CHAPTER 1212.1 (a) When X = 0, the estimated expected value of Y is 2.(b) For each increase in the value X by 1 unit, you can expect an increase by an estimated 5 units in the value of Y .(c) ˆ2525(3)17YX =+=+= 12.2 (a) yes, (b) no,(c) no, (d) yes12.3 (a) When X = 0, the estimated expected value of Y is 16.(b) For each increase in the value X by 1 unit, you can expect a decrease in an estimated 0.5 units in the value of Y .(c) 13)6(5.0165.016ˆ=-=-=X Y12.4(a)(b) For each increase in shelf space of an additional foot, there is an expected increase in weekly sales of an estimated 0.074 hundreds of dollars, or $7.40. (c)042.2)8(074.045.1074.045.1ˆ=+=+=X Y, or $204.2012.5(a)Scatter Diagram0501001502002503003504004500100200300400500600700X, Reported (thousands)Y , A u d i t e d (t h o u s a n d s )(b)For each additional thousand units increase in reported newsstand sales, the mean audited sales will increase by an estimated 0.5719 thousand units.(c)()ˆ26.72400.5719400255.4788Y=+= thousands.12.6(a)(b) Partial Excel output:Coefficients Standard Error t Stat P-value Intercept -2.3697 2.0733 -1.1430 0.2610 Feet 0.0501 0.0030 16.5223 0.0000(c) The estimated mean amount of labor will increase by 0.05 hour for each additional cubic foot moved.(d)()ˆ 2.3697+0.050150022.6705Y=-=Scatter Diagram010********6070809002004006008001000120014001600XY12.7(a)(b) ˆ0.19120.0297YX =+ (c) For each increase of one additional pound, the estimated mean number of orders will increase by 29.7.(d) ()ˆ0.19120.029750015.043Y=+=thousands 12.8(a)Scatter Diagram010020030040050060070080090050100150200250X, Revenue (Millions of dollars)Y , V a l u e (M i l l i o n s o f d o l l a r s )(b)0246.2599b =-, 1 4.1897b =(c)For each additional million dollars increase in revenue, the mean annual value will increase by an estimated 4.1897 million dollars. Since revenue cannot be 0, the Y intercept has no practical interpretation. Also, since the Y intercept is outside the range of the observed values of the X variable, its interpretation should be made cautiously.(d)()ˆ246.2599 4.1897150382.2005Y=-+= million dollars.12.9(a)(b) X Y065.11.177ˆ+= (c) For each increase of 1 square foot in space, the expected monthly rental is estimated to increase by $1.065. Since X cannot be zero, 177.1 has no practical interpretation. (d) =+=+=)1000(065.11.177065.11.177ˆX Y $1242.10(e) An apartment with 500 square feet is outside the relevant range for the independent variable.(f)The apartment with 1200 square feet has the more favorable rent relative to size. Based on the regression equation, a 1200 square foot apartment would have anexpected monthly rent of $1455.10, while a 1000 square foot apartment would have an expected monthly rent of $1242.10.12.10 (a)12.10 (b) ˆ 6.0483 2.0191YX =+ cont. (c) For each increase of one additional Rockwell E unit in hardness, the estimated mean tensile strength will increase by 2.0191 thousand pounds per square inch.(d)()ˆ 6.0483 2.01913066.620Y=+= thousand pounds per square inch.12.11 80% of the variation in the dependent variable can be explained by the variation in the independent variable.12.12 SST = 40 and r 2 = 0.90. So, 90% of the variation in the dependent variable can be explained by the variation in the independent variable.12.13 r 2 = 0.75. So, 75% of the variation in the dependent variable can be explained by the variation in the independent variable.12.14 r 2 = 0.75. So, 75% of the variation in the dependent variable can be explained by the variation in the independent variable.12.15 Since SST = SSR + SSE and since SSE cannot be a negative number, SST must be at least as large as SSR . 12.16 (a)r 2 =2.05353.0025SSR SST == 0.684. So, 68.4% of the variation in the dependent variable can be explained by the variation in the independent variable.(b)0.308YXs ==== (c) Based on (a) and (b), the model should be very useful for predicting sales.12.17 (a) r 2 = 0.9015. So, 90.15% of the variation in audited newsstand sales can be explained by the variation in reported newsstand sales. (b) 42.1859YX s =(c) Based on (a) and (b), the model should be very useful for predicting audited sales.12.18 (a) r 2 = 0.8892. So, 88.92% of the variation in the dependent variable can be explained by the variation in the independent variable. (b) 5.0314YX s =(c) Based on (a) and (b), the model should be very useful for predicting labor hours.12.19 (a) r 2 = 0.9731. So, 97.31% of the variation in the dependent variable can be explained by the variation in the independent variable. (b) 0.7258YX