AP Stats Questions Pls Help!

Question

A program was created to randomly choose customers at a sporting goods store to receive a discount. The program claims 25% of the receipts will get a discount in the long run. The manager of the sporting goods store is skeptical and believes the program's calculations are incorrect. She selects a random sample and finds that 22% received the discount. The confidence interval is 0.22 ± 0.04 with all conditions for inference met. (10 points)

Part A: Using the given confidence interval, is it statistically evident that the program is not working? Explain. (3 points)

Part B: Is it statistically evident from the confidence interval that the program creates the discount with a 0.25 probability? Explain. (2 points)

Part C: Another random sample of receipts is taken. This sample is four times the size of the original. Twenty-two percent of the receipts in the second sample received the discount. What is the value of margin of error based on the second sample with the same confidence level as the original interval? (2 points)

Part D: Using the margin of error from the second sample in part C, is the program working as planned? Explain. 

Answer 1

Part A: The endpoints of the sample confidence interval are 0.18 and 0.26 (0.22 ± 0.04). Since that interval includes 0.25, there is insufficient evidence to reject the null hypothesis that the true proportion is 0.25. So no, it is not statistically evident that the program is not working.

Part B: Again, the endpoints are 0.18 and 0.26. There is insufficient evidence to reject the null hypothesis that the true proportion is 0.25. In this case, the answer to the question as written is "Yes."

Part C: Same scenario as Part A but with 4n as the sample size instead of n. Confidence intervals with proportions are calculated using z-statistics and standard error of means, with the square root of the sample size as the denominator. z*se = z*s/sqrt(n). Taking that formula with 4n in the denominator, we get z*se = z*s/(sqrt(4n) = z*s/[sqrt(4)*sqrt(n)] = z*s/(2*sqrt(n)). So quadrupling the sample size cuts the standard error of mean in half, which in turn cuts the calculated confidence interval in half. So the margin of error would be 0.22 ± 0.02.

Part D: The new endpoints of the confidence interval are 0.20 and 0.24. With the larger sample size, we now reject the null hypothesis that the true proportion is 0.25, as the value of 0.25 falls outside of the interval. In this case, there is sufficient evidence to say that the program is not working as intended.