Question 1

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
$\begin{array}{lrrrrrrr} 
& & & && \text { Odds } & \text { 95: CI } \
\text { Predictor } & {\text { Coef }} & \text { SE Coef } & Z &{\text { P }} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } &-27.118&6 .696& -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$

Log-Likelihood = -21.883
Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000
Goodness-of-Fit Tests

$ \begin{array}{lrcr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046\end{array} $

-Referring to Table 14-11, what is the p-value of the test statistic when testing whether the model is a good-fitting model?

Accepted Answer

The answer of  TABLE 14-11 
 A logistic regression...

Question 2

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

Accepted Answer

A) True 
 B)False

Question 3

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
$\begin{array}{lrrrrrrr} 
& & & && \text { Odds } & \text { 95: CI } \
\text { Predictor } & {\text { Coef }} & \text { SE Coef } & Z &{\text { P }} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } &-27.118&6 .696& -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$

Log-Likelihood = -21.883
Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000
Goodness-of-Fit Tests

$ \begin{array}{lrcr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046\end{array} $

-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?

Accepted Answer

The answer of  TABLE 14-11 
A logistic regression model...

Question 4

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:
 $\begin{array}{lccccccrr} 
& & & & \text { Odds } & \text { 95 \% CI } \
\text {Predictor }& \text {Coef }&\text { SE Coef }&\text { Z }& \text {P} &\text { Ratio} &\text { Lower} & \text {Upper }\
\text {Constant }& -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text {Income }& 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text {Lawn Size} & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text {Attitude} & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text {Teenager} & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text {Age} & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$

Log-Likelihood = -4.890
Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000

Goodness-of-Fit Tests
$ \begin{array}{lrrr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000\end{array} $

-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.

Accepted Answer

A) True 
 B)False

Question 5

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:
 $\begin{array}{lccccccrr} 
& & & & \text { Odds } & \text { 95 \% CI } \
\text {Predictor }& \text {Coef }&\text { SE Coef }&\text { Z }& \text {P} &\text { Ratio} &\text { Lower} & \text {Upper }\
\text {Constant }& -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text {Income }& 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text {Lawn Size} & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text {Attitude} & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text {Teenager} & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text {Age} & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$

Log-Likelihood = -4.890
Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000

Goodness-of-Fit Tests
$ \begin{array}{lrrr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000\end{array} $

-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?

Accepted Answer

The answer of TABLE 14-17
The marketing manager for a nationally...

Question 6

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:
 $\begin{array}{ll} 
& \text { Regression Stuistics } \
\hline \text { Multiple R } & 0.865 \
\text { R Square } & 0.748 \
\text { Adjusted R Square } & 0.726 \
\text { Standard Error } & 5.195 \
\text { Observations } &50  \
\hline
\end{array}$
 ANOVA
$\begin{array}{lrrrrr}
\hline & \text { d f }& \text {S S } & \text {M S } & \text {F } & \text {Significance F } \
\hline \text {Regression} & & 3605.7736 & 1201.9245 & &0.0000 \
\text {Residual} & & 1214.2264 & 26.3962 & \
Total & 49 & 4820.0000 & & & \
\hline
\end{array}$

$\begin{array}{lcccc}
\hline & \text { Coefficients} &  \text {Standard Error} &  \text {t Stat }&  \text {p -value} \
\hline  \text {Intercept }& -1.6335 & 5.8078 & -0.281 & 0.7798 \
 \text {Income} & 0.4485 & 0.1137 & 3.9545 & 0.0003 \
 \text {Size} & 4.2615 & 0.8062 & 5.286 & 0.0001 \
 \text {School }& -0.6517 & 0.4319 & -1.509 & 0.1383 \
\hline
\end{array}$

-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?

