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 140 of the recipients. Use this information to complete parts a through d.

​a) Create a 90​% confidence interval for the percentage of people the company contacts who may buy something.

(___%, ___%)

​(Round to one decimal place as​ needed.)

​b) Explain what this interval means.

A. The values for the interval bounds should be subtracted from 90​% to obtain the​ company's true confidence level.

B. The company is 90​% confident that the percentage of people who will respond to the flyer falls within the confidence interval bounds.

C. The company is 90​% confident that the probability a randomly sampled person will respond to the flyer falls within the confidence interval bounds.

D.The company is 90​% confident that the percentage of people who will not respond to the flyer falls within the confidence interval bounds.

​c) Explain what "90​% ​confidence" means.

A. About 90​% of all randomly sampled people will respond to the new flyer.

B. About 90​% of all random samples will produce intervals that contain the true proportion of people who will respond to the new flyer.

C. About 90​% of all random samples of size 1160 will produce intervals that contain the true proportion of people who will respond to the new flyer.

D. About 90​% of all random samples of size 1160 will produce intervals that do not contain the true proportion of people who will respond to the new flyer.

​d) The company must decide whether to do a mass mailing. The mailing​ won't be​ cost-effective unless it produces at least a 4​% return. What does your confidence interval​ suggest?

A. Do the mass mailing.

B. Do not do the mass mailing.

Answer 1

a)
sample proportion, = 0.1207
sample size, n = 1160
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.1207 * (1 - 0.1207)/1160) = 0.0096

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.645

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.1207 - 1.645 * 0.0096 , 0.1207 + 1.645 * 0.0096)
CI = (0.105 , 0.136)

(10.5% , 13.6%)

b)
B. The company is 90​% confident that the percentage of people who will respond to the flyer falls within the confidence interval bounds.

c)
C. About 90​% of all random samples of size 1160 will produce intervals that contain the true proportion of people who will respond to the new flyer.

d)
Do the mass mailing