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 to1030 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 104 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.)

Answer 1

We are asked to find 90​% confidence interval for the percentage of people the company contacts who may buy something.

Let x be the number of people the company contacts who may buy something.

n is the total number of  people randomly selected from their mailing list.

We are given x = 104 and n = 1030

Therefore = x/n = 0.101

Lower bound = - E

Upper bound = + E

E is margin of error =

z is critical value follows standard normal distribution , we can find its value using z score table.

We are given confidence level = 0.90

Therefore α = 1 - 0.90 = 0.1 , 1 - (α/2) = 0.95

So we have to find z score corresponding to area 0.9500 on z score table

So z = 1.645

E = 1.645*

E = 0.0154

Lower bound = - E = 0.101 - 0.0154 = 0.0856 or 8.6%

Upper bound = + E = 0.101 + 0.0154 = 0.1164 or 11.6%

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

( 8.6% , 11.6%)