A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis $\begin{array}{rc} \mathrm{r}^{2} 0.762 & \mathrm{n} 7 \\ \mathrm{R} 0.873 & \mathrm{k} \text { 1 } \\ \text { Std. Error 11.547 } & \text { Dep. Var. Sales } \end{array}$ ANOVA table $\begin{array}{rrrrrr} \hline \text { Source } & S S & d f & M S & F & p \text {-value } \\ \hline \text { Regression } & 2,133.3333 & 1 & 2,133.3333 & 16.00 & .0103 \\ \text { Residual } & 666.6667 & 5 & 133.3333 & & \\ \hline \text { Total } & 2,800.0000 & 6 & & &\\\hline \end{array}$ $\text { Regression output }$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\text { Confidence interval }$ $\begin{array}{rrrrrrrr} \hline\text { Variables} & \text { Coefficients } & \text { std. error } & t(d f=5) & \text { p-value } & 95 \% &95 \% \text { upper } \\ & & & & & \text { lower } & \\ \hline \text { Intercep } & 63.3333 & 7.9682 & 7.948 & .0005 & 42.8505 & 83.8162 \\ \text { Advertising } & 6.6667 & 1.6667 & 4.000 & .0103 & & \end{array}$ -Determine a 95% confidence interval estimate of the daily average store sales based on $3000 advertising expenditures? The distance value for this particular prediction is reported as .164.

A) $64,496 to $102.170 B) $33,108 to $133,558 C) $71,312 to $95,356 D) $51,314 to $115,353 E) $42,851 to $83,816 A) $64,496 to $102.170 B) $33,108 to $133,558 C) $71,312 to $95,356 D) $51,314 to $115,353 E) $42,851 to $83,816 C

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis $\begin{array}{rc} \mathrm{r}^{2} 0.762 & \mathrm{n} 7 \\ \mathrm{R} 0.873 & \mathrm{k} \text { 1 } \\ \text { Std. Error 11.547 } & \text { Dep. Var. Sales } \end{array}$ ANOVA table $\begin{array}{rrrrrr} \hline \text { Source } & S S & d f & M S & F & p \text {-value } \\ \hline \text { Regression } & 2,133.3333 & 1 & 2,133.3333 & 16.00 & .0103 \\ \text { Residual } & 666.6667 & 5 & 133.3333 & & \\ \hline \text { Total } & 2,800.0000 & 6 & & &\\\hline \end{array}$ $\text { Regression output }$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\text { Confidence interval }$ $\begin{array}{rrrrrrrr} \hline\text { Variables} & \text { Coefficients } & \text { std. error } & t(d f=5) & \text { p-value } & 95 \% &95 \% \text { upper } \\ & & & & & \text { lower } & \\ \hline \text { Intercep } & 63.3333 & 7.9682 & 7.948 & .0005 & 42.8505 & 83.8162 \\ \text { Advertising } & 6.6667 & 1.6667 & 4.000 & .0103 & & \end{array}$ -What is the least-square prediction equation?

A) $\hat { y }$ = 7.9682 + 1.667 x B) $\hat { y }$ = 63.333 + 6.667 x C) $\hat { y }$ = 7.948 + 4.000 x D) $\hat { y }$ = 11.547 + 1.667 x E) $\hat { y }$ = 6.667 + 63.333 x A) $\hat { y }$ = 7.9682 + 1.667 x B) $\hat { y }$ = 63.333 + 6.667 x C) $\hat { y }$ = 7.948 + 4.000 x D) $\hat { y }$ = 11.547 + 1.667 x E) $\hat { y }$ = 6.667 + 63.333 x

Consider the following partial computer output from a simple linear regression analysis: \[\begin{array} { l l c c c } \text { Predictor } & \text { Coef } & \text { SE Coef } & \mathrm { T } & \mathrm { P } \\ \text { Constant } & 5566.1 & 254.0 & 21.91 & 0.000 \\ \text { Independent Var } & - 210.35 & 24.19 & - & \end{array}\] S = _________ R-Sq = Analysis of Variance $\begin{array}{lccrcc} \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & 3963719 & 3963719 & 75.59 & 0.000 \\ \text { Residual Error } & 14 &\_\_\_ & 52439 & &\\ \text { Total }&15&\_\_\_ \end{array}$ -What is the estimated y-intercept?

The answer of Consider the following partial computer output from...

