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 110of the recipients. Use this information to complete parts a through d.

a) 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.)

Answer 1

Solution :

Given that,

n = 1140

x = 110

Point estimate = sample proportion = = x / n = 110 / 1140 = 0.096

1 - = 1 - 0.096 = 0.904

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 (((0.096 * 0.904) / 1140 )

= 0.014

A 90% confidence interval for population proportion p is ,

± E

= 0.096 ± 0.014

= ( 0.082, 0.110 )

= ( 8.2% , 11.0% )