Question

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%

Answer 1

8.04% to 15.96%

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

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
`n = 259, \pi = 0.12`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a p-value of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.12 - 1.96\sqrt{(0.12*0.88)/(259)} = 0.0804`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.12 + 1.96\sqrt{(0.12*0.88)/(259)} = 0.1596`

So 8.04% to 15.96%