"Constructing and Interpreting a Prediction Interval In Exerci...

Question

"Constructing and Interpreting a Prediction Interval In Exercises 21-30, construct the indicated prediction interval and interpret the results.

29. New Vehicle Sales Construct a 95% prediction interval for new vehicle sales for General Motors in Exercise 19 when the number of new vehicles sold by Ford is 2028 thousand."

Accepted Answer

Hello everyone. Let's take a look at this question together. A company uses a regression model to predict monthly sales in thousands of units based on advertising expenditure in thousands of dollars. The regression equation is given as Y hat equals 5 + 3 X. The standard error of estimate is equal to 2, the sample size is N equals 20, the mean advertising expenditure is equal to 8. And the sum of squares of X is equal to 200. Construct a 99% prediction interval for a month when the advertising expenditure is X subzero equals $12,000. Is it answer choice A 41 to 55? Answer choice B, 34.88 to 47.12? Answer choice C 39.5 to 52.5, or answer choice D 38.25 to 49.75. So, in order to solve this question, we have to recall how to construct a 99% prediction interval, so that we can construct one for a month when the advertising expenditure is equal to $12,000 given our regression equation and our standard error of estimate, our sample size, our Mean advertising expenditure and our sum of squares of X. And so the first step in solving this problem is computing the predicted value, where we use the regression equation, and plug in our advertising expenditure. So we have Y hat equals 5 + 3 multiplied by 12, which can be simplified. S 5 + 36 or 41, so our predicted value is equal to 41, and then we compute the standard error of the prediction, which is equal to 2 multiplied by the square root of 1 plus 1 divided by 20 plus 12 minus 8 squared divided by 200. Which can be simplified further to give us to multiplied by 1.0630, which is equal to 2.1260. So our standard error of the prediction is equal to 2.1260. And then our degrees of freedom, we know is N subtracted by 2, which is equal to 18, and using a T table, we know for 99% confidence, our T critical value is equal to 2.878. And then we can compute the margin of error, which is equal to E equals 2.878, multiplied by 2.1260, which is approximately 6.1191. So our prediction interval is given as 41 plus the margin of error, which is 6.1191. So, our prediction interval is calculated as 41, subtracted by 6.1191 to 41 plus 6.1191, or 34.8809 to 47.1191. And when rounded to two decimals, our prediction interval is 34.88 to 47.12. And so looking at our answer choices, we can identify that answer choice B is the correct answer, as answer choice. contains the correct 99% prediction interval for a month when the advertising expenditure is $12,000 and all other intervals are incorrect. I hope you found this video to be helpful. Thank you and goodbye.

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