s =(c)Based on (a) and (b), the model should be very useful for predicting the number of orders.12.20 (a) r 2 = 0.9424. So, 94.24% of the variation in value of a baseball franchise can be explained by the variation in its annual revenue. (b) 33.7876YX s =(c)Based on (a) and (b), the model should be very useful for predicting the value of a baseball franchise.12.21 (a) r 2 = 0.723. So, 72.3% of the variation in the dependent variable can be explained by the variation in the independent variable. (b) 6.194=YX s(c) Based on (a) and (b), the model should be very useful for predicting monthly rent.12.22 (a) r 2 = 0.4613. So, 46.13% of the variation in the dependent variable can be explained by the variation in the independent variable. (b) 9.0616YX s =(c)Based on (a) and (b), the model is only marginally useful for predicting tensile strength.12.23 A residual analysis of the data indicates no apparent pattern. The assumptions of regression appear to be met.12.24A residual analysis of the data indicates a pattern, with sizeable clusters ofconsecutive residuals that are either all positive or all negative. If the data is cross-sectional, this pattern indicates a violation of the assumption of linearity and a quadratic model should be investigated. If the data is time-series, the pattern indicates a violation of the assumption of independence of errors.12.25 (a)Reported Residual Plot-100-80-60-40-2002040600.0100.0200.0300.0400.0500.0600.0700.0ReportedR e s i d u a l sBased on the residual plot, there does not appear to be a pattern in the residual plot.(b)Normal Probability Plot-100-80-60-40-200204060Z ValueR e s i d u a l sBased on the residual plot, there appears to be some heteroscedasticity effect. The normal probability plot of the residuals indicates a departure from the normality assumption. The error distribution appears to be left-skewed.12.26 (a)Space Residual Plot-0.5-0.4-0.3-0.2-0.100.10.20.30.40.50510152025SpaceR e s i d u a l sBased on the residual plot, there does not appear to be a pattern in the residual plot. (b)Normal Probability Plot-0.5-0.4-0.3-0.2-0.100.10.20.30.40.5Z ValueR e s i d u a l sBased on the residual plot, there is not apparent heteroscedasticity effect. The normal probability plot of the residuals indicates a departure from the normality assumption.12.27 (a)Weight Residual Plot-2-1.5-1-0.500.511.50100200300400500600700800WeightR e s i d u a l sThe residual plot does not reveal any obvious pattern. So a linear fit appears to be adequate.(b)Normal Probability Plot-2-1.5-1-0.500.511.5Z ValueR e s i d u a l sThe residual plot does not reveal any possible violation of the homoscedasticity assumption. This is not a time series data, so you do not need to evaluate theindependence assumption. The normal probability plot shows that the distribution has a thicker left tail than a normal distribution but there is no sign of severe skewness.12.28 (a)Feet Residual Plot-15-10-5051015050010001500FeetR e s i d u a l sBased on the residual plot, there appears to be a nonlinear pattern in the residuals. A quadratic model should be investigated.(b)Normal Probability PlotZ ValueR e s i d u a l sThe assumptions of normality and equal variance do not appear to be seriously violated.12.29 (a)Size Residual Plot-500-400-300-200-100010020030040050005001000150020002500SizeR e s i d u a l sBased on a residual analysis of the residuals versus size, the model appears to be adequate. (b)Normal Probability Plot-500-400-300-200-1000100200300400500Z ValueR e s i d u a l sThe normal probability plot shows that the distribution has a thicker left tail than a normal distribution but there is no sign of severe skewness. The assumptions of regression do not appear to be seriously violated.12.30 (a)Revenue Residual Plot-80-60-40-20020406080100050100150200250RevenueR e s i d u a l sBased on the residual plot, there appears to be a nonlinear pattern in the residuals. A quadratic model should be investigated.(b)Normal Probability PlotZ ValueR e s i d u a l sThe normal probability plot of the residuals does not reveal significant departure from the normality assumption. The assumption of equal variance does not appear to be seriously violated.12.31 (a)The residual plot does not reveal any obvious pattern. So a linear fit appears to be adequate.(b)Normal Probability PlotZ ValueR e s i d u a l sThe residual plot does not reveal any possible violation of the homoscedasticity assumption. This is not a time series data, so you do not need to evaluate theindependence assumption. The normal probability plot shows that the distribution has a slightly thinner right tail than a normal distribution but there is no sign of severe skewness.12.32 (a)An increasing linear relationship exists.