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 1140 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 108 of the recipients. Complete parts a through d below.

​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. Use ascending​ order.)

Answer 1

Solution:

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

Number of favourable observations = x = 108

Sample size = n = 1140

P = x/n = 108/1140 = 0.094736842

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.094736842 ± 1.6449* sqrt(0.094736842*(1 – 0.094736842)/1140)

Confidence Interval = 0.094736842 ± 1.6449*0.0087

Confidence Interval = 0.094736842 ± 0.0143

Lower limit = 0.094736842 - 0.0143 = 0.0805

Upper limit = 0.094736842 + 0.0143 = 0.1090

Confidence interval = (8.1%, 10.9%)