Exam 14: Introduction to Multiple Regression

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 determine whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables?

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 is the value of the test statistic when testing whether instructional spending per pupil has any effect on percentage of students passing the proficiency test?

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 value of the adjusted coefficient of multiple determination, r²adj, is ______.

Consider a regression in which b₂ = - 1.5 and the standard error of this coefficient equals 0.3. To determine whether X₂ is a significant explanatory variable, you would compute an observed t-value of - 5.0.

$\text { TABLE 14-17 }$ 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: Logistic Regression Table 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 should be the decision ('reject' or 'do not reject') on the null hypothesis when testing whether LawnSize 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, 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 negative attitude toward outdoor recreation, and one teenager in the household?

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 your decision and conclusion for the test H₀ : ?₂ = 0 versus H₁ : ?₂ < 0 at the ? = 0.01 level of significance using Model 1?

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: $\text {SUMMARY OUTPUT}$ 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 14-17, what is the estimated probability that 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 will purchase a lawn service?

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 alternative hypothesis H1 : At least one of þj × 0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is affected by all of the explanatory variables.

Multiple regression is the process of using several independent variables to predict a number of dependent variables.

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 should be the decision ('reject' or 'do not reject') on the null hypothesis when testing whether Teenager makes a significant contribution to the model in the presence of the other independent variables at a 0.05 level of significance?

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 : ?₁ = ?₂ = ?₃ = 0 implies that percentage of students passing the proficiency test is not affected by any of the explanatory variables.

To explain personal consumption (CONS) measured in dollars, data is collected for INC: $\quad$ $\quad$ $\quad$ personal income in dollars CRDTLIM: $\quad$ $1 plus the credit limit in dollars available to the individual APR: $\quad$ $\quad$ $\quad$ average annualized percentage interest rate for borrowing for the individual ADVT: $\quad$ $\quad$ per person advertising expenditure in dollars by manufacturers in the city where the individual lives SEX: $\quad$ $\quad$ $\quad$ gender of the individual; 1 if female, 0 if male A regression analysis was performed with CONS as the dependent variable and ln(CRDTLIM), ln(APR), ln(ADVT), and SEX as the independent variables. The estimated model was Y^{^} = 2.28 - 0.29 ln(CRDTLIM) + 5.77 ln(APR) + 2.35 ln(ADVT) + 0.39 SEX What is the correct interpretation for the estimated coefficient for SEX?

TABLE 14-3 An economist is interested to see how consumption for an economy (in $ billions) is influenced by gross domestic product ($ billions) and aggregate price (consumer price index). The Microsoft Excel output of this regression is partially reproduced below. $\text {SUMMARY OUTPUT}$ Regression Statistics Multiple R 0.991 R Square 0.982 Adjusted R Square 0.976 Standard Error 0.299 Observations 10 ANOVA d f SS MS F Significance F Regression 2 33.4163 16.7082 186.325 0.0001 Residual 7 0.6277 0.0897 Total 9 34.0440 Coefficients Standard Error t Stat p -value Intercept -0.0861 0.5674 -0.152 0.8837 GDP 0.7654 0.0574 13.340 0.0001 Price -0.0006 0.0028 -0.219 0.8330 -Referring to Table 14-3, to test for the significance of the coefficient on aggregate price index, the value of the relevant t-statistic 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, to test the significance of the multiple regression model, the value of the test statistic 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 alternative hypothesis H1 : At least one of þj × 0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is affected by at least one of the explanatory variables.

A multiple regression is called "multiple" because it has several data points.

TABLE 14-13 As a project for his business statistics class, a student examined the factors that determined parking meter rates throughout the campus area. Data were collected for the price per hour of parking, blocks to the quadrangle, and one of the three jurisdictions: on campus, in downtown and off campus, or outside of downtown and off campus. The population regression model hypothesized is $Y_{i}=\alpha+\beta_{1} X_{1 i}+\beta_{2} X_{2 i}+\beta_{3} X_{3 i}+\varepsilon$ where Y is the meter price; ^X1 ^{is the number of blocks to the quad; X}2 ^{is a dummy variable that takes the value 1 if the meter} is located in downtown and off campus and the value 0 otherwise; ^X3 ^{is a dummy variable that takes the value 1 if the meter} is located outside of downtown and off campus, and the value 0 otherwise. The following Excel results are obtained. Regression Statistics Multiple R 0.9659 R Square 0.9331 Adjusted R Square 0.9294 Standard Error 0.0327 Observations 58 ANOVA d f SS M S F Significance F Regression 3 0.8094 0.2698 251.1995 1.0964-31 Residual 54 0.0580 0.0010 Total 57 0.8675 Coefficients Standard Error t Stat p-value Intercept 0.5118 0.0136 37.4675 2.4904 -0.0045 0.0034 -1.3276 0.1898 -0.2392 0.0123 -19.3942 5.3581-26 -0.0002 0.0123 -0.0214 0.9829 -Referring to Table 14-13, what is the correct interpretation for the estimated coefficient for X₂?