Suppose that you take a random sample of 259 people leaving a grocery store over the course of a day and find that 12% of these people were overcharged. Find a 95% confidence interval for the actual percentage of shoppers who were overcharged. a. 5.7% to 18.3%
b. 8.85 to 15.15%
c. 7% to 17%
d. 9.5% to 14.5%

8.04% to 15.96%

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

Suppose that you take a random sample of 259 people leaving a grocery store over the course of a day and find that 12% of these people were overcharged.

This means that

95% confidence level

So , z is the value of Z that has a p-value of , so .

The lower limit of this interval is:

The upper limit of this interval is:

So 8.04% to 15.96%

can you put the picture up?

step-by-step explanation: