Basic Business Statistics, 9th edition



CHAPTER 13

13.1 (a) When X = 0, the estimated expected value of Y is 2.

(b) For increase in the value X by 1 unit, we can expect an increase by an estimated 5 units in the value of Y.

(c) [pic]

(d) yes, (e) no, (f) no, (g) yes, (h) no

13.2 (a) When X = 0, the estimated expected value of Y is 16.

(b) For increase in the value X by 1 unit, we can expect a decrease in an estimated 0.5 units in the value of Y.

(c) [pic]

13.3 (a)

(b),(c) [pic], [pic]

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.

(d) [pic], or $204.20

(e) [pic], [pic]

For each increase in shelf space of an additional foot, there is an expected increase in weekly sales of an estimated 0.064 hundreds of dollars, or $6.40.

[pic], or $204.53

(f) The best allocation to pet food depends on the profit made per foot of shelf space. The expected weekly sales (and profits) per foot of shelf space actually declines at the amount of allocated shelf space increases from 5 to 20 feet, however, if the profitability is still high enough, it will be worthwhile assigning a higher amount to pet food.

13.4 (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 average amount of labor will increase by 0.05 hour for each additional cubic foot moved.

(d) [pic]

(e) Other factors that might affect labor hours are size of the movers, how accessible is the building to the moving truck, etc.

13.5 (a)

[pic]

13.5 (b) [pic]

cont. (c) For each increase of one additional pound, the estimated average number of order will increase by 29.7.

(d) [pic]

13.6 (a)

[pic]

(b),(c) [pic]

(d) For each increase of 1 million dollars in box office gross, expected home video units sold are estimated to increase by 4.3331 thousand, or 4333.1 units.

(e) [pic]163.202 or 163,202 units.

(f) Some other factors that might be useful in predicting video unit sales are (i) the number of days the movie was screened, (ii) the rating of the movie by critics, (iii) the amount of advertisement spent on the video release, etc.

13.7 (a)

[pic]

(b),(c) [pic]

13.7 (d) For each increase of 1 square foot in space, the expected monthly rental is estimated

cont. to increase by $1.065. 177.1 represents the portion of apartment monthly rental that is not affected by square footage.

(e) [pic] $1242.10

(f) An apartment with 500 square feet is outside the relevant range for the

independent variable.

(g) 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 an expected monthly rent of $1455.10, while a 1000 square foot apartment would have an expected monthly rent of $1242.10.

13.8 (a)

[pic]

(b) [pic]

(c) For each increase of one additional Rockwell E unit in hardness, the estimated average tensile strength will increase by 2.0191 thousand pounds per square inch.

(d) [pic] thousand pounds per square inch.

13.9 80% of the variation in the dependent variable can be explained by the variation in the independent variable.

13.10 SST = 40 and r2 = 0.90. So, 90% of the variation in the dependent variable can be explained by the variation in the independent variable.

13.11 r2 = 0.75. So, 75% of the variation in the dependent variable can be explained by the variation in the independent variable.

13.12 r2 = 0.75. So, 75% of the variation in the dependent variable can be explained by the variation in the independent variable.

13.13 Since SST = SSR + SSE and since SSE cannot be a negative number, SST must be at least as large as SSR.

13.14 (a) r2 = 0.684. So, 68.4% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b) [pic]

(c) Based on (a) and (b), the model should be very useful for predicting sales.

13.15 (a) r2 = 0.8892. So, 88.92% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b) [pic]

(c) Based on (a) and (b), the model should be very useful for predicting the number of order.

13.16 (a) r2 = 0.9731. So, 97.31% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b) [pic]

(c) Based on (a) and (b), the model should be very useful for predicting the number of order.

13.17 (a) r2 = 0.728. So, 72.8% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b) [pic]

(c) Based on (a) and (b), the model should be very useful for predicting sales.

13.18 (a) r2 = 0.723. So, 72.3% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b) [pic]

(c) Based on (a) and (b), the model should be very useful for predicting monthly rent.