Accepted Answer

A)  0.0005 
B)  0.001 
C)  0.05 
D)  0.01

Question 7

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.
  $$\begin{array}{l}
\text { Regression Statistics }\
\begin{array} { l r } 
\hline \text { Multiple R } & 0.830 \
\text { R Square } & 0.689 \
\text { Adjusted R Square } & 0.662 \
\text { Standard Error } & 17501.643 \
\text { Observations } & 26
\end{array}
\end{array}$$

ANOVA
$\begin{array}{lcrrrr}
\hline & \text { d f }&\text { S S } & \text { M S } & \text { F }&\text { Significance  F } \
\hline \text {Regression} & 2 & 15579777040 & 7789888520 & 25.432 & 0.0001 \
\text {Residual }& 23 & 7045072780 & 306307512 & & \
\text {Total }& 25 & 22624849820 & & & \
\hline
\end{array}$

$$\begin{array} { l c c c c } 
& \text { Coefficients } & \text { Standard Error} & \text {  t Stat } & \text {  p-value } \
\text { Intercept } & 15800.0000 & 6038.2999 & 2.617 & 0.0154 \
\text { C apital } & 0.1245 & 0.2045 & 0.609 & 0.5485 \
\text { W ages } & 7.0762 & 1.4729 & 4.804 & 0.0001 \
\hline
\end{array}$$

-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?

Accepted Answer

A)  15,800.00 
B)  17,277.49 
C)  16,520.07 
D)  20,455.98

Question 8

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:
 $\begin{array}{lccccccrr} 
& & & & \text { Odds } & \text { 95 \% CI } \
\text {Predictor }& \text {Coef }&\text { SE Coef }&\text { Z }& \text {P} &\text { Ratio} &\text { Lower} & \text {Upper }\
\text {Constant }& -70.49 & 47.22 & -1.49 & 0.135 & & & \
\text {Income }& 0.2868 & 0.1523 & 1.88 & 0.060 & 1.33 & 0.99 & 1.80 \
\text {Lawn Size} & 1.0647 & 0.7472 & 1.42 & 0.154 & 2.90 & 0.67 & 12.54 \
\text {Attitude} & -12.744 & 9.455 & -1.35 & 0.178 & 0.00 & 0.00 & 326.06 \
\text {Teenager} & -0.200 & 1.061 & -0.19 & 0.850 & 0.82 & 0.10 & 6.56 \
\text {Age} & 1.0792 & 0.8783 & 1.23 & 0.219 & 2.94 & 0.53 & 16.45
\end{array}$

Log-Likelihood = -4.890
Test that all slopes are zero: G = 31.808, DF = 5, P-Value = 0.000

Goodness-of-Fit Tests
$ \begin{array}{lrrr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 9.313 & 24 & 0.997 \ \text { Deviance } & 9.780 & 24 & 0.995 \ \text { Hosmer-Lemeshow } & 0.571 & 8 & 1.000\end{array} $

-Referring to Table 14-15, the predicted mileage for a 200 horsepower, 4-cylinder car is_______

Accepted Answer

The answer of TABLE 14-17 
The marketing manager for a...

Question 9

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: $\text {SUMMARY OUTPUT}$ $\begin{array}{ll} \hline \text { Regression Statistics } \ \hline \text { Multiple R }& 0.916 \ \text { R Square} & 0.839 \ \text { Adjusted R Square} & 0.732 \ \text { Standard Error }& 0.24685 \ \text { Observations} & 6 \ \hline \end{array}$ ANOVA $\begin{array}{lccclc} \hline & \text {d f }& \text { SS }& \text { M S } & \text { F } & \text {Significance F } \ \hline \text { Regression} & 2 & 0.95219 & 0.47610 & 7.813 & 0.0646 \ \text {Residual} & 3 & 0.18281 & 0.06094 & & \ \text {Total} & 5 & 1.13500 & & & \ \hline \end{array}$ $\begin{array}{lrcrr} \hline & \text {Coefficients }& \text {Standard Error} & \text {t Stat } & \text { p -value} \ \hline \text { Intercept }& 4.593897 & 1.13374542 & 4.052 & 0.0271 \ \text {Units }& -0.247270 & 0.06268485 & -3.945 & 0.0290 \ \text {SAT Total }& 0.001443 & 0.00101241 & 1.425 & 0.2494 \ \hline \end{array}$ -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_____ .