(b) There appears to be strong positive autocorrelation among the residuals.12.33 (a) There is no apparent pattern in the residuals over time.(b) D = 1.661>1.36. There is no evidence of positive autocorrelation among the residuals. (c) The data are not positively autocorrelated.12.34(a) No, it is not necessary to compute the Durbin-Watson statistic since the data have been collected for a single period for a set of stores.(b)If a single store was studied over a period of time and the amount of shelf space varied over time, computation of the Durbin-Watson statistic would be necessary.12.35 (a) b 0 = 169.455, b 1 = –1.8579(b) =-=-=)50(8579.1455.1698579.1455.169ˆX Y 76.56(c)(d) D = 1.18<1.27. There is evidence of positive autocorrelation among the residuals.(e)The plot of the residuals versus temperature indicates that positive residuals tend to occur for the lowest and highest temperatures in the data set. A nonlinear modelmight be more appropriate. The evidence of positive autocorrelation is another reason to question the validity of the model.12.36 (a) b 1 =201399.050.016112495626SSXY SSX == b 0 =()171.26210.01614393Y b X -=-= 0.458 (b)=+=+=)4500(0161.0458.00161.0458.0ˆX Y 72.908 or $72,908(c)(d)D =()212211243.2244599.0683ni i i nii e e e-==-=∑∑= 2.08>1.45. There is no evidence of positiveautocorrelation among the residuals.(e)Based on a residual analysis, the model appears to be adequate.12.37 (a)ˆ17.08335YX =- (b) ()ˆ17.083350.514.5833Y=-= seconds(c)There is no noticeable pattern in the plot.Residuals Plot-8-6-4-20246Time OrderR e s i d u a l s12.37 (d) H 0: There is no autocorrelation.cont. H 1: There is positive autocorrelation.PHStat output:Durbin-Watson CalculationsSum of Squared Difference of Residuals 238.4375L U H 0. There is no evidence of autocorrelation.(e)Based on the results of (c) and (d), there is no reason to question the validity of the model.12.38 (a) b 0 = –2.535, b 1 = 0.060728(b) =+-=+-=)83(060728.0535.2060728.0535.2ˆX Y 2.5054 or $2505.40(c)(d) D = 1.64>1.42. There is no evidence of positive autocorrelation among the residuals.(e)The plot of the residuals versus time period shows some clustering of positive and negative residuals for intervals in the domain, suggesting a nonlinear model might be better. Otherwise, the model appears to be adequate.12.39 (a)Scatter Diagram0501001502002501020304050X, Crude Oi Price (dollars/barrel)Y , G a s o l i n e P r i c e (c e n t s /g a l l o n )There appears to be a positive curvilinear relationship between crude oil price and gasoline price.(b) ()ˆ60.6944 2.8864YX =+ (c)Crude Oil Residual Plot-25-20-15-10-50510152025202530354045Crude OilR e s i d u a l sThere appears to be positive autocorrelation where clusters of positive residuals can be seen at the lower and upper range of the X value, and a cluster of negative residuals is apparent near the center of the X range. There also appears to be non-linear relationship between crude oil price and gasoline price.(d) 0.3161D =. 1.65L d =, 1.69U d =. Since D < 1.65, there is evidence of positive autocorrelation at 5% level of significance.(e)Based on the result of (c)-(d), the linear model and independent of errors assumptions do not appear to be valid.12.40 (a) 01:0H β= 11:0H β≠Test statistic: ()110/ 4.5/1.5 3.00b t b s =-==(b) With n = 18, df = 18 – 2 =16, 1199.216±=t .(c) Reject H 0. There is evidence that the fitted linear regression model is useful. (d) 111161116b b b t s b t s β-≤≤+)5.1(1199.25.4)5.1(1199.25.41+≤≤-β 68.732.11≤≤β12.41 (a) /60/160MSR SSR k ===/(1)40/18 2.222MSE SSE n k =--== 27222.2/60/===MSE MSR F (b) 1,18 4.41F =(c) Reject H 0. There is evidence that the fitted linear regression model is useful.(d) 2600.6100SSR r SST ===0.7746r ==-(e) 0:0H ρ=There is no correlation between X and Y . 1:0H ρ≠There is correlation between X and Y .d.f. = 18.Decision rule: Reject 0H if cal t >2.1009.Test statistic: 5.196t ===-.Since cal t = 5.196 >2.1009, reject 0H . There is enough evidence to conclude that the correlation between X and Y is significant.12.42 (a) 01:0H β= 11:0H β≠111100.074 4.65 2.22810.0159b b t t S β-===>=with 10 degrees of freedom for05.0=α. Reject H 0. There is enough evidence to conclude that the fitted linearregression model is useful.