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

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1160, \pi = \frac{144}{1160} = 0.1241$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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