Accepted Answer

The answer of  TABLE 14-7
The department head of the...

Question 10

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, XS1U1B11S1U1B0S1U1B1 S1U1B0= % Attendance, XS1U1B12S1U1B0S1U1B1 S1U1B0= Salaries and
XS1U1B13 S1U1B0= Spending:
 $\begin{array}{lr}
\hline\text {Regression Statistics } \
\hline \text {Multiple R }& 0.7930 \
\text {R Square} & 0.6288 \
\text {Adjusted R Square} & 0.6029 \
\text {Standard Error }& 10.4570 \
\text {Observations }& 47 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccccc}
\hline &\text {  d f } &\text {  SS }& \text { MS }& \text { F } & \text { Significance F} \
\hline \text { Regression }& 3 & 7965.08 & 2655.03 & 24.2802 & 2.3853 \mathrm{E}-09  \
\text { Residual} & 43 & 4702.02 & 109.35 & & \
\text { Total }& 46 & 12667.11 & & & \
\hline
\end{array}$

$\begin{array}{lrrrrrr}
\hline &\text {  Coeffs} & \text { Stnd Err} &\text {  t Stat} &\text { p -value }&\text {  Lower 95\%} \text { Upper 95\%} \
\hline\text {  Intercept }& -753.4225 & 101.1149 & -7.4511 & 2.88 \mathrm{E}-09  & -957.3401 & -549.5050 \
\%\text {  Attend }& 8.5014 & 1.0771 & 7.8929 &  6.73 \mathrm{E}-10  & 6.3292 & 10.6735 \
\text { Salary} &  6.85 \mathrm{E}-07  & 0.0006 & 0.0011 & 0.9991 & -0.0013 & 0.0013 \
\text { Spending }& 0.0060 & 0.0046 & 1.2879 & 0.2047 & -0.0034 & 0.0153 \
\hline
\end{array}$

-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.

Accepted Answer

A) True 
 B)False

Question 11

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, XS1U1B11S1U1B0S1U1B1 S1U1B0= % Attendance, XS1U1B12S1U1B0S1U1B1 S1U1B0= Salaries and
XS1U1B13 S1U1B0= Spending:
 $\begin{array}{lr}
\hline\text {Regression Statistics } \
\hline \text {Multiple R }& 0.7930 \
\text {R Square} & 0.6288 \
\text {Adjusted R Square} & 0.6029 \
\text {Standard Error }& 10.4570 \
\text {Observations }& 47 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccccc}
\hline &\text {  d f } &\text {  SS }& \text { MS }& \text { F } & \text { Significance F} \
\hline \text { Regression }& 3 & 7965.08 & 2655.03 & 24.2802 & 2.3853 \mathrm{E}-09  \
\text { Residual} & 43 & 4702.02 & 109.35 & & \
\text { Total }& 46 & 12667.11 & & & \
\hline
\end{array}$

$\begin{array}{lrrrrrr}
\hline &\text {  Coeffs} & \text { Stnd Err} &\text {  t Stat} &\text { p -value }&\text {  Lower 95\%} \text { Upper 95\%} \
\hline\text {  Intercept }& -753.4225 & 101.1149 & -7.4511 & 2.88 \mathrm{E}-09  & -957.3401 & -549.5050 \
\%\text {  Attend }& 8.5014 & 1.0771 & 7.8929 &  6.73 \mathrm{E}-10  & 6.3292 & 10.6735 \
\text { Salary} &  6.85 \mathrm{E}-07  & 0.0006 & 0.0011 & 0.9991 & -0.0013 & 0.0013 \
\text { Spending }& 0.0060 & 0.0046 & 1.2879 & 0.2047 & -0.0034 & 0.0153 \
\hline
\end{array}$

-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?

Accepted Answer

The answer of TABLE 14-16 
The superintendent of a school...