Test to determine if there is a significant correlation between x and y Use H0: ρ = 0 versus Ha: ρ ≠ 0 by setting α = .01

Reject the null hypothesis,there is a si

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis $\begin{array}{rc} \mathrm{r}^{2} 0.762 & \mathrm{n} 7 \\ \mathrm{R} 0.873 & \mathrm{k} \text { 1 } \\ \text { Std. Error 11.547 } & \text { Dep. Var. Sales } \end{array}$ ANOVA table $\begin{array}{rrrrrr} \hline \text { Source } & S S & d f & M S & F & p \text {-value } \\ \hline \text { Regression } & 2,133.3333 & 1 & 2,133.3333 & 16.00 & .0103 \\ \text { Residual } & 666.6667 & 5 & 133.3333 & & \\ \hline \text { Total } & 2,800.0000 & 6 & & &\\\hline \end{array}$ $\text { Regression output }$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\text { Confidence interval }$ $\begin{array}{rrrrrrrr} \hline\text { Variables} & \text { Coefficients } & \text { std. error } & t(d f=5) & \text { p-value } & 95 \% &95 \% \text { upper } \\ & & & & & \text { lower } & \\ \hline \text { Intercep } & 63.3333 & 7.9682 & 7.948 & .0005 & 42.8505 & 83.8162 \\ \text { Advertising } & 6.6667 & 1.6667 & 4.000 & .0103 & & \end{array}$ -What is the value of the coefficient of determination?

A) 11.547 B) 762 C) 873 D) 6.6667 E) 1.6667 A) 11.547 B) 762 C) 873 D) 6.6667 E) 1.6667

Consider the following partial computer output from a simple linear regression analysis. \[\begin{array} { l c c c c } \text { Predictor } & \text { Coef } & \text { SE Coef } & \text { T } & \text { P } \\ \text { Constant } & 67.05 & 20.90 & 3.21 & 0.012 \\ \text { Independent Var } & 5.8167 & 0.7085 & \_\_\_& 0.000 \end{array}\] S = _________ R-Sq = _______ Analysis of Variance \[\begin{array} { l c c c c c } \text { Source } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\ \text { Regression } & 1 & - & 34920 & 67.39 & 0.000 \\ \text { Residual Error } & 8 & - & 518 & & \\ \text { Total } & 9 & 39065 & & & \end{array}\] -What is the estimated slope?

The answer of Consider the following partial computer output from...

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis $\begin{array}{rc} \mathrm{r}^{2} 0.762 & \mathrm{n} 7 \\ \mathrm{R} 0.873 & \mathrm{k} \text { 1 } \\ \text { Std. Error 11.547 } & \text { Dep. Var. Sales } \end{array}$ ANOVA table $\begin{array}{rrrrrr} \hline \text { Source } & S S & d f & M S & F & p \text {-value } \\ \hline \text { Regression } & 2,133.3333 & 1 & 2,133.3333 & 16.00 & .0103 \\ \text { Residual } & 666.6667 & 5 & 133.3333 & & \\ \hline \text { Total } & 2,800.0000 & 6 & & &\\\hline \end{array}$ $\text { Regression output }$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\text { Confidence interval }$ $\begin{array}{rrrrrrrr} \hline\text { Variables} & \text { Coefficients } & \text { std. error } & t(d f=5) & \text { p-value } & 95 \% &95 \% \text { upper } \\ & & & & & \text { lower } & \\ \hline \text { Intercep } & 63.3333 & 7.9682 & 7.948 & .0005 & 42.8505 & 83.8162 \\ \text { Advertising } & 6.6667 & 1.6667 & 4.000 & .0103 & & \end{array}$ -What are the limits of the 95% confidence interval for the population slope?

A) 5.00 to 8.333 B) 2.667 to 10.667 C) 4.096 to 9.238 D) 2.382 to 10.952 E) 3.308 to 10.025 A) 5.00 to 8.333 B) 2.667 to 10.667 C) 4.096 to 9.238 D) 2.382 to 10.952 E) 3.308 to 10.025

Exam 11: Correlation Coefficient and Simple Linear Regression Analysis

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis 0.762 7 0.873 1 Std. Error 11.547 Dep. Var. Sales ANOVA table Source SS df MS F p -value Regression 2,133.3333 1 2,133.3333 16.00 .0103 Residual 666.6667 5 133.3333 Total 2,800.0000 6 $\text { Regression output }$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\text { Confidence interval }$ Variables Coefficients std. error t(df=5) p-value 95\% 95\% upper lower Intercep 63.3333 7.9682 7.948 .0005 42.8505 83.8162 Advertising 6.6667 1.6667 4.000 .0103 -Determine a 95% confidence interval estimate of the daily average store sales based on $3000 advertising expenditures? The distance value for this particular prediction is reported as .164.

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Question 1

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C

What is the explained variation?

