In: Statistics and Probability
Dixie Showtime Movie Theaters, Inc., owns and operates a chain of cinemas in several markets in the southern U.S. The owners would like to estimate weekly gross revenue as a function of advertising expenditures. Data for a sample of eight markets for a recent week follow. Market Weekly Gross Revenue ($100s) Television Advertising ($100s) Newspaper Advertising ($100s) Mobile 103.5 4.9 1.7 Shreveport 51.6 3.3 3 Jackson 75.8 4 1.5 Birmingham 127.8 4.2 4 Little Rock 134.8 3.2 4.3 Biloxi 101.4 3.6 2.2 New Orleans 237.8 5 8.5 Baton Rouge 219.6 6.7 5.9 (a) Use the data to develop an estimated regression with the amount of television advertising as the independent variable. Let x represent the amount of television advertising. If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300) = -45.432 + 40.064 x Test for a significant relationship between television advertising and weekly gross revenue at the 0.05 level of significance. What is the interpretation of this relationship? The input in the box below will not be graded, but may be reviewed and considered by your instructor. blank (b) How much of the variation in the sample values of weekly gross revenue does the model in part (a) explain? If required, round your answer to two decimal places. 56 % (c) Use the data to develop an estimated regression equation with both television advertising and newspaper advertising as the independent variables. Let x1 represent the amount of television advertising. Let x2 represent the amount of newspaper advertising. If required, round your answers to three decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300) = -42.57 + 22.402 x1 + 19.5 x2 Test whether each of the regression parameters β0, β1, and β2 is equal to zero at a 0.05 level of significance. What are the correct interpretations of the estimated regression parameters? Are these interpretations reasonable? The input in the box below will not be graded, but may be reviewed and considered by your instructor. blank (d) How much of the variation in the sample values of weekly gross revenue does the model in part (c) explain? If required, round your answer to two decimal places. 93.22 %
Solution:
a. The following is the excel output used to estimate the regression equation.
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.719757 | |||||
R Square | 0.518051 | |||||
Adjusted R Square | 0.437726 | |||||
Standard Error | 49.30337 | |||||
Observations | 8 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 1 | 15677.45 | 15677.45 | 6.449443 | 0.044106 | |
Residual | 6 | 14584.93 | 2430.822 | |||
Total | 7 | 30262.38 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | -46.8238 | 72.36359 | -0.64706 | 0.541539 | -223.891 | 130.2435 |
X Variable 1 | 40.88511 | 16.09919 | 2.539575 | 0.044106 | 1.491808 | 80.27842 |
The estimated regression equation is Y = -46.8238 + 40.8851X
The p-value associated with b1 is 0.0441. We reject the hypothesis that 1 because the p-value is less than 0.05 significance level. We can conclude that there is a relationship between television advertising and weekly gross revenue at 0.05 level of significance. If we hold newspaper advertising constant, a $100 increase in television advertising corresponds to the increase of $4088.51 in weekly gross revenue. It seems to be reasonable.
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b. The coefficient of determination, R-square us 0.5181 so the regression model computed in part c explains approximately 51.81% of the variation in the weekly gross revenue.
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c. The following is the excel output used to estimate the regression equation.
SUMMARY OUTPUT | ||||||
Regression Statistics | ||||||
Multiple R | 0.959913 | |||||
R Square | 0.921433 | |||||
Adjusted R Square | 0.890007 | |||||
Standard Error | 21.80647 | |||||
Observations | 8 | |||||
ANOVA | ||||||
df | SS | MS | F | Significance F | ||
Regression | 2 | 27884.77 | 13942.38 | 29.32015 | 0.00173 | |
Residual | 5 | 2377.612 | 475.5223 | |||
Total | 7 | 30262.38 | ||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | |
Intercept | -40.2681 | 32.03196 | -1.25712 | 0.264235 | -122.609 | 42.07272 |
X Variable 1 | 21.66028 | 8.068409 | 2.684578 | 0.043581 | 0.91977 | 42.40078 |
X Variable 2 | 19.88749 | 3.925141 | 5.066694 | 0.003878 | 9.797592 | 29.97738 |
The estimated multiple linear regression equation is
y = -40.2681 + 21.6603 x1 + 19.8875 x2
The p-value associated with b1 is 0.0436. We reject the hypothesis that 1 because the p-value is less than 0.05 significance level. We can conclude that there is a relationship between television advertising and weekly gross revenue at 0.05 level of significance. If we hold newspaper advertising constant, a $100 increase in television advertising corresponds to the increase of $2216.03 in weekly gross revenue. It seems to be reasonable.
The p-value associated with b2 is 0.0039. We reject the hypothesis that 2 because the p-value is less than 0.05 significance level. We can conclude that there is a relationship between newspaper advertising and weekly gross revenue at 0.05 level of significance. If we hold television advertising constant, a $100 increase in newspaper advertising corresponds to the increase of $1988.75 in weekly gross revenue. It seems to be reasonable.
The regression parameter b0 indicates that when television advertising and newspaper advertising are both zero, the predicted weekly gross revenue is $4026.81. This result is not real because this estimate and test of the hypothesis that 0 have no meaning because the intercept is estimated using extrapolation.
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d. The coefficient of determination, R-square us 0.9214, so the regression model computed in part c explains approximately 92.14% of the variation in the weekly gross revenue.