Question 1

TABLE 14-8 A financial analyst wanted to examine the relationship between salary (in $1,000) and 4 variables: age (XS1U1B11S1U1B0S1U1B1 S1U1B0= Age), experience in the field (XS1U1B12S1U1B0S1U1B1 S1U1B0= Exper), number of degrees (XS1U1B13S1U1B0S1U1B1 S1U1B0= Degrees), and number of previous jobs in the field (XS1U1B14S1U1B0S1U1B1 S1U1B0= Prevjobs). He took a sample of 20 employees and obtained the following Microsoft Excel output: $\text {SUMMARY OUTPUT}$ $\begin{array}{ll} \hline \text { Regression Statistics } \ \hline \text { Multiple R }& 0.992 \ \text { R Square} & 0.984 \ \text { Adjusted R Square} & 0.979 \ \text { Standard Error }& 2.26743 \ \text { Observations} & 20 \ \hline \end{array}$ ANOVA $\begin{array}{lccclc} \hline & \text {d f }& \text { SS }& \text { M S } & \text { F } & \text {Significance F } \ \hline \text { Regression} & 4 & 4609.83164& 1152.45791 & 224.160& 0.0001 \ \text {Residual} &15& 77.11836 & 5.14122& & \ \text {Total} & 19 & 4686.95000 & & & \ \hline \end{array}$ $\begin{array}{lrrrr} \hline & \text {Coefficients} &\text { Standard Error} & \text { t Stat } & \text { p -value} \ \hline \text { Intercept }& -9.611198 & 2.77988638 & -3.457 & 0.0035 \ \text { Age }& 1.327695 & 0.11491930 & 11.553 & 0.0001 \ \text { Exper }& -0.106705 & 0.14265559 & -0.748 & 0.4660 \ \text { Degrees} & 7.311332 & 0.80324187 & 9.102 & 0.0001 \ \text { Prevjobs }& -0.504168 & 0.44771573 & -1.126 & 0.2778 \ \hline \end{array}$ -Referring to Table 14-8, the value of the coefficient of multiple determination, r² Y.1234, is____

Accepted Answer

The answer of  TABLE 14-8 
A financial analyst wanted...

Question 2

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, to test the significance of the multiple regression model, the null hypothesis should be rejected while allowing for 1% probability of committing a type I error.

Accepted Answer

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

Question 3

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 : þ1 = þ2 = 0. At a level of significance of 0.05, the null hypothesis is rejected.

Accepted Answer

The significance F value is 0.0646, which is greater than the 0.05 level of significance, so the null hypothesis is not rejected.

Question 4

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 are the degrees of freedom for the chi-square distribution when testing whether the model is a good-fitting model?

Accepted Answer

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

Question 5

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 use a t test to test for the significance of the coefficient of X1. The value of the test statistic is_____ .

Accepted Answer

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

Question 6

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 value of the coefficient of multiple determination, r² Y.12, is_____

Accepted Answer

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

Question 7

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

The p-value for the average teacher salary is 0.9991, which is much greater than the 0.05 significance level, indicating that it is not statistically significant. Additionally, the 95% confidence interval for the salary coefficient includes zero (-0.0013 to 0.0013), further supporting the conclusion that salary has no significant impact on the percentage of students passing the test.

Question 8

The slopes in a multiple regression model are called net regression coefficients.

Accepted Answer

In a multiple regression model, the slopes are indeed referred to as net regression coefficients because they measure the average change in the dependent variable for one unit of change in the predictor variable, holding all other predictors constant.

Question 9

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 is the standard error of estimate?

Accepted Answer

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

Question 10

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 p-value of the test statistic when testing whether LawnSize makes a significant contribution to the model in the presence of the other independent variables?

Accepted Answer

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

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, the null hypothesis should be rejected at a 5% level of significance when testing whether there is a significant relationship between percentage of students passing the proficiency test and the entire set of explanatory variables.

Accepted Answer

The null hypothesis for the overall significance of the regression model is that all coefficients are equal to zero. The F-statistic is 24.2802 with a significance F (p-value) of 2.3853E-09, which is much less than 0.05. Therefore, we reject the null hypothesis at the 5% level of significance, indicating a significant relationship between the percentage of students passing and the explanatory variables.

Question 12

In calculating the standard error of the estimate, $ S_{Y X}=\sqrt{\mathrm{MSE}} $,, there are n - k - 1 degrees of freedom, where n is the sample size and k represents the number of independent variables in the model.

Accepted Answer

This statement is true. The standard error of the estimate (SYX) is equal to the mean squared error (MSE) and both are calculated using n - k - 1 degrees of freedom, where n is the sample size and k is the number of independent variables in the model.

