the same quantities we get = [98.971, 120.775] which also compares within rounding. d. 113 minutes 11.48 [S 11.7] a. 8.0806; [7.948, 8.213] b. 8.0806; [7.419, 8.743] c. dv = (.0648/.316561)2 = .0419 d. 99% C.I.: [8.0806 ± 2.763(.065)] = [7.9016, 8.2596] 99% P.I.: [ ] = [ ] e. (1) 8.4804; [8.360, 8.600] (2) 8.4804; [7.821, 9.140] (3) 99% C.I.: [8.4804 ± 2.763(.059)] = [8.3857, 8.5251] 99% P.I.: [ ] = [ ] 11.49 [S 11.7] a. 627.26, [621.05, 633.47] b. 627.26, [607.03, 647.49] c. 99% C.I.: [627.26 ± 3.169(2.79)] = [(618.42, 636.10)] 99% P.I.: [ ] = [ ] 11.50 [S 11.7] a. 162.27, [154.31, 170.22] b. 162.27, [136.69, 187.84] 11.51 [S 11.7] From Megastat Predicted values for: Sales confidence interval 95% Confidence Intervals 95% Prediction Intervals Advertising Predicted lower upper lower upper Leverage 11 114.945 110.604 119.287 102.024 127.867 0.127 95% C.I.: [110.604, 119.287] 11.52 [S 11.8] Total variation: measures the total amount of variation exhibited by the observed values of y. Unexplained variation: measures the amount of variation in the values of y that is not explained by the model (predictor variable). Explained variation: measures the amount of variation in the values of y that is explained by the predictor variable. 11.53 [S 11.8] Proportion of the total variation in the n observed values of y that is explained by the simple linear regression model. 11.54 [S 11.8] Explained variation = total variation – SSE = 20,110.5455 – 191.7017 = 19918.8438 r2 = 19918.8438 / 20110.5455 = 0.990 r = +sqrt(0.990) = 0.995 99% of the variation in service time can be explained by variation in number of copiers repaired. 11.55 [S 11.8] Explained variation = 13.459 – 2.806 = 10.653 r2 = 10.653 / 13.459 = 0.792 r = +sqrt(0.792) = 0.890 79.2% of the variation in demand can be explained by variation in price differential. 11.56 [S 11.8] Explained variation = 1,025,339.6667 – 746.7624 = 1,024,592.904 r2 = 1024592.904 / 1025339.6667 = 0.999 r = +sqrt(0.999) = 0.9995 99.9% of the variation in direct labor can be explained by variation in batch size. 11.57 [S 11.8] Explained variation = 7492.4 – 888.96 = 6603.4 r2 = 6603.4 / 7492.4 = 0.88 r = +sqrt(0.88) = 0.939 88% of the variation in sales can be explained by variation in coupons issued. 11.58 [LO 7] . 11.59 [LO 7] a two-tailed t–test on 11.60 [LO 7] a. F = 19918.844 / (191.7017 / 9) = 935.149 b. F.05 = 5.12 df1 = 1, df2 = 9 Since 935.149 > 5.12, reject H0 with strong evidence of a significant relationship between x and y. c. F.01 = 10.56 df1 = 1, df2 = 9 Since 935.149 > 10.56, reject H0 with very strong evidence of a significant relationship between x and y. d. p value =less than .001; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y. e. t2 = (30.58)2 = 935.14 (approximately equals F = 935.149) (t.025)2 = (2.262)2 = 5.12 = F.05 11.61 [LO 7] a. F = 10.65268 / (2.805902 / 28) = 106.303 b. F.05 = 4.20, reject (df1 = 1, df2 = 28). Strong evidence of a significant relationship between x and y. c. F.01 =7.64, reject (df1 = 1, df2 = 28). Very strong evidence of a significant relationship between x and y. d. p-value = less than .001, Reject at all levels of . Extremely strong evidence of a significant relationship between x and y. e. t2 = (within rounding error) (t.025)2 = 4.19 = F.05 11.62 [LO 7] a. F = 1,024,592.9043/ (746.7624 / 10) = 13,720.47. b. F.05 =4.96, reject (df1 = 1, df2 = 10). Strong evidence of a significant relationship between x and y. c. F.01 =10.04, reject (df1 = 1, df2 = 10). Very strong evidence of a significant relationship between x and y. d. p-value = .000; reject at all levels of . extremely strong evidence of a significant relationship between x and y. e. t2 = (within rounding error); F at .05 = 4.96 = (2.28)2 = (t)2 at .025 11.63 [LO 7] a. F = 6603.445/(888.9553571 / 8) = 59.4266 b. F.05 = 5.32, df1 = 1, df2 = 8 Since 59.43 > 5.32, reject H0 at .05, strong evidence of a significant relationship between x and y. c. F.01 = 11.26 df1 = 1, df2 = 8 Since 59.43 > 11.3, reject H0 at .01, very strong evidence of a significant relationship between x and y. d. p value =.000; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y. e. t2 = (7.71)2 = 59.44 (approximately equals F = 59.43) (t.025)2 = (2.306)2 = 5.32 = F.05 11.64 [S 11.10] They should be plotted against the independent variable and against . Funneling or curved patterns indicate violations of the regression assumptions. 