Question 1

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

Accepted Answer

The adjusted r2 can increase when an explanatory variable is dropped from a multiple regression model if the variable was not contributing significantly enough to the model or was causing multicollinearity. This increases the weight of the other variables and can lead to a higher adjusted r2.

Question 2

If we have taken into account all relevant explanatory factors, the residuals from a multiple regression should be random.

Accepted Answer

If all relevant explanatory factors have been included in the model, the residuals should reflect only random variation and should not show any pattern or systematic variation.

Question 3

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 F test for the significance of the entire regression performed at a level of significance of 0.01 leads to a rejection of the null hypothesis.

Accepted Answer

The p-value for the F test is less than 0.01, indicating that there is significant evidence to reject the null hypothesis that all regression coefficients are zero. Therefore, the regression model as a whole is significant.

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, the null hypothesis that the model is a good-fitting model cannot be rejected when allowing for a 5% probability of making a type I error.

Accepted Answer

The Pearson chi-square goodness-of-fit test has a p-value of 0.000, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis that the model is a good-fitting model.

Question 5

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 multiple regression model is significant at a 10% level of significance.

Accepted Answer

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

Question 6

In a particular model, the sum of the squared residuals was 847. If the model had 5 independent variables, and the data set contained 40 points, the value of the standard error of the estimate is 24.911.

Accepted Answer

The formula for the standard error of the estimate is:
SE = sqrt(SSres/(n-k-1))
where SSres is the sum of squared residuals, n is the number of data points, and k is the number of independent variables.

Plugging in the given values:
SE = sqrt(847/(40-5-1))
SE = 4.954

Therefore, the statement is false and the standard error of the estimate is not 24.911.

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 instructional spending per pupil 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 þ3.

Accepted Answer

The p-value for Spending is 0.2047, which is greater than 0.05, indicating that it is not statistically significant at the 5% level. The 95% confidence interval for Spending also includes zero (-0.0034 to 0.0153), supporting the conclusion that it has no significant impact.

Question 8

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

Accepted Answer

The null hypothesis H0: β1 = β2 = β3 = 0 suggests that none of the explanatory variables (% Attendance, Salaries, Spending) have a significant effect on the dependent variable (% Passing).

Question 9

The coefficient of multiple determination r²Y.12 measures the proportion of variation in Y that is explained by X₁ and X₂.

Accepted Answer

The statement is true. The coefficient of multiple determination r2Y.12 measures the proportion of variation in Y that is explained by X1 and X2.

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, the null hypothesis H₀ : ?₁ = ?₂ = ?₃ = 0 implies that percentage of students passing the proficiency test is not affected by some of the explanatory variables.

Accepted Answer

The null hypothesis H₀: β₁ = β₂ = β₃ = 0 implies that none of the explanatory variables (Attendance, Salaries, Spending) have a significant effect on the percentage of students passing the proficiency test.

Question 11

A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 63.84% of the variation in the dependent variable is explained by the independent variables in the regression.

Accepted Answer

The coefficient of determination (R-squared) can be calculated as 1 - (SSE/SST). Plugging in the given values: R-squared = 1 - (29.85/82.55) = 0.6384 or 63.84%. This means that 63.84% of the variation in the dependent variable is explained by the independent variables in the regression. Therefore, the statement is true.

Question 12

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 Income 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 Income is 0.060, which is greater than the 0.05 significance level, indicating that Income is not statistically significant in the presence of other variables.

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, X1 = % Attendance, X2 = Salaries and
X3 = 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 alternative hypothesis H1 : At least one of þj × 0 for j = 1, 2, 3 implies that percentage of students passing the proficiency test is related to at least one of the explanatory variables.

Accepted Answer

Table 14-16 deals with ANOVA for a multiple regression model, and the alternative hypothesis H1 states that at least one of the regression coefficients (þj) is not equal to zero, implying that there is a relationship between the response variable and at least one of the explanatory variables. In the context of the table, this translates to the percentage of students passing the proficiency test being related to at least one of the explanatory variables.

Question 14

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

The null hypothesis H0: β1 = β2 = β3 = 0 suggests that none of the explanatory variables (% Attendance, Salaries, Spending) have a statistically significant relationship with the dependent variable (% Passing).

Question 15

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, there is sufficient evidence that all of the explanatory variables is related to the percentage of students passing the proficiency test.

Accepted Answer

The p-values for Salary (0.9991) and Spending (0.2047) are not statistically significant, indicating insufficient evidence that these variables are related to the percentage of students passing the test.

Question 16

The interpretation of the slope is different in a multiple linear regression model as compared to a simple linear regression model.

Accepted Answer

In a multiple linear regression model, the slope represents the change in the dependent variable associated with a unit change in the independent variable while holding all other independent variables constant. In contrast, in a simple linear regression model, the slope represents the change in the dependent variable associated with a unit change in the independent variable.

Question 17

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. At a level of significance of 0.05, the department head would decide that ?₁ ? 0.

Accepted Answer

The p-value for the coefficient of X₁ is 0.0290, which is less than the level of significance of 0.05. Therefore, the department head would reject the null hypothesis that ?₁ = 0 and conclude that ?₁ ? 0.

Question 18

A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 73.4% of the variation in the dependent variable is explained by the independent variables in the regression.

Accepted Answer

The answer of A regression had the following results: SST...

Quiz 14: Introduction to Multiple Regression

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

If we have taken into account all relevant explanatory factors, the residuals from a multiple regression should be random.

In a particular model, the sum of the squared residuals was 847. If the model had 5 independent variables, and the data set contained 40 points, the value of the standard error of the estimate is 24.911.

The coefficient of multiple determination r2Y.12 measures the proportion of variation in Y that is explained by X1 and X2.

A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 63.84% of the variation in the dependent variable is explained by the independent variables in the regression.

The interpretation of the slope is different in a multiple linear regression model as compared to a simple linear regression model.

A regression had the following results: SST = 82.55, SSE = 29.85. It can be said that 73.4% of the variation in the dependent variable is explained by the independent variables in the regression.

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

The coefficient of multiple determination r²Y.12 measures the proportion of variation in Y that is explained by X₁ and X₂.