Final Answer:
(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.