Chapter 17
17.1
a
b The standard error of estimate is = 40.24. It is an estimate of the standard deviation of the error variable.
c The coefficient of determination is = .2425; 24.25% of the variation in prices is explained by the model.
d The coefficient of determination adjusted for degrees of freedom is .2019. It differs from because it includes an adjustment for the number of independent variables.
e 0
At least one is not equal to zero
F = 5.97, p-value = .0013. There is enough evidence to conclude that the model is valid.
f = .700; for each addition thousand square feet the price on average increases by .700 thousand dollars provided that the other variables remain constant.
= .679; for each addition tree the price on average increases by .679 thousand dollars provided that the other variables remain constant.
= –.378; for each addition foot from the lake the price on average decreases by .378 thousand dollars provided that the other variables remain constant.
g 0
0
Lot size: t = 1.25, p-value = .2156
Trees: t = 2.96, p-value = .0045
Distance: t = –1.94, p-value = .0577
Only for the number of trees is there enough evidence to infer a linear relationship with price.
h
We predict that the lot in question will sell for between $35,500 and $172,240
i
Estimated average price lies between $39,290 and $90,300.
17.2
a
b The standard error of estimate is = 3.75. It is an estimate of the standard deviation of the error variable.
c The coefficient of determination is = .7629; 76.29% of the variation in final exam marks is explained by the model.
d 0
At least one is not equal to zero
F = 43.43, p-value = 0. There is enough evidence to conclude that the model is valid.
e= .194; for each addition mark on assignments the final exam mark on average increases by .194 provided that the other variable remains constant.
= 1.112; for each addition midterm mark the final exam mark on average increases by 1.112 provided that the other variable remains constant.
f0
0
t = .97, p-value = .3417. There is not enough evidence to infer that assignment marks and final exam marks are linearly related.
g0
0
t = 9.12, p-value = 0. There is sufficient evidence to infer that midterm marks and final exam marks are linearly related.
h
Pat’s final exam mark is predicted to lie between 23 and 39
i Pat’s predicted final grade: LCL = 12 + 14 + 23 = 49, UCL = 12 + 14 + 39 = 65
17.3a
b The standard error of estimate is = 40.13. It is an estimate of the standard deviation of the error variable.
c The coefficient of determination is = .8935; 89.35% of the variation in monthly sales of drywall is explained by the model.
d0
At least one is not equal to zero
F = 39.86, p-value = 0. There is enough evidence to conclude that the model is valid.
e= 4.76; for each addition building permit monthly sales on average increase by 4.76 hundred sheets provided that the other variables remain constant.
= 16.99; for each addition one point increase in mortgage rates monthly sales on average increase by 16.99 hundred sheets provided that the other variables remain constant.
= –10.53; for each one percentage point increase in the apartment vacancy rate monthly sales decrease on average by 10.53 hundred sheets provided that the other variables remain constant.
= 1.31; for each one percentage point increase in the office vacancy rate monthly sales increase on average by 1.31 hundred sheets provided that the other variables remain constant.
f0
0
Permits: t = 12.06, p-value = 0
Mortgage: t = 1.12, p-value = .2764
A Vacancy: t = –1.65, p-value = .1161
O Vacancy: t = .47, p-value = .6446
Only the number of building permits is linearly related to monthly sales.
g
Next month’s drywall sales are predicted to lie between 16,710 and 35,290 sheets.
17.4a
b = .666; for each additional minor league home run the number of major league home runs increases on average by .666 provided that the other variables remain constant.
= .136; for each additional year of age the number of major league home runs increases on average by .14 provided that the other variables remain constant.
= 1.18; for each additional year as a professional the number of major league home runs increases on average by 1.18 provided that the other variables remain constant.
c = 6.99 and = .3511; the model's fit is not very good.
d 0
At least one is not equal to zero
F = 22.01, p-value = 0. There is enough evidence to conclude that the model is valid.
e0
0
Minor league home runs: t = 7.64, p-value = 0
Age: t = .26, p-value = .7961
Years professional: t = 1.75, p-value = .0819
At the 5% significance level only the number of minor league home runs is linearly related to the number of major league home runs.
f
We predict that the player will hit between 9.86 (rounded to 10) and 38.76 (rounded to 39) home runs.
g
It is estimated that the average player will hit between 14.66 and 24.47 home runs.