Question 12

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 - 0.05XS1U1B11S1U1B0 + 20XS1U1B12S1U1B0 - 0.1XS1U1B11S1U1B0XS1U1B12S1U1B0
where XS1U1B11S1U1B0 = horsepower
XS1U1B12S1U1B0 = 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 .

Accepted Answer

A)  60 - 0.15X1 
B)  40 - 0.10X1 
C)  40 - 0.05X1 
D)  60 - 0.10X1

Question 13

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, XS1U1B11S1U1B0S1U1B1 S1U1B0= % Attendance, XS1U1B12S1U1B0S1U1B1 S1U1B0= Salaries and
XS1U1B13 S1U1B0= Spending:
  $\begin{array}{lr}
\hline\text {Regression Statistics } \
\hline \text {Multiple R }& 0.7930 \
\text {R Square} & 0.6288 \
\text {Adjusted R Square} & 0.6029 \
\text {Standard Error }& 10.4570 \
\text {Observations }& 47 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccccc}
\hline &\text {  d f } &\text {  SS }& \text { MS }& \text { F } & \text { Significance F} \
\hline \text { Regression }& 3 & 7965.08 & 2655.03 & 24.2802 & 2.3853 \mathrm{E}-09  \
\text { Residual} & 43 & 4702.02 & 109.35 & & \
\text { Total }& 46 & 12667.11 & & & \
\hline
\end{array}$

$\begin{array}{lrrrrrr}
\hline &\text {  Coeffs} & \text { Stnd Err} &\text {  t Stat} &\text { p -value }&\text {  Lower 95\%} \text { Upper 95\%} \
\hline\text {  Intercept }& -753.4225 & 101.1149 & -7.4511 & 2.88 \mathrm{E}-09  & -957.3401 & -549.5050 \
\%\text {  Attend }& 8.5014 & 1.0771 & 7.8929 &  6.73 \mathrm{E}-10  & 6.3292 & 10.6735 \
\text { Salary} &  6.85 \mathrm{E}-07  & 0.0006 & 0.0011 & 0.9991 & -0.0013 & 0.0013 \
\text { Spending }& 0.0060 & 0.0046 & 1.2879 & 0.2047 & -0.0034 & 0.0153 \
\hline
\end{array}$

-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.

Accepted Answer

A) True 
 B)False

Question 14

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 (XS1U1B11S1U1B0), the amount of insulation in inches (XS1U1B12S1U1B0), the number of windows in the house (XS1U1B13S1U1B0), and the age of the furnace in years (XS1U1B14S1U1B0). Given below are the EXCEL outputs of two regression models.
 $\begin{array}{ll}
\text { Model } 1 \
\text { Regression Statistics } \
\hline \text { R Square }& 0.8080 \
\text { AdjustedR S quare }& 0.7568 \
\text { Observations } &20 
\end{array}$
 
ANOVA
$\begin{array}{lrrccc} 
& \text { df } &{S S} & M S & F & \text { Signuficance F } \
\hline \text { Regression } & 4 & 169503.4241 & 42375.86 & 15.7874 & 2.96869 E-05 \
\text { Residual } & 15 & 40262.3259 & 2684.155 & & \
\text { Total } & 19 & 209765.75 & & & \
\hline
\end{array}$
 
$\begin{array}{lrrrrrrr} 
&&   \text { Standard } & &   \text {Lower} &   \text {Upper }\
&   \text {Coefficients} &   \text {Error} &   \text { t Stat } &   \text {p -value} &  90.0 \%  &  90.0 \%  \
 \hline \text { Intercept }& 421.4277 & 77.8614 & 5.4125 & 7.2 \mathrm{E}-05& 284.9327 & 557.9227 \
  \text { X{1}  (Temperature)} & -4.5098 & 0.8129 & -5.5476 &5.58 \mathrm{E}-05 & -5.9349 & -3.0847 \
  \text {X{2}  (Insulation) }& -14.9029 & 5.0508 & -2.9505 & 0.0099 & -23.7573 & -6.0485 \
  \text { X{3}  (Windows) }& 0.2151 & 4.8675 & 0.0442 & 0.9653 & -8.3181 & 8.7484 \
  \text { X{4} (Furnace Age)} & 6.3780 & 4.1026 & 1.5546 & 0.1408 & -0.8140 & 13.5702 \
\hline
\end{array}$ 
 