(b)()1120.074 2.22810.0159n b b t S -±=± 1094.00386.01≤≤β12.43 (a) 01:0H β=11:0H β≠PHStat output:CoefficientsStandard Errort Stat P-value Lower 95% Upper 95% Intercept26.7240 26.5425 1.00680.3435 -34.4832 87.9311 Reported0.5719 0.06688.55670.00000.41780.7260Since the p -value is essentially zero, reject H 0 at 5% level of significance. There is evidence of a linear relationship between reported sales and audited sales.(b)10.41780.7260β≤≤12.44 (a) 3416.5223 2.0322t t =>= for 05.0=α. Reject H 0. There is evidence that thefitted linear regression model is useful. (b) 10.04390.0562β≤≤12.45 (a) p -value is virtually 0 < 0.05. Reject H 0. There is evidence that the fitted linear regression model is useful. (b) 10.02760.0318β≤≤12.46 (a) 01:0H β=11:0H β≠PHStat output:CoefficientsStandard Errort Stat P-value Lower 95% Upper 95% Intercept-246.2599 26.0405 -9.4568 0.0000 -299.6015 -192.9183 Revenue4.1897 0.195721.40750.00003.78884.5906Since the p -value is essentially zero, reject H 0 at 5% level of significance. There is evidence of a linear relationship between annual revenue sales and franchise value. (b)13.7888 4.5906β≤≤12.47 (a) 0687.274.723=>=t t with 23 degrees of freedom for 05.0=α. Reject H 0. Thereis evidence that the fitted linear regression model is useful. (b) 10.7803 1.3497β≤≤12.48 (a) p -value = 7.26497E-06 < 0.05. Reject H 0. There is evidence that the fitted linear regression model is useful. (b) 11.2463 2.7918β≤≤12.49 (a)Sears, Roebuck and Company’s stock moves only 60.3% as much as the overall market and is much less volatile. LSI Logic’s stock moves 142.1% more than the overall market and is considered as extremely volatile. The stocks of Disney Company, Ford Motor Company and IBM, move 10.9%, 34% and 44.9%respectively more than the overall market and are considered somewhat more volatile than the market.(b)Investors can use the beta value as a measure of the volatility of a stock to assess its risk.12.50 (a)()()0% daily change in ULPIX 2.00% daily change in S&P 500 Index b =+ (b) If the S&P gains 30% in a year, the ULPIX is expected to gain an estimated 60%. (c) If the S&P loses 35% in a year, the ULPIX is expected to lose an estimated 70%.(d)Since the leverage funds have higher volatility and, hence, higher risk than the market, risk averse investors should stay away from these funds. Risk takers, on the other hand, will benefit from the higher potential gain from these funds.12.51 (a) r = -0.2810.(b) t = -1.0556, p -value = 0.3104 > 0.05. Do not reject H 0. There is not enoughevidence to conclude that there is a significant linear relationship between the retail price and the energy cost per year of medium-size top-freezer refrigerators.12.52 (a)r = –0.4014.(b)t = –1.8071, p -value = 0.089 > 0.05. Do not reject H 0. At the 0.05 level ofsignificance, there is no linear relationship between the turnover rate of pre-boarding screeners and the security violations detected.(c)There is not sufficient evidence to conclude that there is a linear relationship between the turnover rate of pre-boarding screeners and the security violations detected.12.53 (a)r = 0.3409.(b)t = 1.4506, p -value = 0.1662 > 0.05. Do not reject H 0. At the 0.05 level of significance, there is no linear relationship between the battery capacity and the digital-mode talk time.(c) There is not sufficient evidence to conclude that there is a linear relationship between the battery capacity and the digital-mode talk time.(d)No, the expectation that the cellphones with higher battery capacity have a higher talk time is not borne out by the data.12.54 (a)r = 0.4838(b)t = 2.5926, p -value = 0.0166 < 0.05. Reject H 0. At the 0.05 level of significance, there is a linear relationship between the cold-cranking amps and the price. (c) The higher the price of a battery is, the higher is its cold-cranking amps.(d)Yes, the expectation that batteries with higher cranking amps to have a higher price is borne out by the data.12.55 (a) When X = 2, 11)2(3535ˆ=+=+=X Y05.020)22(201)()(12122=-+=--+=∑=n i i i X X X X n h95% confidence interval: 05.011009.211ˆ18⋅⋅±=±h s t Y Y X|10.5311.47Y X μ≤≤(b)95% prediction interval: 05.111009.2111ˆ18⋅⋅±=+±h s t Y Y X28.84713.153X Y =≤≤12.56 (a) When X = 4, 17)4(3535ˆ=+=+=X Y25.020)24(201)()(12122=-+=--+=∑=n i i i X X X X n h95% confidence interval: 18ˆ17 2.10091YXY t s ±=±⋅|415.9518.05Y X μ=≤≤(b)95% prediction interval: 18ˆ17 2.10091YX Y t s ±=±⋅414.65119.349X Y =≤≤(c)The intervals in this problem are wider because the value of X is farther from X .12.57 (a)|400223.5000287.4577Y X μ=≤≤ (b)400153.0765357.8812X Y =≤≤ (c)Part (b) provides an interval prediction for the individual response given a specific value of the independent variable, and part (a) provides an interval estimate for the mean value given a specific value of the independent variable. Since there is much more variation in predicting an individual value than in estimating a mean value, a prediction interval is wider than a confidence interval estimate holding everything else fixed.12.58 (a)(2ˆ 2.042 2.22810.3081i n YX Y t S -±=±|81.7876 2.2964Y X μ=≤≤(b)(2ˆ 2.042 2.22810.3081i n YX Y t S -±=±81.3100 2.7740X Y =≤≤(c)Part (b) provides an interval prediction for the individual response given a specific value of the independent variable, and part (a) provides an interval estimate for the mean value given a specific value of the independent variable. Since there is much more variation in predicting an individual value than in estimating a mean value, a prediction interval is wider than a confidence interval estimate holding everything else fixed.12.59 (a)|50014.715015.3701Y X μ=≤≤ (b)50013.505916.5793X Y =≤≤ (c)Part (b) provides an interval prediction for the individual response given a specific value of the independent variable, and part (a) provides an interval estimate for the mean value given a specific value of the independent variable. Since there is much more variation in predicting an individual value than in estimating a mean value, a prediction interval is wider than a confidence interval estimate holding everything else fixed.12.60 (a)|50020.799024.5419Y X μ=≤≤ (b)50012.275533.0654X Y =≤≤ (c)Part (b) provides an interval prediction for the individual response given a specific value of the independent variable, and part (a) provides an interval estimate for the mean value given a specific value of the independent variable. Since there is much more variation in predicting an individual value than in estimating a mean value, a prediction interval is wider than a confidence interval estimate holding everything else fixed.12.61 (a)|10001153.01331.5Y X μ=≤≤(b) 1000829.91654.6X Y =≤≤(c) Part (b) provides an interval prediction for the individual response given a specificvalue of the independent variable, and part (a) provides an interval estimate for themean value given a specific value of the independent variable. Since there is muchmore variation in predicting an individual value than in estimating a mean value, aprediction interval is wider than a confidence interval estimate holding everythingelse fixed.12.62 (a)|150367.0757397.3254Y X μ=≤≤(b) 150311.3562453.0448X Y =≤≤(c) Part (b) provides an interval prediction for the individual response given a specificvalue of the independent variable, and part (a) provides an interval estimate for themean value given a specific value of the independent variable. Since there is muchmore variation in predicting an individual value than in estimating a mean value, aprediction interval is wider than a confidence interval estimate holding everythingelse fixed.12.63 (a)|30116.7082178.0564Y X μ=≤≤(b) 30111.5942183.1704X Y =≤≤(c) Part (b) provides an interval prediction for the individual response given a specificvalue of the independent variable, and part (a) provides an interval estimate for themean value given a specific value of the independent variable. Since there is muchmore variation in predicting an individual value than in estimating a mean value, aprediction interval is wider than a confidence interval estimate holding everythingelse fixed.12.64 The slope of the line, b 1, represents the estimated expected change in Y per unit change in X .It represents the estimated mean amount that Y changes (either positively or negatively) for a particular unit change in X . The Y intercept b 0 represents the estimated mean value of Y when X equals 0.12.65 The coefficient of determination measures the proportion of variation in Y that is explainedby the independent variable X in the regression model.12.66 The unexplained variation or error sum of squares (SSE) will be equal to zero only when theregression line fits the data perfectly and the coefficient of determination equals 1.12.67 The explained variation or regression sum of squares (SSR) will be equal to zero only whenthere is no relationship between the Y and X variables, and the coefficient of determination equals 0.12.68 Unless a residual analysis is undertaken, you will not know whether the model fit isappropriate for the data. In addition, residual analysis can be used to check whether theassumptions of regression