Exam 14: Introduction to Multiple Regression

TABLE 14-11 A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on average total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise). The Minitab output is given below: Logistic Regression Table Odds 95: CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -27.118 6.696 -4.05 0.000 SAT 0.015 0.004666 3.17 0.002 1.01 1.01 1.02 Toefl550 -0.390 0.9538 -0.41 0.682 0.68 0.10 4.39 Room/Brd 2.078 0.5076 4.09 0.000 7.99 2.95 21.60 Log-Likelihood = -21.883 Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000 Goodness-of-Fit Tests Method Chi-Square DF P Pearson 143.551 76 0.000 Deviance 43.767 76 0.999 Hosmer-Lemeshow 15.731 8 0.046 -Referring to Table 14-11, what is the p-value of the test statistic when testing whether the model is a good-fitting model?

When an explanatory variable is dropped from a multiple regression model, the adjusted r² can increase.

TABLE 14-11 A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on average total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise). The Minitab output is given below: Logistic Regression Table Odds 95: CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -27.118 6.696 -4.05 0.000 SAT 0.015 0.004666 3.17 0.002 1.01 1.01 1.02 Toefl550 -0.390 0.9538 -0.41 0.682 0.68 0.10 4.39 Room/Brd 2.078 0.5076 4.09 0.000 7.99 2.95 21.60 Log-Likelihood = -21.883 Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000 Goodness-of-Fit Tests Method Chi-Square DF P Pearson 143.551 76 0.000 Deviance 43.767 76 0.999 Hosmer-Lemeshow 15.731 8 0.046 -Referring to Table 14-11, what should be the decision ('reject' or 'do not reject') on the null hypothesis when testing whether SAT makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

TABLE 14-17 The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Attitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age). The Minitab output is given below: Odds 95 \% CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -70.49 47.22 -1.49 0.135 Income 0.2868 0.1523 1.88 0.060 1.33 0.99 1.80 Lawn Size 1.0647 0.7472 1.42 0.154 2.90 0.67 12.54 Attitude -12.744 9.455 -1.35 0.178 0.00 0.00 326.06 Teenager -0.200 1.061 -0.19 0.850 0.82 0.10 6.56 Age 1.0792 0.8783 1.23 0.219 2.94 0.53 16.45 Log-Likelihood = -4.890 Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000 Goodness-of-Fit Tests Method Chi-Square DF P Pearson 9.313 24 0.997 Deviance 9.780 24 0.995 Hosmer-Lemeshow 0.571 8 1.000 -Referring to Table 14-17, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance.

TABLE 14-17 The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Attitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age). The Minitab output is given below: Odds 95 \% CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -70.49 47.22 -1.49 0.135 Income 0.2868 0.1523 1.88 0.060 1.33 0.99 1.80 Lawn Size 1.0647 0.7472 1.42 0.154 2.90 0.67 12.54 Attitude -12.744 9.455 -1.35 0.178 0.00 0.00 326.06 Teenager -0.200 1.061 -0.19 0.850 0.82 0.10 6.56 Age 1.0792 0.8783 1.23 0.219 2.94 0.53 16.45 Log-Likelihood = -4.890 Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000 Goodness-of-Fit Tests Method Chi-Square DF P Pearson 9.313 24 0.997 Deviance 9.780 24 0.995 Hosmer-Lemeshow 0.571 8 1.000 -Referring to Table 14-17, what is the estimated odds ratio for a 48-year-old home owner with a family income of $100,000, a lawn size of 5,000 square feet, a positive attitude toward outdoor recreation, and two teenagers in the household?

TABLE 14-4 A real estate builder wishes to determine how house size (House) is influenced by family income (Income), family size (Size), and education of the head of household (School). House size is measured in hundreds of square feet, income is measured in thousands of dollars, and education is in years. The builder randomly selected 50 families and ran the multiple regression. Microsoft Excel output is provided below: Regression Stuistics Multiple R 0.865 R Square 0.748 Adjusted R Square 0.726 Standard Error 5.195 Observations 50 ANOVA d f S S M S F Significance F Regression 3605.7736 1201.9245 0.0000 Residual 1214.2264 26.3962 Total 49 4820.0000 Coefficients Standard Error t Stat p -value Intercept -1.6335 5.8078 -0.281 0.7798 Income 0.4485 0.1137 3.9545 0.0003 Size 4.2615 0.8062 5.286 0.0001 School -0.6517 0.4319 -1.509 0.1383 -Referring to Table 14-4, which of the following values for the level of significance is the smallest for which the regression model as a whole is significant?