$\begin{array}{ll}
 \text { Model 2} \
\hline  \text {Regression Statistics} \
\hline \text {R Square }& 0.7768\
 \text {Adjusted R Square }&0.7506 \
 \text {Observations }& 20 \
\hline
\end{array}$
ANOVA
$\begin{array}{lrrrcc}
\hline &  \text { d f}  & \text { SS } &  \text { MS } &  \text {SS } &  \text {Significance F } \
\hline  \text {Regression} & 2 & 162958.2277 & 81479.11 & 29.5923 &  2.9036 \mathrm{E}-06  \
 \text {Residual }& 17 & 46807.5222 & 2753.384 & & \
 \text {Total }& 19 & 209765.75 & & & \
\hline
\end{array}$

$\begin{array}{lcccccc} 
&& \text { Standard } & &\text { Lower} &\text { Upper} \
& \text {Coefficients }&\text { Error }&\text { t Stat }& \text { p -value} &95 \%&  95 \% \
\hline \text {Intercept }& 489.3227 & 43.982611 .1253 &3.17 \mathrm{E}-09& 396.5273 & 582.1180 \
\text { X{1} (Temperature) }& -5.1103 & 0.6951-7.3515 &  1.13 \mathrm{E}-06 & -6.5769 & -3.6437 \
\text { X{2} (Insulation) }& -14.7195 &  4.8864-3.0123 & 0.0078 & -25.0290 & -4.4099 \
\hline
\end{array}$

-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?

Accepted Answer

A)  0.820 
B)  1.382 
C)  15.787 
D)  1.219

Question 15

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:
 $\text {SUMMARY OUTPUT}$
$\begin{array}{ll}
\hline  \text {  Regression Statistics } \
\hline  \text { Multiple R }& 0.916 \
 \text { R Square} & 0.839 \
 \text { Adjusted R Square} & 0.732 \
 \text { Standard Error }& 0.24685 \
 \text { Observations} & 6 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccclc}
\hline &  \text {d f }& \text { SS }&  \text { M S } &  \text { F } &  \text {Significance F } \
\hline \text { Regression} & 2 & 0.95219 & 0.47610 & 7.813 & 0.0646 \
 \text {Residual} & 3 & 0.18281 & 0.06094 & & \
 \text {Total} & 5 & 1.13500 & & & \
\hline
\end{array}$

$\begin{array}{lrcrr}
\hline &  \text {Coefficients }&  \text {Standard Error} & \text {t Stat } &  \text { p -value} \
\hline \text { Intercept }& 4.593897 & 1.13374542 & 4.052 & 0.0271 \
 \text {Units }& -0.247270 & 0.06268485 & -3.945 & 0.0290 \
 \text {SAT Total }& 0.001443 & 0.00101241 & 1.425 & 0.2494 \
\hline
\end{array}$

-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______ .

Accepted Answer

The answer of TABLE 14-7 
 The department head of...

Question 16

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 $\begin{array}{lr} \hline\text {Regression Statistics } \ \hline\text { Multiple R} & 0.63 \ \text {R Square }& 0.40 \ \text {Adjusted R Square} & 0.23 \ \text {Standard Error }& 50.00 \ \text {Observations} & 15.00 \ \hline \end{array}$ ANOVA $\begin{array}{lccccc} \hline & \text { d f} & \text {SS} & \text {MS} & \text {F } & \text {Significance F } \ \hline \text {Regression} & 3 & & 5994.24 & 2.40 & 0.12 \ \text {Residual }& 11 & 27496.82 & & & \ \text {Total }& 45479.54 & & & \ \hline \end{array}$ $\begin{array}{lccrrrr} \hline & \text {oefficients}&\text { Standard Error }& \text { t Stat} &\text { p-value }& \text {Lower 99.0\% }& \text {Upper 99.0 \% } \ \hline\text { Intercept} & 123.80 & 48.71 & 2.54 & 0.03 & -27.47 & 275.07 \ \text {AGE }& -0.82 & 0.87 & -0.95 & 0.36 & -3.51 & 1.87 \ \text {TICKETS}& 21.25 & 10.66 & 1.99 & 0.07 & -11.86 & 54.37 \ \text {DENSITY} & -3.14 & 6.46 & -0.49 & 0.64 & -23.19 & 16.91 \ \hline \end{array}$ -Referring to Table 14-10, the adjusted r² is ____ .

