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.
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. (3 points)

Answer 1

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nickhollandmck

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Answer:

A: We know that the confidence interval is 0.22 ± 0.04, this equals .22 plus .04, .22 minus .44, or (.18,.26). Because of the fact that .25 is in the range of (.18,.26), there is no statistical evidence that the program is not working as intended.

B: As we established earlier, .25 is within the range of (.18,.26). This means that there is statistical evidence that the program makes the discount with a .25 probability.

C: To solve this problem, we need to use the margin of error formula. The margin of error formula is z times the square rout of p-hat times 1 minus p hat over n. From this formula we see that the me is inversely proportional to the square rout of the sample size. Because this is the case, the new me value is determined by dividing the old me value by 2 because of the fact that the proportion of p is 22%. This means that the new me is equal to .04/2, which equals .02, which means that the new margin of error is .02.

D: We know using the new me that the new CI is .22 ± .02. This also equals .22 plus .02, .22 minus .02, or (.2,.24) Because .25 is not within the range of (.2,.24), there is not statistical evidence that the program is working as intended.  

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