TABLE 14-5 A microeconomist wants to determine how corporate sales are influenced by capital and wage spending by companies. She proceeds to randomly select 26 large corporations and record information in millions of dollars. The Microsoft Excel output below shows results of this multiple regression. Regression Statistics Multiple R 0.830 R Square 0.689 Adjusted R Square 0.662 Standard Error 17501.643 Observations 26 ANOVA d f S S M S F Significance F Regression 2 15579777040 7789888520 25.432 0.0001 Residual 23 7045072780 306307512 Total 25 22624849820 Coefficients Standard Error t Stat p-value Intercept 15800.0000 6038.2999 2.617 0.0154 C apital 0.1245 0.2045 0.609 0.5485 W ages 7.0762 1.4729 4.804 0.0001 -Referring to Table 14-5, what are the predicted sales (in millions of dollars) for a company spending $500 million on capital and $200 million on wages?

TABLE 14-17 The marketing manager for a nationally franchised lawn service company would like to study the characteristics that differentiate home owners who do and do not have a lawn service. A random sample of 30 home owners located in a suburban area near a large city was selected; 15 did not have a lawn service (code 0) and 15 had a lawn service (code 1). Additional information available concerning these 30 home owners includes family income (Income, in thousands of dollars), lawn size (Lawn Size, in thousands of square feet), attitude toward outdoor recreational activities (Attitude 0 = unfavorable, 1 = favorable), number of teenagers in the household (Teenager), and age of the head of the household (Age). The Minitab output is given below: Odds 95 \% CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -70.49 47.22 -1.49 0.135 Income 0.2868 0.1523 1.88 0.060 1.33 0.99 1.80 Lawn Size 1.0647 0.7472 1.42 0.154 2.90 0.67 12.54 Attitude -12.744 9.455 -1.35 0.178 0.00 0.00 326.06 Teenager -0.200 1.061 -0.19 0.850 0.82 0.10 6.56 Age 1.0792 0.8783 1.23 0.219 2.94 0.53 16.45 Log-Likelihood = -4.890 Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000 Goodness-of-Fit Tests Method Chi-Square DF P Pearson 9.313 24 0.997 Deviance 9.780 24 0.995 Hosmer-Lemeshow 0.571 8 1.000 -Referring to Table 14-15, the predicted mileage for a 200 horsepower, 4-cylinder car is_______

TABLE 14-7 The department head of the accounting department wanted to see if she could predict the GPA of students using the number of course units (credits) and total SAT scores of each. She takes a sample of students and generates the following Microsoft Excel output: $\text {SUMMARY OUTPUT}$ Regression Statistics Multiple R 0.916 R Square 0.839 Adjusted R Square 0.732 Standard Error 0.24685 Observations 6 ANOVA d f SS M S F Significance F Regression 2 0.95219 0.47610 7.813 0.0646 Residual 3 0.18281 0.06094 Total 5 1.13500 Coefficients Standard Error t Stat p -value Intercept 4.593897 1.13374542 4.052 0.0271 Units -0.247270 0.06268485 -3.945 0.0290 SAT Total 0.001443 0.00101241 1.425 0.2494 -Referring to Table 14-7, the department head wants to test H₀ : ?₁ = ?₂ = 0. The critical value of the F test for a level of significance of 0.05 is_____ .