17.5
a The regression equation is = 6.06 –.0078x + .603x–.0702x
b = 1.92, .7020, F = 36.12, p-value = 0. The model is valid and the fit is reasonably good.
c 0
0
Age: t = –.12, p-value = .9069
Years: t = 6.25, p-value = 0
Pay: t = –1.34, p-value = .1864
Only the number of years with the company is linearly related to severance pay
d.
95% prediction interval: Lower prediction limit = 5.64, upper prediction limit = 13.50. The offer of 5 weeks severance pay falls below the prediction interval and thus Bill is correct.
17.6
b The coefficient of determination is = .2882; 28.82% of the variation in university GPAs is explained by the model.
c 0
At least one is not equal to zero
F = 12.96, p-value = 0. There is enough evidence to conclude that the model is valid.
d 0
0
High school GPA: t = 6.06,, p-value = 0
SAT: t = .94, p-value = .3485
Activities: t = .72, p-value = .4720
At the 5% significance level only the high school GPA is linearly related to the university GPA
e
We predict that the student's GPA will fall between 4.45 and 12.00 (12 is the maximum).
f
The mean GPA is estimated to lie between 6.90 and 8.22.
17.7
a The regression equation is = 12.31 + .570x + 3.32x + .732x
b The coefficient of determination is = .1953; 19.53% of the variation in sales is explained by the model. The coefficient of determination adjusted for degrees of freedom is .0803. The model fits poorly.
c The standard error of estimate is = 2.59. It is an estimate of the standard deviation of the error variable
d 0
At least one βiis not equal to zero
F = 1.70, p-value = .1979. There is not enough evidence to conclude that the model is valid.
e0
0
Direct: t = .33, p-value = .7437
Newspaper: t = 2.16, p-value = .0427
Television: t = .37, p-value = .7123
Only expenditures on newspaper advertising is linearly related to sales.
f & g
f We predict that sales will fall between $12,270 and $24,150.
g We estimate that mean sales will fall between $15,700 and $20,730.
h The interval in part f predicts one week’s gross sales, whereas the interval in part h estimates the mean weekly gross sales.
17.8 a
b0
At least one βiis not equal to zero
F = 29.80, p-value = 0. There is enough evidence to conclude that the model is valid.
cb1= .0019; for each additional square foot the amount of garbage increases on average by .0019 pounds holding the other variables constant.
b2= 1.10; for each additional child in the home the amount of garbage increases on average by 1.10pounds holding the other variables constant.
b3= 1.04; for each additional adult at home during the day the amount of garbage increases on average by 1.04holding the other variables constant.
d0
0
House size : t = 3.21, p-value = .0014
Number of children: t = 7.84 p-value = 0
Number of adults at home: t = 4.48, p-value = 0
All three independent variable are linearly related to the amount of garbage.
17.9a
a = 35.68 + .247x+ .245x+ .133x
b 0
At least one is not equal to zero
F = 6.66, p-value = .0011. There is enough evidence to conclude that the model is valid.
c= .247; for each one percentage point increase in the proportion of teachers with mathematics degrees the test score increases on average by .247 provided the other variables are constant.
= .245; for each one year increase in mean age test score increases on average by .245 provided the other variables are constant .
= .135; for each one thousand dollar increase in salary test score increases on average by .135 provided the other variables are constant.
0
0
Proportion of teachers with at least one mathematics degree: t = 3.54, p-value = .0011
Age: t = 1.32, p-value = .1945
Income: t = .87, p-value = .3889.
The proportion of teachers with at least one mathematics degree is linearly related to test scores. The other two variables appear to be unrelated.
d
The school's test score is predicted to fall between 49.02 and 81.02.
