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Coupons driving visits. A store randomly samples 603 shoppers over the course of a year and nds that 142 of them made their visit because of a coupon they'd received in the mail. Construct a 95% con dence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail.

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Answer:

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:


n = 603, \pi = (142)/(603) = 0.2355

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.2355 - 1.96\sqrt{(0.2355*0.7645)/(603)} = 0.2016

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2355 + 1.96\sqrt{(0.2355*0.7645)/(603)} = 0.2694

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

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