Question

Direct mail advertisers send solicitations​ ("junk mail") to thousands of potential customers in the hope that some will buy the​ company's product. The response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1160 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 144 of the recipients. Create a 90% confidence interval for the percentage of people the company contacts who may buy something.

Answer 1

The 90% confidence interval for the percentage of people the company contacts who may buy something is between 10.82% and 14%

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 zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

`n = 1160, \pi = (144)/(1160) = 0.1241`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1241 - 1.645\sqrt{(0.1241*0.8759)/(1160)} = 0.1082`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1241 + 1.645\sqrt{(0.1241*0.8759)/(1160)} = 0.14`

The 90% confidence interval for the percentage of people the company contacts who may buy something is between 10.82% and 14%