Mathematics, 05.12.2019 03:31 Bradgarner772

Astore randomly samples 640 shoppers over the course of a year and finds that 173 of them made their visit because of a coupon they'd receive in the mail.

( 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 ( )

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We are confident at 95% that the true proportion of people who made their visit because of a coupon they'd receive in the mail is between (0.236;0.304).

Step-by-step explanation:

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

represent the real population proportion of people who made their visit because of a coupon they'd receive in the mail.

$\hat p$ represent the estimated proportion of people who made their visit because of a coupon they'd receive in the mail.

n=640 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.

$\hat p=\frac{173}{640}=0.270$ represent the estimated proportion of people who made their visit because of a coupon they'd receive 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:

$0.270 - 1.96 \sqrt{\frac{0.270(1-0.270)}{640}}=0.236$

$0.270 + 1.96 \sqrt{\frac{0.270(1-0.270)}{640}}=0.304$

And the 95% confidence interval would be given (0.236;0.304).

We are confident at 95% that the true proportion of people who made their visit because of a coupon they'd receive in the mail is between (0.236;0.304).

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