have been seriously violated.12.69 The assumptions of regression are normality of error, homoscedasticity, and independence oferrors.12.70 The normality of error assumption can be evaluated by obtaining a histogram, box-and-whisker plot, and/or normal probability plot of the residuals. The homoscedasticityassumption can be evaluated by plotting the residuals on the vertical axis and the X variable on the horizontal axis. The independence of errors assumption can be evaluated by plotting the residuals on the vertical axis and the time order variable on the horizontal axis. This assumption can also be evaluated by computing the Durbin-Watson statistic.12.71 If the data in a regression analysis has been collected over time, then the assumption ofindependence of errors needs to be evaluated using the Durbin-Watson statistic.12.72 The confidence interval for the mean response estimates the mean response for a given Xvalue. The prediction interval estimates the value for a single item or individual.12.73 (a) There is a strong positive correlation between course average and cumulative GPA,and between total hits and hit consistency. There is a moderate positive correlationbetween course average and hit consistency, and between cumulative GPA and hitconsistency.(b) It is not surprising that cumulative GPA is strongly and positively related to courseaverage because course average in a course contributes to the cumulative GPA andboth measure the performance of a student. It is also not surprising that total hits andhit consistency are highly positively related because hit consistency contributes tototal hits. Hit consistency is positively related to both cumulative GPA and courseaverage because the more frequently a student visits the Internet site supporting acourse, the more current the student is in the course and presumably the better thestudent performs in the course.12.74 (a) b 0 = 24.84, b 1 = 0.14(b) 24.84 is the portion of estimated mean delivery time that is not affected by thenumber of cases delivered. For each additional case, the estimated mean deliverytime increases by 0.14 minutes.(c) 84.45)150(14.084.2414.084.24ˆ=+=+=X Y(d) No, 500 cases is outside the relevant range of the data used to fit the regressionequation.(e) r 2 = 0.972. So, 97.2% of the variation in delivery time can be explained by thevariation in the number of cases.(f) Based on a visual inspection of the graphs of the distribution of residuals and theresiduals versus the number of cases, there is no pattern. The model appears to beadequate.(g) 1009.288.2418=>=t t with 18 degrees of freedom for 05.0=α. Reject H 0.There is evidence that the fitted linear regression model is useful.(h) |15044.8846.80Y X μ=≤≤(i) 15041.5650.12X Y =≤≤(j) 1518.01282.01≤≤β12.75 (a)b 0 = –63.02, b 1 = 0.189 (b)For each additional incoming call, the estimated mean number of trade executions increases by 0.189. – 63.02 is the portion of the estimated mean number of trade executions that is not affected by the number of incoming calls. Since the Y intercept is outside the range of the observed values of the X variable, its interpretation should be made cautiously. (c)99.314)2000(189.002.63189.002.63ˆ=+-=+-=X Y (d)No, 5000 incoming calls is outside the relevant range of the data used to fit the regression equation. (e)r 2 = 0.630. So, 63.0% of the variation in the number of trade executions can be explained by the variation in the number of incoming calls. (f)Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the number of cases, there is no pattern. The model appears to be adequate. (g)D = 1.96 (h)D = 1.96>1.52. There is no evidence of positive autocorrelation. The model appears to be adequate. (i)0345.250.733=>=t t with 33 degrees of freedom for 05.0=α. Reject H 0. There is evidence that the fitted linear regression model is useful. (j)|2000302.07327.91Y X μ=≤≤ (k)2000253.76376.22X Y =≤≤ (l)2403.01377.01≤≤β12.76 (a)Scatter Diagram 020406080100120140020406080100Assessed Value ($000)S e l l i n g P r i c e ($000)b 0 = –44.172, b 1 = 1.78171 (b)Since the assessed value of a home cannot be zero, the intercept has no practical interpretation. -44.172 is the portion of estimated mean selling price that is not affected by the assessed value. For each additional dollar in assessed value, the estimated mean selling price increases by $1.78. (c)458.80)70(78171.1172.4478171.1172.44ˆ=+-=+-=X Y or $80,458 (d)r 2 = 0.926. 92.6% of the variation in selling price can be explained by the variation in the assessed value.。