17.10a
b 0
At least one is not equal to zero
F = 67.97, p-value = 0. There is enough evidence to conclude that the model is valid.
c= .451; for each one year increase in the mother's age the customer's age increases on average by .451 provided the other variables are constant (which may not be possible because of the multicollinearity).
= .411; for each one year increase in the father's age the customer's age increases on average by .411 provided the other variables are constant.
= .0166; for each one year increase in the grandmothers' mean age the customer's age increases on average by .0166 provided the other variables are constant.
= .0869; for each one year increase in the grandfathers' mean age the customer's age increases on average by .0869 provided the other variables are constant.
0
0
Mothers: t = 8.27, p-value = 0
Fathers: t = 8.26, p-value = 0
Grandmothers: t = .25, p-value .8028
Grandfathers: t = 1.32, p-value = .1890
The ages of mothers and fathers are linearly related to the ages of their children. The other two variables are not.
d
The man is predicted to live to an age between 65.54 and 77.31
e
The mean longevity is estimated to fall between 68.75 and 74.66.
17.11
Assessing the Model:
= 7.01 and = .7209; the model fits well.
Testing the validity of the model:
0
At least one is not equal to zero
F = 60.70, p-value = 0. There is enough evidence to conclude that the model is valid.
Drawing inferences about the independent variables:
0
0
Evaluations: t = .60, p-value = .5529
Articles: t = 8.08, p-value = 0.
The number of articles a professor publishes is linearly related to salary. Teaching evaluations are not.
17.12
a = –28.43 + .604x+ .374x
b = 7.07 and = .8072; the model fits well.
c= .604; for each one additional box, the amount of time to unload increases on average by .604 minutes provided the weight is constant.
= .374; for each additional hundred pounds the amount of time to unload increases on average by .374 minutes provided the number of boxes is constant.
0
0
Boxes: t = 10.85, p-value = 0
Weight: t = 4.42, p-value = .0001
Both variables are linearly related to time to unload.
de
d It is predicted that the truck will be unloaded in a time between 35.16 and 66.24 minutes.
e The mean time to unload the trucks is estimated to lie between 44.43 and 56.96 minutes.
17.13
b0
At least one is not equal to zero
F = 18.17, p-value = 0. There is enough evidence to conclude that the model is valid.
e 0
< 0 (for beliefs 1 and 4)
> 0 for beliefs 2 and 3)
Belief 1: t = –3.26, p-value = .0016/2 = .0008
Belief 2: t = 1.16, p-value = .2501/2 = .1251
Belief 3: t = .42, p-value = .6780/2 = .3390
Belief 4: t = –2.69, p-value = .0085/2 = .0043
There is enough evidence to support beliefs 1 and 4. There is no evidence to support beliefs 2 and 3.
17.14 a
The normality requirement is satisfied.
b
The variance of the error variable appears to be constant.
17.15
There does not appear to be a multicollinearity problem. The t–tests are valid.
17.16b
The normality requirement has not been violated.
c
The variance of the error variable appears to be constant.
d
The lack of multicollinearity means that the t–tests were valid.
17.17a
The error variable may be normally distributed.
b
The variance of the error variable is constant.
c
Multicollinearity is not a problem.
17.18a
Age and years as a professional are highly correlated. The correlations of the other combinations are small.
b The t–tests may not be valid.
17.19 a
The error variable is approximately normally distributed.
b
The variance of the error variable is constant.
c
The correlation between age and years is high indicating that multicollinearity is a problem.
17.20
The error variable is approximately normally distributed and the variance is constant.
17.21a
The error variable is approximately normal. However, the variance is not constant.
b
Multicollinearity is not a problem.
c There is one observation whose standardized residual exceeds 2.0 that should be checked
17.22
The error appears to be normally distributed.
There are indications that the error grows smaller as the predicted value increases. However, overall the requirement of constant variance may be valid.
17.23 a
The normality requirement is satisfied.
