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

Part a)

Confidence Interval = 0.22 ± 0.04 = (0.22 - 0.04, 0.22 + 0.04) = (0.18, 0.26)

Since, 0.25 lies within the (0.18, 0.26), so it is not statistically evident that the program is not working

Part b)

Since, 0.25 lies within the (0.18, 0.26), so it is statistically evident that the program creates the discount with a 0.25 probability.

Part c)

Margin of Error (ME) Formula = zα/2•√p̂(1 - p̂)/n

From formula of margin of error it can be seen that ME is inversely proportional to square root of n

So if n increase by 4 times, new ME value can be obtained by dividing ole ME value with 2 because proportion p is same that is 22%.

New ME = 0.04/2 = 0.02

Part d)

New Confidence Interval = 0.22 ± 0.02 = (0.22 - 0.02, 0.22 + 0.02) = (0.20, 0.24)

Since, 0.25 lies within the (0.20, 0.24), so it is not statistically evident that the program working as planned.