Exam 14: Inference for Regression

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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 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 S=1.57429RSq=78.3%RSq(adj)=76.8%\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { 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 Answer:  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  GPA  \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}   \mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%  Analysis of Variance  \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}  Answer:   -What is the independent variable in this regression? Write the null and alternative hypothesis to test the slope of this variable. -What is the independent variable in this regression? Write the null and alternative hypothesis to test the slope of this variable.

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The independent variable is GPA
Ho: There is no association between GPA and sales performance.
HaH _ { a } : There is an association between GPA and sales performance.

A regression equation was fit to the data showing the number of hours of training time received by production line workers and the time it took (in minutes) for them to Trouble shoot their last process problem and the following histogram of residuals Obtained. Based on this histogram of the residuals, we can say that the: A regression equation was fit to the data showing the number of hours of training time received by production line workers and the time it took (in minutes) for them to Trouble shoot their last process problem and the following histogram of residuals Obtained. Based on this histogram of the residuals, we can say that the:

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A

Based on the scatterplot of data of the number of hours of training time received by production line workers and the time it took (in minutes) for them to trouble shoot Their last process problem shown, we can say that Based on the scatterplot of data of the number of hours of training time received by production line workers and the time it took (in minutes) for them to trouble shoot Their last process problem shown, we can say that

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B

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 =30.71.84= 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 S=1.43588RSq=93.2 % RSq(adj)=92.7 % S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% } 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  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  Training  \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}   S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }  Analysis of Variance  \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}     -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. -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.

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A researcher is interested in developing a model that can be used to distribute assistance to low income families for food costs. She used data from a national social Survey to predict weekly amount spent on food using household income (in $1000). The resulting regression equation is How Much money would be needed to feed a family for a week whose household income Is $12,000?

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In a significant regression model determining if there is a relationship between college GPA and sales performance (number of units sold in the previous month), the 95% confidence interval for the number of units sold when GPA = 3.00 was Determined to be 20.914 to 22.657. The correct interpretation is

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A researcher gathers data on the length of essays (number of lines) and the SAT scores received for a sample of students enrolled at his university. Based on his Regression results, the 95% confidence interval for the slope of the regression Equation is -0.88 to 1.34. At α = 0.05, we can say

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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 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 S=1.57429RSq=78.3%RSq(adj)=76.8%\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { 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 Answer:  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  GPA  \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}   \mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%  Analysis of Variance  \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}  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. -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.

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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. 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? -From which of these plots would you expect a more consistent regression slope estimate? Why?

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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 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 S=1.57429RSq=78.3%RSq(adj)=76.8%\mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { 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 Answer:  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  GPA  \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}   \mathrm { S } = 1.57429 \quad \mathrm { R } - \mathrm { Sq } = 78.3 \% \mathrm { \circ } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 76.8 \%  Analysis of Variance  \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}  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. -Circle the standard error of the slope and its components in the output shown. If the information is not in the output, list components.

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Which of the following does NOT affect the standard error of the regression slope?

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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. 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? -Compare the two plots with respect to the aspects that would affect the standard error of the regression slope?

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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 =30.71.84= 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 S=1.43588RSq=93.2 % RSq(adj)=92.7 % S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% } 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  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  Training  \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}   S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }  Analysis of Variance  \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}     -Write a sentence to interpret the coefficient of training in the regression equation. -Write a sentence to interpret the coefficient of training in the regression equation.

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Cars from an online service were examined to see how fuel efficiency (highway mpg) relates to cost (in dollars). According to the regression equation, a used car that costs $13,000 is predicted to get about 30.24 miles per gallon. According to the data, the Car got 35 miles per gallon. What is the value of the residual for this car?

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As the carbon content in steel increases, its ductility tends to decrease. A researcher at a steel company measures carbon content and ductility for a sample of 15 types of Steel. Based on these data he obtained the following regression results, which of the Following statements is NOT true? The regression equation is Ductility =7.673.30= 7.67 - 3.30 Carbon Content Predictor Coef SE Coef T P Constant 7.671 1.507 5.09 0.000 Carbon Content -3.296 1.097 -3.01 0.010 S=2.36317RSq=41.0%RSq(adj)=36.5%\mathrm { S } = 2.36317 \quad \mathrm { R } - \mathrm { Sq } = 41.0 \% \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 36.5 \%

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An estimated regression equation that was fit to estimate ductility in steel using its carbon content was found to be significant at α = 0.05. The 95% prediction interval For the ductility of steel with 0.5% carbon content was determined to be 0.45 to 11.59. The correct interpretation is

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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 =30.71.84= 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 S=1.43588RSq=93.2 % RSq(adj)=92.7 % S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% } 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  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  Training  \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}   S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }  Analysis of Variance  \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}     -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. -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.

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A sample of 15 recently trained line workers was selected to determine if there is a relationship between the number of hours of training time received by production line Workers and the time it took (in minutes) for them to trouble shoot their last process Problem were captured. Use the regression output for the independent variable Shown below to find the 95% confidence interval for the slope of the regression Equation.  Training 1.83600.137613.350.000\text { Training } \quad - 1.8360 \quad 0.1376 \quad - 13.35 \quad 0.000

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As the carbon content in steel increases, its ductility tends to decrease. A researcher at a steel company measures carbon content and ductility for a sample of 15 types ofsteel. Use the following regression results to find the 95% confidence interval for the Slope of the regression equation. Predictor Coef SE Coef T P Constant 7.671 1.507 5.09 0.000 Carbon Content -3.296 1.097 -3.01 0.010

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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 =30.71.84= 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 S=1.43588RSq=93.2 % RSq(adj)=92.7 % S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% } 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  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  Training  \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}   S = 1.43588 \quad \mathrm { R } - \mathrm { Sq } = 93.2 \text { \% } \quad \mathrm { R } - \mathrm { Sq } ( \mathrm { adj } ) = 92.7 \text { \% }  Analysis of Variance  \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}     -Based on the scatterplot, what is the relationship between training and troubleshooting? Is a regression appropriate for this data? Why or why not? -Based on the scatterplot, what is the relationship between training and troubleshooting? Is a regression appropriate for this data? Why or why not?

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