The variance of the error variable is constant.
c
The correlation between age and income is high. Multicollinearity is a problem.
17.24a
The normality requirement is satisfied.
b
The variance of the error variable is constant.
c
The correlations are large enough to cause problems with the t–tests.
17.25
The error variable appears to be normal. The error variable's variance appears to be constant. The required conditions are satisfied.
17.26
c The error variable appears to be normally distributed. The variance of the errors appears to be constant.
17.27a
The errors appear to be normally distributed. The variance of the errors is not constant.
b
There is a strong correlation between income and education. The t–tests of these two coefficients may be distorted.
17.28
The requirements are satisfied.
17.29 = 1.12, = 1.66. There is evidence of positive first–order autocorrelation.
17.30 = 1.16, = 1.59, 4 – = 2.41, 4 –= 2.84. The test is inconclusive.
17.31 = .95, = 1.89, 4 –= 2.11, 4 –= 3.05. There is evidence of first–order autocorrelation.
17.32 = 1.46, = 1.63. There is evidence of positive first–order autocorrelation.
17.33 = 1.41, = 1.64, 4 – = 2.36, 4 –= 2.59. The test is inconclusive
17.34 4 – = 4 – 1.73 = 2.27, 4 –= 4 – 1.19 = 2.81. There is no evidence of negative first–order autocorrelation.
17.35 a The regression equation is = 303.3 + 14.94 + 10.52
b
The graph indicates that autocorrelation exists.
c
= 1.63, = 1.72, 4 – = 2.37, 4 –= 2.28. There is evidence of autocorrelation.
d The model is + + + +
The regression equation is = 10.00 + 6.78 + 9.37+ 9.64t
e
There is no evidence of autocorrelation.
First model: = 348.7 and = .2825. Second model: = 208.1 and = .7471
The second model fits better.
17.36 a The regression equation is = 2260 + .423x
b
There appears to be a strong autocorrelation.
c
1.50, 1.59, 2.41, 2.50. There is evidence of first–order autocorrelation.
d The model is + + +
The regression equation is = 446.2 + 1.10x + 38.92t
e
There is no evidence of autocorrelation.
First model: = 709.7 and = .0146. Second model: = 413.7 and = .6718.
The second model fits better.
17.37
= 1.01, = 1.78. There is no evidence of positive first–order autocorrelation.
17.38
d = 2.2003; = 1.30, = 1.46, 4 –= 2.70, 4 –= 2.54. There is no evidence of first–order autocorrelation.
17.39 a= 898.0 + 11.33x
b
The normality requirement is satisfied. However, the constant variance requirement is not.
= 1.24, = 1.43, 4 –= 2.76, 4 –= 2.57. There is evidence of first–order autocorrelation.
c The problem is that the errors are not independent. We add a time variable (week number) to the model. Thus, the new model is y = + + +.
The regression equation is = 960.6 + 13.88x – 7.69t
= 1.15, = 1.54, 4 –= 2.85, 4 –= 2.46. There is no evidence of first–order autocorrelation.
d = 48.55 and = .7040
Snowfall: for each additional inch of snowfall tire sales increase on average by 13.88 (holding the time period constant).
t = 5.862, p-value = 0.
Week: weekly sales decrease on average by 7.687 (holding snowfall constant).
t = –4.579, p-value = .0002
17.40a
a
For each additional unit of fertilizer crop yield increases on average by .140 (holding the amount of water constant).
For each additional unit of water crop yield increases on average by .0313 (holding the fertilizer constant).
b0
t =1.72, p-value = .0974. There is not enough evidence to conclude that there is a linear relationship between crop yield and amount of fertilizer.
c0
t =4.64, p-value = .0001. There is enough evidence to conclude that there is a linear relationship between crop yield and amount of water.
d = 63.08 and = .4752; the model fits moderately well.
e
The errors appear to be normal, but the plot of residuals vs predicted aeems to indicate a problem.
f
We predict that the crop yield will fall between 69.2 and 349.3.
