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 1110 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 108 of the recipients.

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

( __ % __ % )

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

Answer 1

Solution :

Given that,

n = 1110

x = 108

Point estimate = sample proportion = = x / n = 108/1110=0.097

1 - = 1-0.097=0.903

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.097*0.903) / 1110)

= 0.0146

A 90% confidence interval for p is ,

- E < p < + E

0.097-0.0146 < p <0.097+0.0146

0.0824< p < 0.1116

8.2%<p<11.2%

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

8.2%,11.2%