Question

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.

Answer 1

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)

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have 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 pvalue 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 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)