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 1010 people randomly selected from their mailing list of over​ 200,000 people. They get orders from 113 of the recipients. Use this information to complete parts a through d.

​a) Create a 95​% confidence interval for the percentage of people the company contacts who may buy something.

Answer 1

Solution :

Given that,

n = 1010

x = 113

Point estimate = sample proportion = = x / n = 113 / 1010 = 0.112

1 - = 1 - 0.112 = 0.888

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025= 1.96}

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

= 1.96(((0.112 * 0.888) / 1010 )

= 0.019

A 90% confidence interval for population proportion p is ,

± E

= 0.112 ± 0.019

= ( 0.131, 0.093)

= ( 13.1% , 9.3% )

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