TABLE 14-16 The superintendent of a school district wanted to predict the percentage of students passing a sixth-grade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state. Following is the multiple regression output with Y = % Passing as the dependent variable, X₁ = % Attendance, X₂ = Salaries and X₃ = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 Adjusted R Square 0.6029 Standard Error 10.4570 Observations 47 ANOVA d f SS MS F Significance F Regression 3 7965.08 2655.03 24.2802 2.3853-09 Residual 43 4702.02 109.35 Total 46 12667.11 Coeffs Stnd Err t Stat p -value Lower 95\% Upper 95\% Intercept -753.4225 101.1149 -7.4511 2.88-09 -957.3401 -549.5050 \% Attend 8.5014 1.0771 7.8929 6.73-10 6.3292 10.6735 Salary 6.85-07 0.0006 0.0011 0.9991 -0.0013 0.0013 Spending 0.0060 0.0046 1.2879 0.2047 -0.0034 0.0153 -Referring to Table 14-16, we can conclude that average teacher salary has no impact on average percentage of students passing the proficiency test at a 5% level of significance using the 95% confidence interval estimate for ?₂.

TABLE 14-16 The superintendent of a school district wanted to predict the percentage of students passing a sixth-grade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state. Following is the multiple regression output with Y = % Passing as the dependent variable, X₁ = % Attendance, X₂ = Salaries and X₃ = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 Adjusted R Square 0.6029 Standard Error 10.4570 Observations 47 ANOVA d f SS MS F Significance F Regression 3 7965.08 2655.03 24.2802 2.3853-09 Residual 43 4702.02 109.35 Total 46 12667.11 Coeffs Stnd Err t Stat p -value Lower 95\% Upper 95\% Intercept -753.4225 101.1149 -7.4511 2.88-09 -957.3401 -549.5050 \% Attend 8.5014 1.0771 7.8929 6.73-10 6.3292 10.6735 Salary 6.85-07 0.0006 0.0011 0.9991 -0.0013 0.0013 Spending 0.0060 0.0046 1.2879 0.2047 -0.0034 0.0153 -Referring to Table 14-16, what are the numerator and denominator degrees of freedom, respectively, for the test statistic to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables?

TABLE 14-15 An automotive engineer would like to be able to predict automobile mileages. She believes that the two most important characteristics that affect mileage are horsepower and the number of cylinders (4 or 6) of a car. She believes that the appropriate model is Y = 40 - 0.05X₁ + 20X₂ - 0.1X₁X₂ where X₁ = horsepower X₂ = 1 if 4 cylinders, 0 if 6 cylinders Y = mileage. -Referring to Table 14-15, the fitted model for predicting mileages for 4-cylinder cars is .

TABLE 14-16 The superintendent of a school district wanted to predict the percentage of students passing a sixth-grade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state. Following is the multiple regression output with Y = % Passing as the dependent variable, X₁ = % Attendance, X₂ = Salaries and X₃ = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 Adjusted R Square 0.6029 Standard Error 10.4570 Observations 47 ANOVA d f SS MS F Significance F Regression 3 7965.08 2655.03 24.2802 2.3853-09 Residual 43 4702.02 109.35 Total 46 12667.11 Coeffs Stnd Err t Stat p -value Lower 95\% Upper 95\% Intercept -753.4225 101.1149 -7.4511 2.88-09 -957.3401 -549.5050 \% Attend 8.5014 1.0771 7.8929 6.73-10 6.3292 10.6735 Salary 6.85-07 0.0006 0.0011 0.9991 -0.0013 0.0013 Spending 0.0060 0.0046 1.2879 0.2047 -0.0034 0.0153 -Referring to Table 14-16, the null hypothesis H0 : þ1 = þ2 = þ3 = 0 implies that percentage of students passing the proficiency test is not related to any of the explanatory variables.

