Question

Coupons Driving Visits: 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. 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. (please round all proportions to four decimal places)

a) Estimate the true proportion of shoppers during the year whose visit was because of a coupon they'd received in the mail.

Answer 1

Answer:

$\hat p=\frac{142}{603}=0.235$ represent the estimated proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail

$0.235 - 1.96 \sqrt{\frac{0.235(1-0.235)}{603}}=0.201$  

$0.235 + 1.96 \sqrt{\frac{0.235(1-0.235)}{603}}=0.269$  

And the 95% confidence interval would be given (0.201;0.269).  

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".  

Description in words of the parameter p

$p$ represent the real population proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail

$\hat p$ represent the estimated proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail

n=603 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 all shoppers during the year whose visit was because of a coupon they'd received in the mail

$\hat p=\frac{142}{603}=0.235$ represent the estimated proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail

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:  

$0.235 - 1.96 \sqrt{\frac{0.235(1-0.235)}{603}}=0.201$  

$0.235 + 1.96 \sqrt{\frac{0.235(1-0.235)}{603}}=0.269$  

And the 95% confidence interval would be given (0.201;0.269).  

We are confident at 98% that the true proportion of people that they were planning to pursue a graduate degree is between (0.616;0.686).  

Answer 2

• The estimate of the true proportion of shoppers during the year whose visit was because of a coupon they'd received in the mail is 0.2355.

• 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).

Considering a sample of n, a proportion 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 $1 - \frac{\alpha}{2}$.

Of the 603 shoppers, 142 made the visit because of a coupon received in the mail.

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

The estimate of the true proportion is 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 is (0.2016, 0.2694).

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