13.19 (a) r2 = 0.4613. So, 46.13% of the variation in the dependent variable can be explained by the variation in the independent variable.

(b) [pic]

(c) Based on (a) and (b), the model is only marginally useful for predicting tensile strength.

13.20 A residual analysis of the data indicates no apparent pattern. The assumptions of regression appear to be met.

13.21 A residual analysis of the data indicates a pattern, with sizeable clusters of consecutive residuals that are either all positive or all negative. This appears to violate the assumption of independence of errors.

13.22 (a)-(b) Based on a residual analysis, the model appears to be adequate.

13.23 (a)-(b)

Based on the residual plot, there appears to be a non-linear relationship between labor hours and cubic feet moved, and some heteroskedasticity effect. The normal probability plot of the residuals does not reveal significant departure from the normality assumption.

13.24 (a)

[pic]

[pic]

The residual plot does not reveal any obvious pattern. So a linear fit appears to be adequate.

(b) The residual plot does not reveal any possible violation of the homoscedasticity assumption. This is not a time series data, so we do not need to evaluate the independence 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.

13.25 (a)-(b)

Based on a residual analysis of the residuals versus box office gross, the model appears to be adequate.

13.26 (a)-(b)

Based on a residual analysis of the residuals versus size, the model appears to be adequate. From the normal probability plot, it appears that the normality assumption is violated.

13.27 (a)

[pic]

The residual plot does not reveal any obvious pattern. So a linear fit appears to be adequate.

(b)

[pic]

The residual plot does not reveal any possible violation of the homoscedasticity assumption. This is not a time series data, so we do not need to evaluate the independence 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.

13.28 (a) An increasing linear relationship exists. The critical values of the Durbin Watson statistic are [pic] and [pic]

(b) D = 0.109

(c) Since D = 0.109 < [pic], there is enough evidence to conclude that there is strong positive autocorrelation among the residuals.

13.29 (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.

13.30 (a) No, 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.

13.31 (a)

[pic]

(b) b0 = 169.455, b1 = –1.8579

(c) For each increase of one degree in average Fahrenheit temperature, the expected kilowatt usage is estimated to decrease by 1.8579.

(d) [pic] 76.56

(e) r2 = 0.894. So, 89.4% of the variation in kilowatt usage can be explained by the variation in the average temperature.

(f) [pic]

(g)

[pic]

13.31 (h)

cont.

[pic]

(i) D = 1.181.45. There is no evidence of positive autocorrelation among the residuals.

(j) Based on a residual analysis, the model appears to be adequate.

10.33 (a) [pic]

(b) [pic] There is no linear relationship between the amount of tamping and the separation time.

[pic] There is a linear relationship between the amount of tamping and the separation time.

PHStat output:

| |Coefficients |Standard Error |t Stat |P-value |

|Intercept |17.08333333 |1.549716223 |11.02352359 |1.99243E-10 |

|Tamp |-5 |4.179273856 |-1.196380082 |0.244283119 |

Since the p-value = 0.2443 > 0.05, do not reject H0. There is no evidence of a linear relationship between the amount of tamping and the separation time. This result does not support the employee’s hypothesis.

(c)

There is no noticeable pattern in the plot.

(d) H0: There is no autocorrelation.

H1: There is positive autocorrelation.

PHStat output:

|Durbin-Watson Calculations | |

|Sum of Squared Difference of Residuals |238.4375 |

|Sum of Squared Residuals |138.3333333 |

|Durbin-Watson Statistic |1.723644578 |

dL = 1.27, dU = 1.45. Since D = 1.7236 > 1.45, do not reject H0. There is no evidence of autocorrelation.

13.34 (a)

[pic]

(b) b0 = –2.535, b1 = 0.060728

(c) For each increase of one degree Fahrenheit in the high temperature, expected sales are estimated to increase by 0.060728 thousand dollars, or $60.73.

(d) [pic] 2.5054 or $2505.40

(e) [pic]

(f) r2 = 0.94. So, 94% of the variation in sales per store can be explained by the variation in the daily high temperature.

(g)

[pic]

13.34 (h)

cont.

[pic]

(i) D = 1.64>1.42. There is no evidence of positive autocorrelation among the residuals.

(j) 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.

(k) b0 = –2.6281, b1 = 0.061713

For each increase of one degree Fahrenheit in the high temperature, expected sales are estimated to increase by 0.061713 thousand dollars, or $61.71.

[pic] 2.4941 or $2494.10

r2 = 0.929. 92.9% of the variation in sales per store can be explained by the variation in the daily high temperature.

[pic]

D = 1.24. The test of the Durbin-Watson statistic is inconclusive as to whether there is positive autocorrelation among the residuals.

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.

The results are similar to those in (a)-(j).

13.35 (a) [pic]

(b) With n = 18, df = 18 – 2 =16. [pic]

(c) Reject H0. There is evidence that the fitted linear regression model is useful.

(d) [pic],[pic],

[pic]

13.36 (a) [pic]

[pic]

[pic]

(b) [pic]

(c) Reject H0. There is evidence that the fitted linear regression model is useful.

(d) [pic] [pic]

(e) [pic] There is no correlation between X and Y.

[pic] There is correlation between X and Y.

d.f. = 18. Decision rule: Reject [pic] if [pic]>2.1009.

Test statistic: [pic].

Since [pic] is below the lower critical bound of –2.1009, reject [pic]. There is enough evidence to conclude that there is a significant correlation between X and Y.

13.37 (a) [pic]with 10 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(b) [pic]

13.38 (a) [pic] with 18 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(b) [pic]

13.39 (a) p-value is virtually 0 < 0.05. Reject H0. There is evidence that the fitted linear regression model is useful.

(b) [pic]

13.40 (a) [pic] with 28 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(b) [pic]

13.41 (a) [pic] with 23 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(b) [pic]

13.42 (a) p-value = 7.26497E-06 < 0.05. Reject H0. There is evidence that the fitted linear regression model is useful.

(b) [pic]

13.43 (a) For Proctor and Gamble, the estimated value of its stock will increase by 0.53% on average when the S & P 500 index increases by 1%.

For Ford Motor Company, the estimated value of its stock will increase by 0.77% on average when the S & P 500 index increases by 1%.

For IBM, the estimated value of its stock will increase by 0.94% on average when the S & P 500 index increases by 1%.

For Immunex Corp, the estimated value of its stock will increase by 1.34% on average when the S & P 500 index increases by 1%.

For LSI Logic, the estimated value of its stock will increase by 2.04% on average when the S & P 500 index increases by 1%.

(b) A stock is riskier than the market if the estimated absolute value of the beta exceeds one. This can be used to gauge the volatility of a stock in relative to how the market behaves in general.

13.44 (a) [pic]

(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.

13.45 (a) r = 0.5777.

(b) t = 3.0852, p-value = 0.0061 < 0.05. Reject H0. There is enough evidence 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.

13.46 (a) r = –0.4014.

(b) t = –1.8071, p-value = 0.0885 > 0.05. Do not reject H0. At 0.05 level of significance, there is no significant 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.

13.47 (a) r = 0.3409.

(b) t = 1.4506, p-value = 0.1662 > 0.05. Do not reject H0. At 0.05 level of significance, there is no significant 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 to have a higher talk time is not borne out by the data.

13.48 (a) r = 0.4838

(b) t = 2.5926, p-value = 0.0166 < 0.05. Reject H0. At 0.05 level of significance, there is significant 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.

13.49 (a) When X = 2, [pic]

[pic]

95% confidence interval: [pic]

[pic]

(b) 95% prediction interval: [pic]

[pic]

13.50 (a) When X = 4, [pic]

[pic]

95% confidence interval: [pic]

[pic]

(b) 95% prediction interval: [pic]

[pic]

(c) The intervals in this problem are wider because the value of X is farther from [pic].

13.51 (a) [pic]

(b) [pic]

(c) Part (b) provides an estimate for an individual response and Part (a) provides an estimate for an average predicted value.

13.52 (a) [pic]

(b) [pic]

(c) Part (b) provides an estimate for an individual response and Part (a) provides an estimate for an average predicted value.

13.53 (a) [pic]

(b) [pic]

(c) Part (b) provides an estimate for an individual response and Part (a) provides an estimate for an average predicted value.

13.54 (a) [pic]

(b) [pic]

(c) Part (b) provides an estimate for an individual response and Part (a) provides an estimate for an average predicted value.

13.55 (a) [pic]

(b) [pic]

(c) Part (b) provides an estimate for an individual response and Part (a) provides an estimate for an average predicted value.

13.56 (a) [pic]

(b) [pic]

(c) Part (b) provides an estimate for an individual response and Part (a) provides an estimate for an average predicted value.

13.57 The slope of the line b1 represents the estimated expected change in Y per unit change in X. It represents the estimated average amount that Y changes (either positively or negatively) for a particular unit change in X. The Y intercept b0 represents the estimated average value of Y when X equals 0.

13.58 The coefficient of determination measures the proportion of variation in Y that is explained by the independent variable X in the regression model.

13.59 The unexplained variation or error sum of squares (SSE) will be equal to zero only when the regression line fits the data perfectly and the coefficient of determination equals 1.

13.60 The explained variation or regression sum of squares (SSR) will be equal to zero only when there is no relationship between the Y and X variables, and the coefficient of determination equals 0.

13.61 Unless a residual analysis is undertaken, you will not know whether the model fit is appropriate for the data. In addition, residual analysis can be used to check whether the assumptions of regression have been seriously violated.

13.62 The assumptions of regression are normality of error, homoscedasticity, and independence of errors. 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 homoscedasticity assumption 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.

13.63 The Durbin-Watson statistic is a measure of the autocorrelation among the residuals. It measures the correlation among consecutive residuals.

13.64 If the data in a regression analysis has been collected over time, then the assumption of independence of errors needs to be evaluated using the Durbin-Watson statistic.

13.65 The confidence interval for the mean response estimates the average response for a given X value. The prediction interval estimates the value for a single item or individual.

13.66 (a)

[pic]

(b) b0 = 24.84, b1 = 0.14

(c) [pic], where X is the number of cases and [pic] is the estimated delivery time.

(d) For each additional case, the estimated delivery time increases by 0.14 minutes.

(e) [pic]

(f) No, 500 cases is outside the relevant range of the data used to fit the regression equation.

(g) r2 = 0.972. So, 97.2% of the variation in delivery time can be explained by the variation in the number of cases.

(h) Since b1 is positive, [pic]

(i) [pic]

(j) 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.

(k) [pic] with 18 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(l) [pic]

(m) [pic]

(n) [pic]

13.67 (a)

[pic]

(b) b0 = –63.02, b1 = 0.189

(c) [pic], where X is the number of incoming calls and [pic] is the estimated number of trade executions.

(d) For each additional incoming call, the estimated number of trade executions increases by 0.189 minutes. – 63.02 is the portion of the estimated delivery time that is not affected by the number of incoming calls.

(e) [pic]

(f) No, 5000 incoming calls is outside the relevant range of the data used to fit the regression equation.

(g) r2 = 0.630. So, 63.0% of the variation in trade executions can be explained by the variation in the number of incoming calls.

(h) Since b1 is positive, [pic]

(i) [pic]

(j) 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.

(k) D = 1.96

(l) D = 1.96>1.52. There is no evidence of positive autocorrelation. The model appears to be adequate.

(m) [pic] with 33 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(n) [pic]

(o) [pic]

(p) [pic]

13.68 (a)

[pic]

b0 = –44.172, b1 = 1.78171

(b) For each additional dollar in assessed value, the estimated selling price increases by $1.78.

(c) [pic] or $80,458

(d) [pic]

(e) r2 = 0.926. 92.6% of the variation in selling price can be explained by the variation in the assessed value.

(f) Since b1 is positive, [pic]

(g) Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the assessed value, there is no pattern. The model appears to be adequate.

(h) [pic] with 28 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(i) [pic]

(j) [pic]

(k) [pic]

13.69 (a)

[pic]

b0 = 51.915, b1 = 16.633

(b) For each additional 1000 square feet in heating area, the estimated assessed value increases by $16,633.

(c) [pic] or $81,024

(d) [pic]

(e) r2 = 0.659. 65.9% of the variation in assessed value can be explained by the variation in heating area.

(f) Since b1 is positive, [pic]

(g) Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the heating area, there is no pattern. The model appears to be adequate.

(h) [pic] with 13 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(i) [pic]

(j) [pic]

(k) [pic]

(l) b0 = 52.805, b1 = 15.849

For each additional 1000 square feet in heating area, the estimated assessed value increases by $15,849.

[pic] or $80,541

[pic]

r2 = 0.689. 68.9% of the variation in assessed value can be explained by the variation in heating area.

Since b1 is positive, [pic]

Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the heating area, there is no pattern. The model appears to be adequate.

13.69 (l) [pic] with 13 degrees of freedom for [pic]. Reject H0. There

cont. is evidence that the fitted linear regression model is useful.

[pic]

[pic]

[pic]

13.70 (a)

[pic]

b0 = 0.30, b1 = 0.00487

(b) For each additional point on the GMAT score, the estimated GPI increases by 0.00487.

(c) [pic]5

(d) [pic]

(e) r2 = 0.7978. 79.78% of the variation in the GPI can be explained by the

variation in the GMAT score.

(f) Since b1 is positive, [pic]

(g) Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the GMAT score, there is no pattern. The model appears to be adequate.

(h) [pic] with 18 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(i) [pic]

(j) [pic]

(k) [pic]

(l) b0 = 0.258, b1 = 0.00494

For each additional point on the GMAT score, the estimated GPI increases by 0.00494.

[pic]

[pic]

r2 = 0.820. 82.0% of the variation in the GPI can be explained by the variation in the GMAT score.

13.70 (l) Since b1 is positive, [pic]

cont. Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the GMAT score, there is no pattern. The model appears to be adequate.

[pic] with 18 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

[pic]

[pic]

[pic]

13.71 (a)

[pic]

(b) b0 = 0.4024, b1 = 0.012607

(c) For each additional invoice processed, the estimated average completion time increases by 0.012607 hours.

(d) [pic]

(e) [pic]

(f) r2 = 0.892. 89.2% of the variation in completion time can be explained by the variation in the number of invoices processed.

(g) Since b1 is positive, [pic]

(i) Based on a visual inspection of the graphs of the distribution of residuals and the residuals versus the number of invoices, there is no pattern. The model appears to be adequate.

(j) D = 1.78

(k) D = 1.78>1.49. There is no evidence of positive autocorrelation. The

model appears to be adequate.

(l) [pic] with 28 degrees of freedom for [pic]. Reject H0. There is evidence that the fitted linear regression model is useful.

(m) [pic]

(n) [pic]

13.72 (a)

[pic]

There is not any clear relationship between atmospheric temperature and O-ring damage from the scatter plot.

(b),(f)

[pic]

(c) In (b), there are 16 observations with an O-ring damage index of 0 for a variety of temperature. If one concentrates on these observations with no O-ring damage, there is obviously no relationship between O-ring damage index and temperature. If all observations are used, the observations with no O-ring damage will bias the estimated relationship. If the intention is to investigate the relationship between the degrees of O-ring damage to atmospheric temperature, it makes sense to focus only on the flight in which there was O-ring damage.

(d) Prediction should not be made for an atmospheric temperature of 31 0F because it is outside the range of the temperature variable in the data. Such prediction will involve extrapolation, which assumes that any relationship between two variables will continue to hold outside the domain of the temperature variable.

(e) [pic]

(g) A nonlinear model is more appropriate for these data.

13.72 (h)

cont.

[pic]

The string of negative residuals and positive residuals that lie on a straight line with a positive slope in the lower-right corner of the plot is a strong indication that a nonlinear model should be used if all 23 observations are to be used in the fit.

13.73 (a)

[pic]

If the outlier () in the upper right corner of the scatter diagram is removed, there is not an obvious linear relationship between page views and gross profits.

(b) [pic]

(c) Since all the companies in the data are internet companies and rely their business on online customers, it is not meaningful to interpret the estimated intercept when there is no online visitor at all. The estimated slope coefficient [pic] = 0.154 means that for each increase in one thousand additional monthly visitors, the average gross product of a company is estimated to increase by 0.154 million dollars.

13.73 (d) [pic]. In the long-run, 95% of all the confidence intervals that are

cont. constructed for the slope parameter will contain the true value of the slope parameter. Since the interval does not contain 0, we are 95% confidence that there is a significant linear relationship between page views and gross profits.

(e) r2 = 0.6183. 61.83% of the total variation in gross profits can be explained by the variation in the number of monthly visitors.

(f) SYX = 13.4245. The standard error of the estimate measures the average squared distance between the values of the dependent variable and its fit on the least squares regression line.

(g) The outliers are , Cheap Tickets, and .

(h) (a)

[pic]

With removed from the data, there is no obvious relationship between page views and gross profits.

(b) [pic]

(c) Since all the companies in the data are internet companies and rely their business on online customers, it is not meaningful to interpret the estimated intercept when there is no online visitor at all. The estimated slope coefficient [pic] = -0.0637 means that for each increase in one thousand additional monthly visitors, the average gross product of a company is estimated to decrease by 0.0637 million dollars.

(d) [pic]. In the long-run, 95% of all the confidence intervals that are constructed for the slope parameter will contain the true value of the slope parameter. Since the interval contains 0, we cannot conclude that there is significant linear relationship between page views and gross profits with 95% level of confidence.

(e) r2 = 0.2259. 22.59% of the total variation in gross profits can be explained by the variation in the number of monthly visitors.

(f) SYX = 6.262. The standard error of the estimate measures the average squared distance between the values of the dependent variable and its fit on the least squares regression line.

13.73 (h) (g) The outliers are Cheap Tickets, and .

cont. The exclusion of changes the estimated slope coefficient from positive to negative. It is extremely influential on the least squares regression estimates. It is called influential point in regression analysis.

13.74 (a)

[pic]

(b) [pic]

(c) For each increase in one additional cm in circumference, the estimated average weight of a pumpkin will increase by 82.4717 grams.

(d) [pic] grams.

(e) There appears to be a positive relationship between weight and circumference of a pumpkin. It is a good idea for the farmer to sell pumpkin by circumference instead of weight for circumference is a good predictor of weight and it is much easier to measure the circumference of a pumpkin than its weight.

(f) r2 = 0.9373. 93.73% of the variation in pumpkin weight can be explained by the variation in circumference.

(g) SYX = 277.7495.

(h)

[pic]

There appears to be a nonlinear relationship between circumference and weight.

13.74 (i) p-value is virtually 0. Reject H0. There is sufficient evidence to conclude that there

cont. is a linear relationship between the circumference and the weight of a pumpkin.

(j) [pic]

(k) [pic]

(l) [pic]

13.75 (a)

[pic]

(b) [pic]

(c) b0 = 167.5399. For a team that has an E.R.A. of 0, the estimated average number of wins is 167.5399. For each additional unit increase in team E.R.A., the estimated average number of wins decreases by 19.6253.

(d) [pic]

(e) SYX = 8.5169.

(f) r2 = 0.5860. So, 58.60% of the variation in number of wins can be explained by the variation in the team E.R.A..

(g) Since b1 is negative, [pic]

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[pic]

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