Question

A statistical program is recommended.

The owner of Showtime Movie Theaters, Inc., would like to predict weekly gross revenue as a function of advertising expenditures. Historical data for a sample of eight weeks follow.
Weekly
Gross
Revenue
($1,000s) Television
Advertising
($1,000s) Newspaper
Advertising
($1,000s)
96 5.0 1.5
90 2.0 2.0
95 4.0 1.5
92 2.5 2.5
95 3.0 3.3
94 3.5 2.3
94 2.5 4.2
94 3.0 2.5
The owner then used multiple regression analysis to predict gross revenue (y), in thousands of dollars, as a function of television advertising (x1), in thousands of dollars, and newspaper advertising (x2), in thousands of dollars. The estimated regression equation was
ŷ = 83.2 + 2.29x1 + 1.30x2.
(a)What is the gross revenue (in dollars) expected for a week when $2,000 is spent on television advertising
(x1 = 2) and $2,000 is spent on newspaper advertising (x2 = 2)? (Round your answer to the nearest dollar.)
$
(b)Provide a 95% confidence interval (in dollars) for the mean revenue of all weeks with the expenditures listed in part (a). (Round your answers to the nearest dollar.)
$ to $
(c)Provide a 95% prediction interval (in dollars) for next week's revenue, assuming that the advertising expenditures will be allocated as in part (a). (Round your answers to the nearest dollar.)
$ to $

Answer 1

(a) The gross revenue expected for a week when $2,000 is spent on television advertising (x₁ = 2) and $2,000 is spent on newspaper advertising (x₂ = 2) is approximately $89,590.

(b) The 95% confidence interval for the mean revenue of all weeks with the expenditures listed in part (a) is approximately $88,076 to $91,103.

(c) The 95% prediction interval for next week's revenue, assuming that the advertising expenditures will be allocated as in part (a), is approximately $85,953 to $92,226.

Step-by-step explanation:

In part (a), we can substitute the given values into the estimated regression equation ŷ = 83.2 + 2.29x₁ + 1.30x₂ to find the gross revenue. Plugging in x₁ = 2 and x₂ = 2, we get ŷ = 83.2 + 2.29(2) + 1.30(2) ≈ $89,590.

For part (b), the 95% confidence interval can be calculated using the formula: Confidence Interval = ŷ ± tα/2 * SE(ŷ), where ŷ is the predicted value, tα/2 is the t-score for a 95% confidence interval (with 6 degrees of freedom in this case), and SE(ŷ) is the standard error of the estimate. Plugging in the values, we get a confidence interval of approximately $88,076 to $91,103.

In part (c), the 95% prediction interval can be calculated using the formula: Prediction Interval = ŷ ± tα/2 * SE(prediction), where SE(prediction) is the standard error of the prediction. Similar to part (b), the values are substituted into the formula, resulting in a prediction interval of approximately $85,953 to $92,226. This interval reflects the range within which we can be 95% confident that the actual revenue for the next week will fall, considering the given advertising expenditures.

Answer 2

The expected weekly gross revenue when $2,000 is spent on both television and newspaper advertising is $90,380. To provide the 95% confidence and prediction intervals, additional information such as standard errors would be required, which is not supplied in the context of the question.

Step-by-step explanation:

The owner of Showtime Movie Theaters, Inc. is seeking to predict weekly gross revenue based on advertising expenditures using multiple regression analysis. Given the regression equation ˅ = 83.2 + 2.29x1 + 1.30x2, we can calculate the predicted values.

(a) To find the expected gross revenue when $2,000 is spent on television advertising (x1 = 2) and $2,000 on newspaper advertising (x2 = 2), we plug these values into the regression equation:

˅ = 83.2 + (2.29 × 2) + (1.30 × 2) = 83.2 + 4.58 + 2.6 = 90.38

Therefore, the expected weekly gross revenue is $90,380 (rounded off to the nearest dollar).

(b) To provide a 95% confidence interval for the mean revenue, and (c) a 95% prediction interval for next week's revenue, additional statistical analysis is required, specifically calculation of the standard error of the estimate for the confidence interval and the inclusion of the prediction error for the prediction interval. Without sufficient additional information such as the standard error or the exact distribution of the residuals, it is not possible to produce these intervals.