TABLE 14-6 One of the most common questions of prospective house buyers pertains to the average cost of heating in dollars (Y). To provide its customers with information on that matter, a large real estate firm used the following 4 variables to predict heating costs: the daily minimum outside temperature in degrees of Fahrenheit (X₁), the amount of insulation in inches (X₂), the number of windows in the house (X₃), and the age of the furnace in years (X₄). Given below are the EXCEL outputs of two regression models. Model 1 Regression Statistics R Square 0.8080 AdjustedR S quare 0.7568 Observations 20 ANOVA df SS MS F Signuficance F Regression 4 169503.4241 42375.86 15.7874 2.96869E-05 Residual 15 40262.3259 2684.155 Total 19 209765.75 Standard Lower Upper Coefficients Error t Stat p -value 90.0\% 90.0\% Intercept 421.4277 77.8614 5.4125 7.2-05 284.9327 557.9227 X 1 (Temperature) -4.5098 0.8129 -5.5476 5.58-05 -5.9349 -3.0847 X 2 (Insulation) -14.9029 5.0508 -2.9505 0.0099 -23.7573 -6.0485 X 3 (Windows) 0.2151 4.8675 0.0442 0.9653 -8.3181 8.7484 X 4 (Furnace Age) 6.3780 4.1026 1.5546 0.1408 -0.8140 13.5702 Model 2 Regression Statistics R Square 0.7768 Adjusted R Square 0.7506 Observations 20 ANOVA d f SS MS SS Significance F Regression 2 162958.2277 81479.11 29.5923 2.9036-06 Residual 17 46807.5222 2753.384 Total 19 209765.75 Standard Lower Upper Coefficients Error t Stat p -value 95\% 95\% Intercept 489.3227 43.982611.1253 3.17-09 396.5273 582.1180 X 1 (Temperature) -5.1103 0.6951-7.3515 1.13-06 -6.5769 -3.6437 X 2 (Insulation) -14.7195 4.8864-3.0123 0.0078 -25.0290 -4.4099 -Referring to Table 14-6, what is the value of the partial F test statistic for H₀ :?₃ = ?₄ = 0 versus H₁ : At least one ?j ?0, j = 3,4?

TABLE 14-7 The department head of the accounting department wanted to see if she could predict the GPA of students using the number of course units (credits) and total SAT scores of each. She takes a sample of students and generates the following Microsoft Excel output: $\text {SUMMARY OUTPUT}$ Regression Statistics Multiple R 0.916 R Square 0.839 Adjusted R Square 0.732 Standard Error 0.24685 Observations 6 ANOVA d f SS M S F Significance F Regression 2 0.95219 0.47610 7.813 0.0646 Residual 3 0.18281 0.06094 Total 5 1.13500 Coefficients Standard Error t Stat p -value Intercept 4.593897 1.13374542 4.052 0.0271 Units -0.247270 0.06268485 -3.945 0.0290 SAT Total 0.001443 0.00101241 1.425 0.2494 -Referring to Table 14-7, the predicted GPA for a student carrying 15 course units and who has a total SAT of 1,100 is______ .

TABLE 14-10 You worked as an intern at We Always Win Car Insurance Company last summer. You notice that individual car insurance premiums depend very much on the age of the individual, the number of traffic tickets received by the individual, and the population density of the city in which the individual lives. You performed a regression analysis in EXCEL and obtained the following information: Regression Analysis Regression Statistics Multiple R 0.63 R Square 0.40 Adjusted R Square 0.23 Standard Error 50.00 Observations 15.00 ANOVA d f SS MS F Significance F Regression 3 5994.24 2.40 0.12 Residual 11 27496.82 Total 45479.54 oefficients Standard Error t Stat p-value Lower 99.0\% Upper 99.0 \% Intercept 123.80 48.71 2.54 0.03 -27.47 275.07 AGE -0.82 0.87 -0.95 0.36 -3.51 1.87 TICKETS 21.25 10.66 1.99 0.07 -11.86 54.37 DENSITY -3.14 6.46 -0.49 0.64 -23.19 16.91 -Referring to Table 14-10, the adjusted r² is ____ .

TABLE 14-16 The superintendent of a school district wanted to predict the percentage of students passing a sixth-grade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state. Following is the multiple regression output with Y = % Passing as the dependent variable, X₁ = % Attendance, X₂ = Salaries and X₃ = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 Adjusted R Square 0.6029 Standard Error 10.4570 Observations 47 ANOVA d f SS MS F Significance F Regression 3 7965.08 2655.03 24.2802 2.3853-09 Residual 43 4702.02 109.35 Total 46 12667.11 Coeffs Stnd Err t Stat p -value Lower 95\% Upper 95\% Intercept -753.4225 101.1149 -7.4511 2.88-09 -957.3401 -549.5050 \% Attend 8.5014 1.0771 7.8929 6.73-10 6.3292 10.6735 Salary 6.85-07 0.0006 0.0011 0.9991 -0.0013 0.0013 Spending 0.0060 0.0046 1.2879 0.2047 -0.0034 0.0153 -Referring to Table 14-16, which of the following is the correct alternative hypothesis to test whether instructional spending per pupil has any effect on percentage of students passing the proficiency test?

