Exam 13: Simple Linear Regression Analysis

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An experiment was performed on a certain metal to determine if the strength is a function of heating time.The simple linear regression equation is  An experiment was performed on a certain metal to determine if the strength is a function of heating time.The simple linear regression equation is   = 1 + 1X and sample coefficient of determination (r<sup>2</sup>)= .7777.The time is in minutes and the strength is measured in pounds per square inch.Test to determine if there is a significant correlation between the heating time and strength of the metal. Using H<sub>0</sub>:  \rho  = 0 vs.H<sub>A</sub>:  \rho   \neq  0 at  \alpha  = .05,determine the test statistic and decision. = 1 + 1X and sample coefficient of determination (r2)= .7777.The time is in minutes and the strength is measured in pounds per square inch.Test to determine if there is a significant correlation between the heating time and strength of the metal. Using H0: ρ\rho = 0 vs.HA: ρ\rho \neq 0 at α\alpha = .05,determine the test statistic and decision.

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An experiment was performed on a certain metal to determine if the strength is a function of heating time.95% prediction interval for the strength of a metal sheet when the average heating time is 4 minutes is from 3.235 to 6.765.We are 95% confident that an individual sheet of metal heated for four minutes will have strength of at least 4 pounds per square inch.Do you agree with this statement?

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All of the following are assumptions of the error terms in the simple linear regression model except

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If there is significant autocorrelation present in a data set the ________________ assumption is violated.

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A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000. = 3 + 1x. A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000. = 24 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000. = 124 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000. = 42 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000. = 338 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000. = 196 MSE = 4 Using the sums of the squares given above,determine the 90% prediction interval for an individual month's tire sales when the advertising expenditure is $5000.

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A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 model yielded the following results. A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 model yielded the following results.   = 24   = 124   = 42   = 338   = 196 Determine the value of the estimated y intercept. = 24 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 model yielded the following results.   = 24   = 124   = 42   = 338   = 196 Determine the value of the estimated y intercept. = 124 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 model yielded the following results.   = 24   = 124   = 42   = 338   = 196 Determine the value of the estimated y intercept. = 42 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 model yielded the following results.   = 24   = 124   = 42   = 338   = 196 Determine the value of the estimated y intercept. = 338 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 model yielded the following results.   = 24   = 124   = 42   = 338   = 196 Determine the value of the estimated y intercept. = 196 Determine the value of the estimated y intercept.

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The residual is the difference between the observed value of the dependent variable and the predicted value of the dependent variable.

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An experiment was performed on a certain metal to determine if the strength is a function of heating time.The sample size consists of ten metal sheets.Residuals are calculated for all ten metal sheets and ordered from smallest to largest. Determine the normal point for the smallest residual.

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For the same set of observations on a specified dependent variable two different independent variables were used to develop two separate simple linear regression models.A portion of the results is presented below. For the same set of observations on a specified dependent variable two different independent variables were used to develop two separate simple linear regression models.A portion of the results is presented below.   Based on the results given above,we can conclude that: Based on the results given above,we can conclude that:

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The ___________ the r2,and the __________ the s (standard error),the stronger the relationship between the dependent variable and the independent variable.

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Regression Analysis Regression Analysis   The 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.He 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 above summarizes the results of the regression model. What are the limits of the 99% prediction interval of the daily sales in dollars of an individual grocery store that has spent $3000 on advertising expenditures? The distance value for this particular prediction is reported as .164. The 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.He 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 above summarizes the results of the regression model. What are the limits of the 99% prediction interval of the daily sales in dollars of an individual grocery store that has spent $3000 on advertising expenditures? The distance value for this particular prediction is reported as .164.

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An experiment was performed on a certain metal to determine if the strength is a function of heating time.Partial results based on a sample of 10 metal sheets are given below.The simple linear regression equation is An experiment was performed on a certain metal to determine if the strength is a function of heating time.Partial results based on a sample of 10 metal sheets are given below.The simple linear regression equation is   = 1 + 1X .The time is in minutes and the strength is measured in pounds per square inch,MSE = 0.5,   = 30,   = 104. Determine the 95% prediction interval for the strength of a metal sheet when the average heating time is 4 minutes.The distance value has been found to be equal to 0.17143. = 1 + 1X .The time is in minutes and the strength is measured in pounds per square inch,MSE = 0.5, An experiment was performed on a certain metal to determine if the strength is a function of heating time.Partial results based on a sample of 10 metal sheets are given below.The simple linear regression equation is   = 1 + 1X .The time is in minutes and the strength is measured in pounds per square inch,MSE = 0.5,   = 30,   = 104. Determine the 95% prediction interval for the strength of a metal sheet when the average heating time is 4 minutes.The distance value has been found to be equal to 0.17143. = 30, An experiment was performed on a certain metal to determine if the strength is a function of heating time.Partial results based on a sample of 10 metal sheets are given below.The simple linear regression equation is   = 1 + 1X .The time is in minutes and the strength is measured in pounds per square inch,MSE = 0.5,   = 30,   = 104. Determine the 95% prediction interval for the strength of a metal sheet when the average heating time is 4 minutes.The distance value has been found to be equal to 0.17143. = 104. Determine the 95% prediction interval for the strength of a metal sheet when the average heating time is 4 minutes.The distance value has been found to be equal to 0.17143.

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The following results were obtained as a part of simple regression analysis: r2= .9162 F statistic from the F table = 3.59 Calculated value of F from the ANOVA table = 81.87 The following results were obtained as a part of simple regression analysis: r<sup>2</sup>= .9162 F statistic from the F table = 3.59 Calculated value of F from the ANOVA table = 81.87   = .05 P-value = .000 The null hypothesis of no linear relationship between the dependent variable and the independent variable = .05 P-value = .000 The null hypothesis of no linear relationship between the dependent variable and the independent variable

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In simple regression analysis,the quantity In simple regression analysis,the quantity   is called the __________ sum of squares. is called the __________ sum of squares.

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The ______ is the range of the previously observed values of x.

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After plotting the data point s on a scatter diagram,we have observed an inverse relationship between the independent variable (X)and the dependent variable (Y).Therefore,we can expect both the sample slope and the __________ to be a negative value.

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Consider the following partial computer output from a simple linear regression analysis. Consider the following partial computer output from a simple linear regression analysis.   What is the predicted value of y when x = 1,000? What is the predicted value of y when x = 1,000?

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Consider the following partial computer output from a simple linear regression analysis. Consider the following partial computer output from a simple linear regression analysis.   S = 0.4862 R-Sq = ______ Analysis of Variance   What is the estimated slope? S = 0.4862 R-Sq = ______ Analysis of Variance Consider the following partial computer output from a simple linear regression analysis.   S = 0.4862 R-Sq = ______ Analysis of Variance   What is the estimated slope? What is the estimated slope?

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A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000. = 3 + 1x. A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000. = 24 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000. = 124 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000. = 42 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000. = 338 A local tire dealer wants to predict the number of tires sold each month.He believes that the number of tires sold is a linear function of the amount of money invested in advertising.He 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   = 3 + 1x.   = 24   = 124   = 42   = 338   = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000. = 196 MSE = 4 Using the sums of the squares given above,determine the 90% confidence interval for the mean value of monthly tire sales when the advertising expenditure is $5000.

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An experiment was performed on a certain metal to determine if the strength is a function of heating time.95% confidence interval for the average strength of a metal sheet when the average heating time is 4 minutes is from 4.325 to 5.675.Therefore,we are confident at β\beta = .05 that the average strength of metal heated for four minutes is between 4.325 and 5.675 pounds per square inch.Do you agree or disagree with this statement?

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