Accepted Answer

The answer of TABLE 14-10 
 You worked as an...

Question 17

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, XS1U1B11S1U1B0S1U1B1 S1U1B0= % Attendance, XS1U1B12S1U1B0S1U1B1 S1U1B0= Salaries and
XS1U1B13 S1U1B0= Spending:
 $\begin{array}{lr}
\hline\text {Regression Statistics } \
\hline \text {Multiple R }& 0.7930 \
\text {R Square} & 0.6288 \
\text {Adjusted R Square} & 0.6029 \
\text {Standard Error }& 10.4570 \
\text {Observations }& 47 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccccc}
\hline &\text {  d f } &\text {  SS }& \text { MS }& \text { F } & \text { Significance F} \
\hline \text { Regression }& 3 & 7965.08 & 2655.03 & 24.2802 & 2.3853 \mathrm{E}-09  \
\text { Residual} & 43 & 4702.02 & 109.35 & & \
\text { Total }& 46 & 12667.11 & & & \
\hline
\end{array}$

$\begin{array}{lrrrrrr}
\hline &\text {  Coeffs} & \text { Stnd Err} &\text {  t Stat} &\text { p -value }&\text {  Lower 95\%} \text { Upper 95\%} \
\hline\text {  Intercept }& -753.4225 & 101.1149 & -7.4511 & 2.88 \mathrm{E}-09  & -957.3401 & -549.5050 \
\%\text {  Attend }& 8.5014 & 1.0771 & 7.8929 &  6.73 \mathrm{E}-10  & 6.3292 & 10.6735 \
\text { Salary} &  6.85 \mathrm{E}-07  & 0.0006 & 0.0011 & 0.9991 & -0.0013 & 0.0013 \
\text { Spending }& 0.0060 & 0.0046 & 1.2879 & 0.2047 & -0.0034 & 0.0153 \
\hline
\end{array}$
-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?

Accepted Answer

A)  $ H_{1}: \beta_{1} \neq 0 $ 
B)  $ H_{1}: \beta_{3} \neq 0 $ 
C)  $ H_{1}: \beta_{0} \neq 0 $ 
D)  $ H_{1}: \beta_{2} \neq 0 $

Question 18

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
$\begin{array}{lr}
\hline\text {Regression Statistics } \
\hline\text { Multiple R} & 0.63 \
\text {R Square }& 0.40 \
\text {Adjusted R Square} & 0.23 \
\text {Standard Error }& 50.00 \
\text {Observations} & 15.00 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccccc}
\hline & \text { d f} & \text {SS} & \text {MS} & \text {F } & \text {Significance  F } \
\hline \text {Regression} & 3 & & 5994.24 & 2.40 & 0.12 \
\text {Residual }& 11 & 27496.82 & & & \
\text {Total }& 45479.54 & & & \
\hline
\end{array}$

$\begin{array}{lccrrrr}
\hline & \text {oefficients}&\text { Standard Error }& \text { t Stat} &\text { p-value }& \text {Lower 99.0\% }& \text {Upper 99.0 \% } \
\hline\text { Intercept} & 123.80 & 48.71 & 2.54 & 0.03 & -27.47 & 275.07 \
\text {AGE }& -0.82 & 0.87 & -0.95 & 0.36 & -3.51 & 1.87 \
\text {TICKETS}& 21.25 & 10.66 & 1.99 & 0.07 & -11.86 & 54.37 \
\text {DENSITY} & -3.14 & 6.46 & -0.49 & 0.64 & -23.19 & 16.91 \
\hline
\end{array}$

-Referring to Table 14-10, the total degrees of freedom that are missing in the ANOVA table should be_____ .