TABLE 14-10 You worked as an intern at We Always Win Car Insurance Company last summer. You notice that individual car insurance premiums depend very much on the age of the individual, the number of traffic tickets received by the individual, and the population density of the city in which the individual lives. You performed a regression analysis in EXCEL and obtained the following information: Regression Analysis Regression Statistics Multiple R 0.63 R Square 0.40 Adjusted R Square 0.23 Standard Error 50.00 Observations 15.00 ANOVA d f SS MS F Significance F Regression 3 5994.24 2.40 0.12 Residual 11 27496.82 Total 45479.54 oefficients Standard Error t Stat p-value Lower 99.0\% Upper 99.0 \% Intercept 123.80 48.71 2.54 0.03 -27.47 275.07 AGE -0.82 0.87 -0.95 0.36 -3.51 1.87 TICKETS 21.25 10.66 1.99 0.07 -11.86 54.37 DENSITY -3.14 6.46 -0.49 0.64 -23.19 16.91 -Referring to Table 14-10, the total degrees of freedom that are missing in the ANOVA table should be_____ .

TABLE 14-16 The superintendent of a school district wanted to predict the percentage of students passing a sixth-grade proficiency test. She obtained the data on percentage of students passing the proficiency test (% Passing), daily average of the percentage of students attending class (% Attendance), average teacher salary in dollars (Salaries), and instructional spending per pupil in dollars (Spending) of 47 schools in the state. Following is the multiple regression output with Y = % Passing as the dependent variable, X₁ = % Attendance, X₂ = Salaries and X₃ = Spending: Regression Statistics Multiple R 0.7930 R Square 0.6288 Adjusted R Square 0.6029 Standard Error 10.4570 Observations 47 ANOVA d f SS MS F Significance F Regression 3 7965.08 2655.03 24.2802 2.3853-09 Residual 43 4702.02 109.35 Total 46 12667.11 Coeffs Stnd Err t Stat p -value Lower 95\% Upper 95\% Intercept -753.4225 101.1149 -7.4511 2.88-09 -957.3401 -549.5050 \% Attend 8.5014 1.0771 7.8929 6.73-10 6.3292 10.6735 Salary 6.85-07 0.0006 0.0011 0.9991 -0.0013 0.0013 Spending 0.0060 0.0046 1.2879 0.2047 -0.0034 0.0153 -Referring to Table 14-16, which of the following is the correct null hypothesis to determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables?

TABLE 14-11 A logistic regression model was estimated in order to predict the probability that a randomly chosen university or college would be a private university using information on average total Scholastic Aptitude Test score (SAT) at the university or college, the room and board expense measured in thousands of dollars (Room/Brd), and whether the TOEFL criterion is at least 550 (Toefl550 = 1 if yes, 0 otherwise.) The dependent variable, Y, is school type (Type = 1 if private and 0 otherwise). The Minitab output is given below: Logistic Regression Table Odds 95: CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -27.118 6.696 -4.05 0.000 SAT 0.015 0.004666 3.17 0.002 1.01 1.01 1.02 Toefl550 -0.390 0.9538 -0.41 0.682 0.68 0.10 4.39 Room/Brd 2.078 0.5076 4.09 0.000 7.99 2.95 21.60 Log-Likelihood = -21.883 Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000 Goodness-of-Fit Tests Method Chi-Square DF P Pearson 143.551 76 0.000 Deviance 43.767 76 0.999 Hosmer-Lemeshow 15.731 8 0.046 -Referring to Table 14-11, there is not enough evidence to conclude that the model is not a good-fitting model at a 0.05 level of significance.