11.65 [S 11.10] Create a histogram, stem-and-leaf, and normal plot. 11.66 [S 11.10] Transforming the dependent variable. 11.67 [S 11.10] Possible violations of the normality and constant variance assumptions. 11.68 [S 11.10] No. 11.69 [S 11.10] a. b. No c. Data: Data z p E = np (obs-exp)sqr/exp lower upper frequency -8 < -6 1 -1.36 0.0869 0.9559 0.002034533 -6 < -4 1 -0.91 0.0945 1.0395 0.001500962 -4 < -2 2 -0.45 0.1450 1.595 0.102836991 -2 < -0 2 0.1736 1.9096 0.004279514 0 < 2 1 0.46 0.1772 1.9492 0.462230987 2 < 4 2 0.92 0.1440 1.584 0.109252525 4 < 6 1 1.38 0.0950 1.045 0.001937799 6 < 8 1 1.83 0.0502 0.5522 0.363138066 Chi-square obtained is 1.0472. Degrees of freedom = k-1-2 = 8-1-2 = 5. Chi-square critical for alpha = .05 is 11.0705. Chi-square obtained is less than the chi-square critical, suggesting that the residual values are normally distributed (cannot reject the hypothesis that the scores are normally distributed). 11.70 [S 11.10] The residual plot has somewhat of a cyclical appearance. Since d =.473 is less than dL, 05 = 1.27, we conclude there is positive autocorrelation and since 4 – .473 = 3.527 and this is greater than dU,.05 = 1.45 we conclude that there is no negative autocorrelation. 11.71 [S 11.10] a. ln yt = 2.07012 + 0.25688t ln y16 = 2.07012 + 0.25688(16) = 6.1802 b. e6.1802 = 483.09 e5.9945 = 401.22 e6.3659 = 581.67 c. d = 1.87643 dL,0.05 = 1.08. Since d >1.10 we fail to reject. d. Growth rate = e0.25688 = 1.293 This means the growth rate is expected to be 29.3% per year. 11.72 [S 11.10] The constant variance assumption seems to be violated. Might also have a slight curve, which may mean a violation of the assumption of correct functional form. 11.73 [S 11.10] a. Yes b. Allow 200 minutes. 11.74 [LO 7] a. Yes; see the plot in part c. b. c. d. p–value = .000, reject , significant e. MegaStat output: Regression Analysis r² 0.990 n 6 r -0.995 k 1 Std. Error 1.191 Dep. Var. y ANOVA table Source SS df MS F p-value Regression 537.6571 1 537.6571 378.89 4.11E-05 Residual 5.6762 4 1.4190 Total 543.3333 5 Regression output confidence interval variables coefficients std. error t (df=4) p-value 95% lower 95% upper Intercept 306.6190 3.5926 85.348 1.13E-07 296.6445 316.5936 x -27.7143 1.4238 -19.465 4.11E-05 -31.6674 -23.7612 Predicted values for: y 95% Confidence Intervals 95% Prediction Intervals x Predicted lower upper lower upper Leverage 2.10 248.419 246.340 250.498 244.512 252.326 0.395 2.75 230.405 228.731 232.078 226.698 234.111 0.256 3.10 220.705 217.975 223.434 216.417 224.993 0.681 11.75 [LO 1,2] a. b1 = -6.44 b. p<.000 c. r = .99, eta2 = .9836, 98.36% of the variance in accidents overlaps with bridge width. 11.76 [LO 1, 2, 3] a. r = .69 b. F(1,23) = 21.41, p<.001 c. r2 = .4821, so 48.21% d. Greater satisfaction with greater perceived control. e. Appears fairly normal f. Chi-square obtained is 4.2061. df = k-1-2 = 4. Chi-square critical at .05 = 11.07. Chi-square obtained is less than the critical value therefore we do not reject the assumption of normality. Following the histogram intervals: p Ep f (obs-E)sqr/E 0.0577 1.4425 3 1.6817 0.0887 2.2175 1 0.6685 00.1529 3.8225 3 0.1770 0.1985 4.9625 4 0.1867 0.2002 5.005 5 0.000005 0.1546 3.865 4 0.004715 0.8531 2.2325 4 1.3994 0.058 1.455 1 0.142285 Sum = 4.2601 11.77 [LO 1, 5] a. r = .86, strong relationship between debt and population b. R = .86, same value c. Smallest = Nigeria, largest = Pakistan 11.78 [S 11.7] For aggressive stocks a 95% confidence interval for is where is based on 7 degrees of freedom. We are 95% confident that the effect of a one-month increase in the return length time for an aggressive stock is to increase the mean value of the average estimate of between .00749 and .02512. For defensive stocks a 95% confidence interval for is [–0.00462 ± 2.365(.00084164)] = [–.00661, –.00263]. We are 95% confident that the effect of a one-month increase in the return length time for a defensive stock is to decrease the mean value of the average estimate of between .00661 and .00263. For neutral stocks a 95% confidence interval for is [.0087255 ± 2.365(.001538)] = [.005088, .01236]. We are 95% confident that the effect of a one-month increase in the return length time for a neutral stock is to increase the mean value of the average estimate of between .00509 and .01236. Internet Exercise 11.79 [LO 1, 2, 6] Results will vary depending on the year the data is accessed.

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