# Relationship

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A movie studio wishes to determine the relationship between the revenue from the rental of comedies on DVD and videotape and the revenue generated from the theatrical release of such movies. The studio has bivariate data from a sample of comedies released over the past five years. These data give the revenue from theatrical release (in millions of dollars) and the revenue from DVD and videotape rentals (in millions of dollars) for each of the movies. The least-squares regression equation computed from the data is , with an error sum of squares of .

The studio has plans for the DVD/video release of a comedy that grossed about million dollars in theaters. The studio's data yield the computation

in which denote the sample data values for theatrical release revenue, and denotes their mean.

Based on this information, and assuming that the regression assumptions hold, answer the questions in the table below.

1. The mean square error (MSE) provides an estimate of an important parameter in the simple linear regression model. What is the value of the MSE for the data? (Round your answer to at least two decimal places) _____

2. What is the lower limit of the 90% prediction interval for an individual value for rental revenue when the theater revenue is 44.7 million dollars? (Carry your intermediate computations to at least 4 decimal places, and round your answer to at least 1 decimal place. _____

3. What is the upper limit of the 90% prediction interval for an individual value for rental revenue when the theater revenue is 44.7 million dollars? (Carry your intermediate computations to at least 4 decimal places, and round your answer to at least 1 decimal place. _____

4. For the theater revenue in this sample, 21.4 million dollars is more extreme than 44.7 million dollars is, that is, 21.4 is farther from the sample mean theater revenue than 44.7 is. How would the 90% prediction interval for the mean rental revenue when the theater revenue is 21.4 million dollars compare to the 90% prediction interval for the mean rental revenue when the theater revenue is 44.7 million dollars?

Choose one response to answer the question below.

a. The intervals would be identical

b. The interval computer from a theater revenue of 21.4 would be wider and have a different center.

c. The interval computer from a theater revenue of 21.4 would be narrower but have the same center.

d. The interval computer from a theater revenue of 21.4 would be wider but have the same center.

e. The interval computer from a theater revenue of 21.4 would be narrower and have a different center.

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##### Solution Summary

A movie studio wishes to determine the relationship between the revenue from the rental of comedies on DVD and videotape and the revenue generated from the theatrical release of such movies. The studio has bivariate data from a sample of comedies released over the past five years. These data give the revenue from theatrical release (in millions of dollars) and the revenue from DVD and videotape rentals (in millions of dollars) for each of the movies. The least-squares regression equation computed from the data is , with an error sum of squares of .

The studio has plans for the DVD/video release of a comedy that grossed about million dollars in theaters. The studio's data yield the computation

(Picture I sent you goes here)

in which denote the sample data values for theatrical release revenue, and denotes their mean.

Based on this information, and assuming that the regression assumptions hold, answer the questions in the table below.

1. The mean square error (MSE) provides an estimate of an important parameter in the simple linear regression model. What is the value of the MSE for the data? (Round your answer to at least two decimal places) _____

2. What is the lower limit of the 90% prediction interval for an individual value for rental revenue when the theater revenue is 44.7 million dollars? (Carry your intermediate computations to at least 4 decimal places, and round your answer to at least 1 decimal place. _____

3. What is the upper limit of the 90% prediction interval for an individual value for rental revenue when the theater revenue is 44.7 million dollars? (Carry your intermediate computations to at least 4 decimal places, and round your answer to at least 1 decimal place. _____

4. For the theater revenue in this sample, 21.4 million dollars is more extreme than 44.7 million dollars is, that is, 21.4 is farther from the sample mean theater revenue than 44.7 is. How would the 90% prediction interval for the mean rental revenue when the theater revenue is 21.4 million dollars compare to the 90% prediction interval for the mean rental revenue when the theater revenue is 44.7 million dollars?

Choose one response to answer the question below.

a. The intervals would be identical

b. The interval computer from a theater revenue of 21.4 would be wider and have a different center.

c. The interval computer from a theater revenue of 21.4 would be narrower but have the same center.

d. The interval computer from a theater revenue of 21.4 would be wider but have the same center.

e. The interval computer from a theater revenue of 21.4 would be narrower and have a different center.

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
- "Thank you"
- "Thank you very much for your valuable time and assistance!"

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