17.41
b
At least one is not equal to zero
F = 7.87, p-value = .0001. There is enough evidence to conclude that the model is valid.
c The regression equation for Exercise 16.12 is = 4040 + 44.97x. The addition of the new variables changes the coefficients of the regression line in Exercise 17.12.
17.42a
b 61.23 % of the variation in rents is explained by the independent variables.
c0
At least one is not equal to zero
F = 21.32, p-value = 0. There is enough evidence to conclude that the model is valid.
d0
0
Vacancy rate: t = –4.58, p-value = .0001
Unemployment rate: t = –4.73, p-value = .0001
Both vacancy and unemployment rates are linearly related to rents.
e
The error is approximately normally distributed with a constant variance.
f
= 1.28, = 1.57, 4 –= 2.72, 4 –= 2.43. There is no evidence of first–order autocorrelation.
g
The city's office rent is predicted to lie between $14.18 and $23.27.
Case 17.1
The model is valid (F = 7.75, p-value = 0) but the model does not fit well (R= .0151; only 1.51% of the variation in returns is explained by the model).
Interpreting the coefficients in this sample:
For each additional one–point increase in the SAT score, returns increase on average by .0051 provided the other variables remain constant.
The returns of mutual funds managed by MBAs are on average .674 larger than the returns of mutual funds managed by people without an MBA
For each additional one–year increase in age of the manager , returns decrease on average by .141 provided the other variables remain constant.
For each additional one–year increase in the manager’s job tenure, returns increase on average by .082 provided the other variables remain constant.
Testing the coefficients:
SAT: t = 3.96, p-value = .0001
MBA: t = 1.79, p-value = .0730
Age: t = –3.31, p-value = .0009
Tenure:t = .47, p-value = .6412
There is overwhelming evidence to infer that SAT scores of the undergraduate university and age of the manager are linearly related to returns. There is weak evidence that MBAs and non–MBAs have different mean returns. There is not enough evidence to conclude that job tenure is linearly related to returns.
Case 17.2
Analysis of Betas
The model is valid (F = 75.20, p-value = 0) with R= .1294; only 12.94% of the variation in betas is explained by the model.
Interpreting the coefficients in this sample:
For each additional one–point increase in the SAT score, betas increase on average by .00050 provided the other variables remain constant.
The betas of mutual funds managed by MBAs are on average .0366 larger than the betas of mutual funds managed by people without an MBA
For each additional one–year increase in age of the manager, betas increase on average by .0088 provided the other variables remain constant.
For each additional one–year increase in the manager’s job tenure, betas decrease on average by .0352 provided the other variables remain constant.
Testing the coefficients:
SAT: t = 14.55, p-value = 0
MBA: t = 3.60, p-value = .0003
Age: t = 7.66, p-value = 0
Tenure:t = –7.42, p-value = 0
There is overwhelming evidence to infer that all four independent variables are linearly related to mutual fund betas.
Analysis of MERs
The model is valid (F = 31.74, p-value = 0) with R= .0728; only 7.28% of the variation in MERs is explained by the model.
Interpreting the coefficients in this sample:
For each additional one–point increase in the SAT score, MERs decrease on average by .00055 provided the other variables remain constant.
The MERs of mutual funds managed by MBAs are on average .082 smaller than the MERs of mutual funds managed by people without an MBA
For each additional one–year increase in age of the manager, MERs increase on average by .0133 provided the other variables remain constant.
For each additional one–year increase in the manager’s job tenure, MERs increase on average by .0375 provided the other variables remain constant.
For each additional one–point increase in the log of the assets, MERs decrease on average by .209 provided the other variables remain constant.
Testing the coefficients:
SAT: t = –5.21, p-value = 0
MBA: t = –2.65, p-value = .0081
Age: t = 3.80, p-value = .0001
Tenure:t = 2.59, p-value = .0097
Log Assets: t = –9.13, p-value = 0
There is overwhelming evidence to infer that all five independent variables are linearly related to mutual fund MERs.
1