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Question 2

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Explained variation = Total variation - Unexplained variation = 11.324 - 3.073 = 8.251

An experiment was performed on a certain metal to determine if the strength is a function of heating time.Results based on 10 metal sheets are given below.Use the simple linear regression model. $\sum X$ = 30 $\sum X ^ { 2 }$ = 104 $\sum Y$ = 40 $\sum Y ^ { 2 }$ = 178 $\sum X Y$ = 134 -Calculate the correlation coefficient.

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Question 3

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.882 $\begin{array} { l } S S _ { X Y } = 134 - \frac { ( 30 ) ( 40 ) } { 10 } = 14 \\S S _ { Y Y } = 178 - \frac { ( 40 ) ^ { 2 } } { 10 } = 18 \\S S _ { X X } = 104 - \frac { ( 30 ) ^ { 2 } } { 10 } = 14 \\r = \frac { 14 } { \sqrt { ( 18 ) ( 14 ) } } = .882\end{array}$

Calculate the t statistic and then using appropriate rejection point,test H₀:

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Question 4

An experiment was performed on a certain metal to determine if the strength is a function of heating time.Results based on 10 metal sheets are given below.Use the simple linear regression model. $\sum X$ = 30 $\sum X ^ { 2 }$ = 104 $\sum Y$ = 40 $\sum Y ^ { 2 }$ = 178 $\sum X Y$ = 134 -Determine the standard error.

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Question 5

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis 0.762 7 0.873 1 Std. Error 11.547 Dep. Var. Sales ANOVA table Source SS df MS F p -value Regression 2,133.3333 1 2,133.3333 16.00 .0103 Residual 666.6667 5 133.3333 Total 2,800.0000 6 $\text { Regression output }$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\text { Confidence interval }$ Variables Coefficients std. error t(df=5) p-value 95\% 95\% upper lower Intercep 63.3333 7.9682 7.948 .0005 42.8505 83.8162 Advertising 6.6667 1.6667 4.000 .0103 -What is the least-square prediction equation?

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Question 6

The sample ______________________ measures the strength and direction of the linear relationship between two quantitative variables.

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Question 7

The experimental region is the range of the previously observed values of the dependent variable.

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Question 8

Consider the following partial computer output from a simple linear regression analysis: Predictor Coef SE Coef Constant 5566.1 254.0 21.91 0.000 Independent Var -210.35 24.19 - S = _________ R-Sq = Analysis of Variance Source DF SS MS F P Regression 1 3963719 3963719 75.59 0.000 Residual Error 14 \_\_\_ 52439 Total 15 \_\_\_ -What is the estimated y-intercept?

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Question 9

A local tire dealer wants to predict the number of tires sold each month.The dealer believes that the number of tires sold is a linear function of the amount of money invested in advertising.The dealer randomly selects 6 months of data consisting of tire sales (in thousands of tires)and advertising expenditures (in thousands of dollars).Based on the data set with 6 observations,the simple linear regression equation of the least squares line is $\hat { y }$ = 3 + 1x. $\sum X$ = 24 $\sum X ^ { 2 }$ = 124 $\sum Y$ = 42 $\sum Y ^ { 2 }$ = 338 $\sum X Y$ = 196 MSE = 4 -Use the least squares regression equation and estimate the monthly tire sales when advertising expenditures is $4000

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Question 10

Determine the 95% confidence interval for the mean value of metal strength when the average heating time is 4 minutes.Provide an interpretation of this interval.

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Question 11

A data set with 7 observed pairs of data (x, y) yielded the following statistics. $\sum X$ =21.57 $\sum X ^ { 2 }$ =68.31 $\sum Y$ =188.9 $\sum Y ^ { 2 }$ =5140.23 $\sum X Y$ =590.83 SSE = unexplained variation = 1.06 -You wish to perform a simple linear regression analysis using x as the independent variable and y as the dependent variable.What is the standard error?

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Question 12

Test to determine if there is a significant correlation between x and y Use H₀: ρ = 0 versus H_a: ρ ≠ 0 by setting α = .01

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Question 13

The coefficient of determination is the proportion of total variation explained by the regression line.

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Question 14

In a simple linear regression analysis,if the correlation coefficient is positive,then:

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Question 15

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis 0.762 7 0.873 1 Std. Error 11.547 Dep. Var. Sales ANOVA table Source SS df MS F p -value Regression 2,133.3333 1 2,133.3333 16.00 .0103 Residual 666.6667 5 133.3333 Total 2,800.0000 6 $\text { Regression output }$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\text { Confidence interval }$ Variables Coefficients std. error t(df=5) p-value 95\% 95\% upper lower Intercep 63.3333 7.9682 7.948 .0005 42.8505 83.8162 Advertising 6.6667 1.6667 4.000 .0103 -What is the value of the coefficient of determination?

