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Business Statistics Study Set 2
Quiz 14: Inference for Regression
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Question 1
Essay
Use the following for questions A sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and college GPA and the number of units sold last month recorded. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Units Sold
=
−
0.48
+
7.42
= - 0.48 + 7.42
=
−
0.48
+
7.42
GPA
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\ \text { GPA } & 7.423 & 1.044 & 7.11 & 0.000 \end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
S
=
1.57429
R
−
S
q
=
78.3
%
∘
R
−
S
q
(
a
d
j
)
=
76.8
%
\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%
S
=
1.57429
R
−
Sq
=
78.3%
∘
R
−
Sq
(
adj
)
=
76.8%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
125.30
125.30
50.56
0.000
Residual Error
14
34.70
2.48
Total
15
160.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 125.30 & 125.30 & 50.56 & 0.000 \\ \text { Residual Error } & 14 & 34.70 & 2.48 & & \\ \text { Total } & 15 & 160.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
14
15
SS
125.30
34.70
160.00
MS
125.30
2.48
F
50.56
P
0.000
Answer:
-List each of the four conditions for regression and inference and describe whether or not they are satisfied.
Question 2
Essay
Use the following information for questions Nutritional information was collected for 77 breakfast cereals including the amount of fiber (in grams), potassium (in mg), and the number of calories per serving. The data resulted in the following scatterplots.
-Compare the two plots with respect to the aspects that would affect the standard error of the regression slope?
Question 3
Essay
Use the following for questions A sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and college GPA and the number of units sold last month recorded. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Units Sold
=
−
0.48
+
7.42
= - 0.48 + 7.42
=
−
0.48
+
7.42
GPA
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\ \text { GPA } & 7.423 & 1.044 & 7.11 & 0.000 \end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
S
=
1.57429
R
−
S
q
=
78.3
%
∘
R
−
S
q
(
a
d
j
)
=
76.8
%
\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%
S
=
1.57429
R
−
Sq
=
78.3%
∘
R
−
Sq
(
adj
)
=
76.8%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
125.30
125.30
50.56
0.000
Residual Error
14
34.70
2.48
Total
15
160.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 125.30 & 125.30 & 50.56 & 0.000 \\ \text { Residual Error } & 14 & 34.70 & 2.48 & & \\ \text { Total } & 15 & 160.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
14
15
SS
125.30
34.70
160.00
MS
125.30
2.48
F
50.56
P
0.000
Answer:
-Circle the standard error of the slope and its components in the output shown. If the information is not in the output, list components.
Question 4
Essay
Data on labor productivity and unit labor costs were obtained for the retail industry from 1987 through 2006 (Bureau of Labor Statistics). A regression was estimated to describe the linear relationship between the two variables. Based on the plot of residuals versus predicted values, is the linear model appropriate? Explain.
Question 5
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-Is there a significant relationship between time it takes to troubleshoot the process(minutes) and training received (use α = .05)? Give the appropriate test statistic,associated P-value, and conclusion.
Question 6
Essay
Use the following for questions A sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and college GPA and the number of units sold last month recorded. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Units Sold
=
−
0.48
+
7.42
= - 0.48 + 7.42
=
−
0.48
+
7.42
GPA
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\ \text { GPA } & 7.423 & 1.044 & 7.11 & 0.000 \end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
S
=
1.57429
R
−
S
q
=
78.3
%
∘
R
−
S
q
(
a
d
j
)
=
76.8
%
\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%
S
=
1.57429
R
−
Sq
=
78.3%
∘
R
−
Sq
(
adj
)
=
76.8%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
125.30
125.30
50.56
0.000
Residual Error
14
34.70
2.48
Total
15
160.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 125.30 & 125.30 & 50.56 & 0.000 \\ \text { Residual Error } & 14 & 34.70 & 2.48 & & \\ \text { Total } & 15 & 160.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
14
15
SS
125.30
34.70
160.00
MS
125.30
2.48
F
50.56
P
0.000
Answer:
-The confidence interval and prediction interval for the number of units sold per month when GPA = 3.00 are shown below. Write a sentence to interpret each interval in this context.
Question 7
Essay
Use the following for questions A sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and college GPA and the number of units sold last month recorded. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Units Sold
=
−
0.48
+
7.42
= - 0.48 + 7.42
=
−
0.48
+
7.42
GPA
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\ \text { GPA } & 7.423 & 1.044 & 7.11 & 0.000 \end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
S
=
1.57429
R
−
S
q
=
78.3
%
∘
R
−
S
q
(
a
d
j
)
=
76.8
%
\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%
S
=
1.57429
R
−
Sq
=
78.3%
∘
R
−
Sq
(
adj
)
=
76.8%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
125.30
125.30
50.56
0.000
Residual Error
14
34.70
2.48
Total
15
160.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 125.30 & 125.30 & 50.56 & 0.000 \\ \text { Residual Error } & 14 & 34.70 & 2.48 & & \\ \text { Total } & 15 & 160.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
14
15
SS
125.30
34.70
160.00
MS
125.30
2.48
F
50.56
P
0.000
Answer:
-What percentage of the variability in sales performance (units sold per month) can be accounted for by college GPA?