Accepted Answer

The answer of TABLE 14-10 
You worked as an intern...

Question 19

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, XS1U1B11S1U1B0S1U1B1 S1U1B0= % Attendance, XS1U1B12S1U1B0S1U1B1 S1U1B0= Salaries and
XS1U1B13 S1U1B0= Spending:
  $\begin{array}{lr}
\hline\text {Regression Statistics } \
\hline \text {Multiple R }& 0.7930 \
\text {R Square} & 0.6288 \
\text {Adjusted R Square} & 0.6029 \
\text {Standard Error }& 10.4570 \
\text {Observations }& 47 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccccc}
\hline &\text {  d f } &\text {  SS }& \text { MS }& \text { F } & \text { Significance F} \
\hline \text { Regression }& 3 & 7965.08 & 2655.03 & 24.2802 & 2.3853 \mathrm{E}-09  \
\text { Residual} & 43 & 4702.02 & 109.35 & & \
\text { Total }& 46 & 12667.11 & & & \
\hline
\end{array}$

$\begin{array}{lrrrrrr}
\hline &\text {  Coeffs} & \text { Stnd Err} &\text {  t Stat} &\text { p -value }&\text {  Lower 95\%} \text { Upper 95\%} \
\hline\text {  Intercept }& -753.4225 & 101.1149 & -7.4511 & 2.88 \mathrm{E}-09  & -957.3401 & -549.5050 \
\%\text {  Attend }& 8.5014 & 1.0771 & 7.8929 &  6.73 \mathrm{E}-10  & 6.3292 & 10.6735 \
\text { Salary} &  6.85 \mathrm{E}-07  & 0.0006 & 0.0011 & 0.9991 & -0.0013 & 0.0013 \
\text { Spending }& 0.0060 & 0.0046 & 1.2879 & 0.2047 & -0.0034 & 0.0153 \
\hline
\end{array}$

-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?

Accepted Answer

A)  $ H_{0}: \beta_{0}=\beta_{1}=\beta_{2}=\beta_{3}=0 $ 
B)  $ H_{0}: \beta_{1}=\beta_{2}=\beta_{3} \neq 0 $ 
C)  $ H_{0}: \beta_{1}=\beta_{2}=\beta_{3}=0 $ 
D)  $ H_{0}: \beta_{0}=\beta_{1}=\beta_{2}=\beta_{3} \neq 0 $

Question 20

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
$\begin{array}{lrrrrrrr} 
& & & && \text { Odds } & \text { 95: CI } \
\text { Predictor } & {\text { Coef }} & \text { SE Coef } & Z &{\text { P }} & \text { Ratio } & \text { Lower } & \text { Upper } \
\text { Constant } &-27.118&6 .696& -4.05 & 0.000 & & & \
\text { SAT } & 0.015 & 0.004666 & 3.17 & 0.002 & 1.01 & 1.01 & 1.02 \
\text { Toefl550 } & -0.390 & 0.9538 & -0.41 & 0.682 & 0.68 & 0.10 & 4.39 \
\text { Room/Brd } & 2.078 & 0.5076 & 4.09 & 0.000 & 7.99 & 2.95 & 21.60
\end{array}$

Log-Likelihood = -21.883
Test that all slopes are zero: G = 62.083, DF = 3, P-Value = 0.000
Goodness-of-Fit Tests

$ \begin{array}{lrcr}\text { Method } & \text { Chi-Square } & \text { DF } & \text { P } \ \text { Pearson } & 143.551 & 76 & 0.000 \ \text { Deviance } & 43.767 & 76 & 0.999 \ \text { Hosmer-Lemeshow } & 15.731 & 8 & 0.046\end{array} $

-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.

Accepted Answer

A) True 
 B)False

Exam 14: Introduction to Multiple Regression

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