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

Step-by-step 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.

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

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that $n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

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

The upper limit of this interval is:

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

The 95% confidence interval for the fraction 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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3 years ago