Question 8
Essay
Use the following for questions A sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and college GPA and the number of units sold last month recorded. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Units Sold
=
−
0.48
+
7.42
= - 0.48 + 7.42
=
−
0.48
+
7.42
GPA
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\ \text { GPA } & 7.423 & 1.044 & 7.11 & 0.000 \end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
S
=
1.57429
R
−
S
q
=
78.3
%
∘
R
−
S
q
(
a
d
j
)
=
76.8
%
\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%
S
=
1.57429
R
−
Sq
=
78.3%
∘
R
−
Sq
(
adj
)
=
76.8%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
125.30
125.30
50.56
0.000
Residual Error
14
34.70
2.48
Total
15
160.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 125.30 & 125.30 & 50.56 & 0.000 \\ \text { Residual Error } & 14 & 34.70 & 2.48 & & \\ \text { Total } & 15 & 160.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
14
15
SS
125.30
34.70
160.00
MS
125.30
2.48
F
50.56
P
0.000
Answer:
-Predict the units sold per month for a new hire whose college GPA is 3.00.
Question 9
Multiple Choice
A sales manager claims that there is a relationship between college GPA and sales performance (number of units sold) among salespeople hired within the last year. Use the regression results are shown below and set α = .05 to test his claim.
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & T & \mathrm {~P} \\\text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\\text { GPA } & 7.423 & 1.044 & 7.11 & 0.000\end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
Question 10
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-Predict the troubleshooting time for a line worker who received 8 hours of training.
Question 11
Essay
The following plots show (1) world population (millions) plotted against 5-year intervals from 1950 through 2000 and (2) residual vs. fitted value for a linear regression model estimated to describe the trend in world population over time. Based on these plots, would you consider this model appropriate? Explain.
Question 12
Essay
Use the following information for questions Nutritional information was collected for 77 breakfast cereals including the amount of fiber (in grams), potassium (in mg), and the number of calories per serving. The data resulted in the following scatterplots.
-From which of these plots would you expect a more consistent regression slope estimate? Why?
Question 13
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-According to the data, a worker who received 8 hours of training had a troubleshooting time of 15 minutes. What is the value of the residual for this worker? Explain what the residual means.
Question 14
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-From the output, write the equation of the regression equation that can be used to predict troubleshooting time.
Question 15
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-Based on the scatterplot, what is the relationship between training and troubleshooting? Is a regression appropriate for this data? Why or why not?
Question 16
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-The 95% confidence interval for troubleshooting time with 8 hours of training is (15.180, 16.903). Interpret this interval with respect to the estimated troubleshooting time.
Question 17
Essay
Use the following for questions A sales manager was interested in determining if there is a relationship between college GPA and sales performance among salespeople hired within the last year. A sample of recently hired salespeople was selected and college GPA and the number of units sold last month recorded. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Units Sold
=
−
0.48
+
7.42
= - 0.48 + 7.42
=
−
0.48
+
7.42
GPA
Predictor
Coef
SE Coef
T
P
Constant
−
0.484
3.256
−
0.15
0.884
GPA
7.423
1.044
7.11
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & - 0.484 & 3.256 & - 0.15 & 0.884 \\ \text { GPA } & 7.423 & 1.044 & 7.11 & 0.000 \end{array}
Predictor
Constant
GPA
Coef
−
0.484
7.423
SE Coef
3.256
1.044
T
−
0.15
7.11
P
0.884
0.000
S
=
1.57429
R
−
S
q
=
78.3
%
∘
R
−
S
q
(
a
d
j
)
=
76.8
%
\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%
S
=
1.57429
R
−
Sq
=
78.3%
∘
R
−
Sq
(
adj
)
=
76.8%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
125.30
125.30
50.56
0.000
Residual Error
14
34.70
2.48
Total
15
160.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 125.30 & 125.30 & 50.56 & 0.000 \\ \text { Residual Error } & 14 & 34.70 & 2.48 & & \\ \text { Total } & 15 & 160.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
14
15
SS
125.30
34.70
160.00
MS
125.30
2.48
F
50.56
P
0.000
Answer:
-What is the independent variable in this regression? Write the null and alternative hypothesis to test the slope of this variable.
Question 18
Essay
Use the following for questions An operations manager was interested in determining if there is a relationship between the amount of training received by production line workers and the time it takes for them to troubleshoot a process problem. A sample of recently trained line workers was selected. The number of hours of training time received and the time it took (in minutes) for them to troubleshoot their last process problem were captured. Below are the scatterplot, regression results, and residual plots for these data. The regression equation is Trouble Shooting
=
30.7
−
1.84
= 30.7 - 1.84
=
30.7
−
1.84
Training
Predictor
Coef
SE Coef
T
P
Constant
30.729
1.023
30.03
0.000
Training
−
1.8360
0.1376
−
13.35
0.000
\begin{array} { l r r r r } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 30.729 & 1.023 & 30.03 & 0.000 \\ \text { Training } & - 1.8360 & 0.1376 & - 13.35 & 0.000 \end{array}
Predictor
Constant
Training
Coef
30.729
−
1.8360
SE Coef
1.023
0.1376
T
30.03
−
13.35
P
0.000
0.000
S
=
1.43588
R
−
S
q
=
93.2
%
R
−
S
q
(
a
d
j
)
=
92.7
%
S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }
S
=
1.43588
R
−
Sq
=
93.2
%
R
−
Sq
(
adj
)
=
92.7
%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
367.20
367.20
178.10
0.000
Residual Error
13
26.80
2.06
Total
14
394.00
\begin{array} { l r r r r r } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 367.20 & 367.20 & 178.10 & 0.000 \\ \text { Residual Error } & 13 & 26.80 & 2.06 & & \\ \text { Total } & 14 & 394.00 & & & \end{array}
Source
Regression
Residual Error
Total
DF
1
13
14
SS
367.20
26.80
394.00
MS
367.20
2.06
F
178.10
P
0.000
-Write a sentence to interpret the coefficient of training in the regression equation.
Question 19
Multiple Choice
A sales manager was interested in determining if there is a relationship between college GPA and sales performance (number of units sold) among salespeople hired Within the last year. The correct null hypothesis is