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Question 16

Consider the following partial computer output from a simple linear regression analysis. Predictor Coef SE Coef T P Constant 67.05 20.90 3.21 0.012 Independent Var 5.8167 0.7085 \_\_\_ 0.000 S = ___ R-Sq = _ Analysis of Variance Source DF SS MS F P Regression 1 - 34920 67.39 0.000 Residual Error 8 - 518 Total 9 39065 -What is the estimated slope?

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Question 17

For simple linear regression model,the least-squares line is the equation that minimizes the sum of the squared deviations between each observed value of y and the line.

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Question 18

A local grocery store wants to predict the daily sales in dollars. The manager believes that the amount of newspaper advertising significantly affects the store sales. The manager randomly selects 7 days of data consisting of daily grocery store sales (in thousands of dollars) and advertising expenditures (in thousands of dollars). The Excel/Mega-Stat output given below summarizes the results of fitting a simple linear regression model using this data. Regression Analysis 0.762 7 0.873 1 Std. Error 11.547 Dep. Var. Sales ANOVA table Source SS df MS F p -value Regression 2,133.3333 1 2,133.3333 16.00 .0103 Residual 666.6667 5 133.3333 Total 2,800.0000 6 $\text { Regression output }$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\text { Confidence interval }$ Variables Coefficients std. error t(df=5) p-value 95\% 95\% upper lower Intercep 63.3333 7.9682 7.948 .0005 42.8505 83.8162 Advertising 6.6667 1.6667 4.000 .0103 -What are the limits of the 95% confidence interval for the population slope?

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Question 19

When using simple linear regression,we would like to use confidence intervals for the _ and prediction intervals for the _ at a given value of x.

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Question 20

What is the explained variation?

Calculate the t statistic and then using appropriate rejection point,test H₀:

The sample ______________________ measures the strength and direction of the linear relationship between two quantitative variables.

The experimental region is the range of the previously observed values of the dependent variable.

Determine the 95% confidence interval for the mean value of metal strength when the average heating time is 4 minutes.Provide an interpretation of this interval.

Test to determine if there is a significant correlation between x and y Use H₀: ρ = 0 versus H_a: ρ ≠ 0 by setting α = .01

The coefficient of determination is the proportion of total variation explained by the regression line.

In a simple linear regression analysis,if the correlation coefficient is positive,then:

For simple linear regression model,the least-squares line is the equation that minimizes the sum of the squared deviations between each observed value of y and the line.

When using simple linear regression,we would like to use confidence intervals for the _ and prediction intervals for the _ at a given value of x.

An Introduction to Business Statistics

Descriptive Statistics

Probability

Discrete Random Variables

Continuous Random Variables

Sampling Distributions

Confidence Intervals

Hypothesis Testing

Statistical Inferences Based on Two Samples

Experimental Design and Analysis of Variance

Multiple Regression and Model Building

Nonparametric Methods

Chi-Square Tests

Decision Theory

Time Series Forecasting

Filters

Exam 11: Correlation Coefficient and Simple Linear Regression Analysis

What is the explained variation?

Calculate the t statistic and then using appropriate rejection point,test H0:

The sample ______________________ measures the strength and direction of the linear relationship between two quantitative variables.

The experimental region is the range of the previously observed values of the dependent variable.

Determine the 95% confidence interval for the mean value of metal strength when the average heating time is 4 minutes.Provide an interpretation of this interval.

Test to determine if there is a significant correlation between x and y Use H0: ρ = 0 versus Ha: ρ ≠ 0 by setting α = .01

The coefficient of determination is the proportion of total variation explained by the regression line.

In a simple linear regression analysis,if the correlation coefficient is positive,then:

For simple linear regression model,the least-squares line is the equation that minimizes the sum of the squared deviations between each observed value of y and the line.

When using simple linear regression,we would like to use confidence intervals for the _____ and prediction intervals for the _____ at a given value of x.

An Introduction to Business Statistics

Descriptive Statistics

Probability

Discrete Random Variables

Continuous Random Variables

Sampling Distributions

Confidence Intervals

Hypothesis Testing

Statistical Inferences Based on Two Samples

Experimental Design and Analysis of Variance

Multiple Regression and Model Building

Nonparametric Methods

Chi-Square Tests

Decision Theory

Time Series Forecasting

Filters

Calculate the t statistic and then using appropriate rejection point,test H₀:

Test to determine if there is a significant correlation between x and y Use H₀: ρ = 0 versus H_a: ρ ≠ 0 by setting α = .01

When using simple linear regression,we would like to use confidence intervals for the _ and prediction intervals for the _ at a given value of x.