Question

5.14 Coupons Driving Visits: A store randomly samples 488 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.
(please round all proportions to four decimal places)


a) Construct a 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.


_____ to _____

Answer 1

Answer:

$0.291 - 1.96 \sqrt{\frac{0.291(1-0.291)}{488}}=0.251$  

$0.291 + 1.96 \sqrt{\frac{0.291(1-0.291)}{488}}=0.331$  

And the 95% confidence interval would be given (0.251;0.331).  

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion of interest

n=488 is the sample size required  

$z_{\alpha/2}$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Numerical estimate for p

In order to estimate a proportion we use this formula:

$\hat p =\frac{X}{n}$ where X represent the number of people with a characteristic and n the total sample size selected.

$\hat p=\frac{142}{488}=0.291$ represent the estimated proportion of interest

Confidence interval

The confidence interval for a proportion is given by this formula  

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$0.291 - 1.96 \sqrt{\frac{0.291(1-0.291)}{488}}=0.251$  

$0.291 + 1.96 \sqrt{\frac{0.291(1-0.291)}{488}}=0.331$  

And the 95% confidence interval would be given (0.251;0.331).