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

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

represent the real population proportion of interest

represent the estimated proportion of interest

n=488 is the sample size required  

represent the critical value for the margin of error  

The population proportion have the following distribution  

Numerical estimate for p

In order to estimate a proportion we use this formula:

where X represent the number of people with a characteristic and n the total sample size selected.

represent the estimated proportion of interest

Confidence interval

The confidence interval for a proportion is given by this formula  

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.  

And replacing into the confidence interval formula we got:  

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

b.-4 units

all you   do is count how many times you move

answerim :

n

step-by-step explanation:

im not sure but my dad knows