Question 13

In trying to obtain a model to estimate grades on a statistics test, a professor wanted to include, among other factors, whether the person had taken the course previously. To do this, the professor included a dummy variable in her regression that was equal to 1 if the person had previously taken the course, and 0 otherwise. The interpretation of the coefficient associated with this dummy variable would be the average amount the repeat students tended to be above or below
non-repeaters, with all other factors the same.

Accepted Answer

This is the correct interpretation of the coefficient associated with a dummy variable in a regression.

Question 14

An interaction term in a multiple regression model may be used when the relationship between X1 and Y changes for differing values of X₂.

Accepted Answer

This is correct, an interaction term allows for the relationship between X1 and Y to differ based on the values of another variable (X₂ in this case).

Question 15

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

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 14-17, 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?

Accepted Answer

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

Question 16

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 net regression coefficient of X₂ is ______ .

Accepted Answer

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

Question 17

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 estimated odds ratio for a school with an average SAT score of 1100, a TOEFL criterion that is not at least 550, and the room and board expense of 7 thousand dollars?

Accepted Answer

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

Question 18

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 Teenager 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 p-value for the Teenager variable is 0.850, which is greater than the 0.05 level of significance, indicating that it does not make a significant contribution to the model.

Question 19

TABLE 14-8
 A financial analyst wanted to examine the relationship between salary (in $1,000) and 4 variables: age (XS1U1B11S1U1B0S1U1B1 S1U1B0= Age), experience in the field (XS1U1B12S1U1B0S1U1B1 S1U1B0= Exper), number of degrees (XS1U1B13S1U1B0S1U1B1 S1U1B0= Degrees), and number of previous jobs in the field (XS1U1B14S1U1B0S1U1B1 S1U1B0= Prevjobs). He took a sample of 20 employees and obtained the following Microsoft Excel output:
  $\text {SUMMARY OUTPUT}$
$\begin{array}{ll}
\hline  \text {  Regression Statistics } \
\hline  \text { Multiple R }& 0.992 \
 \text { R Square} & 0.984 \
 \text { Adjusted R Square} & 0.979 \
 \text { Standard Error }& 2.26743 \
 \text { Observations} & 20 \
\hline
\end{array}$

ANOVA
$\begin{array}{lccclc}
\hline &  \text {d f }& \text { SS }&  \text { M S } &  \text { F } &  \text {Significance F } \
\hline \text { Regression} & 4 & 4609.83164& 1152.45791 & 224.160& 0.0001 \
 \text {Residual} &15& 77.11836 & 5.14122& & \
 \text {Total} & 19 & 4686.95000 & & & \
\hline
\end{array}$

$\begin{array}{lrrrr}
\hline & \text {Coefficients} &\text { Standard Error} & \text { t Stat } & \text {  p -value} \
\hline \text { Intercept }& -9.611198 & 2.77988638 & -3.457 & 0.0035 \
\text { Age }& 1.327695 & 0.11491930 & 11.553 & 0.0001 \
\text { Exper }& -0.106705 & 0.14265559 & -0.748 & 0.4660 \
\text { Degrees} & 7.311332 & 0.80324187 & 9.102 & 0.0001 \
\text { Prevjobs }& -0.504168 & 0.44771573 & -1.126 & 0.2778 \
\hline
\end{array}$

-Referring to Table 14-8, the critical value of an F test on the entire regression for a level of significance of 0.01 is _____.

Accepted Answer

The answer of TABLE 14-8
 A financial analyst wanted to...

Question 20

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 p-value of the test statistic when testing whether Attitude makes a significant contribution to the model in the presence of the other independent variables?

Accepted Answer

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

Quiz 14: Introduction to Multiple Regression

The slopes in a multiple regression model are called net regression coefficients.

In calculating the standard error of the estimate, $S_{Y X}=\sqrt{\mathrm{MSE}}$ ,, there are n - k - 1 degrees of freedom, where n is the sample size and k represents the number of independent variables in the model.

An interaction term in a multiple regression model may be used when the relationship between X1 and Y changes for differing values of X₂.

Quiz 14: Introduction to Multiple Regression

The slopes in a multiple regression model are called net regression coefficients.

In calculating the standard error of the estimate, SYX=MSE S_{Y X}=\sqrt{\mathrm{MSE}} SYX​=MSE​ ,, there are n - k - 1 degrees of freedom, where n is the sample size and k represents the number of independent variables in the model.

An interaction term in a multiple regression model may be used when the relationship between X1 and Y changes for differing values of X2.

In calculating the standard error of the estimate, $S_{Y X}=\sqrt{\mathrm{MSE}}$ ,, there are n - k - 1 degrees of freedom, where n is the sample size and k represents the number of independent variables in the model.

An interaction term in a multiple regression model may be used when the relationship between X1 and Y changes for differing values of X₂.