posted by Nick .
Direct mail advertisers send solicitations (that is "junk mail") to thousands of potential customers in the hope that some will buy the company's products. The response rate is usually quite low. Suppose a company wants to test the response rate for a new flyer format. The company randomly selects 1000 from its mailing list of over 200,000 people. Their sample results in orders from 112 people.
Create a 90% confidence interval for the percentage of people the company contacts who may buy something.
The 90% confidence interval is % to %.
What is the margin of error? The margin of error is %. Show your answers as percents rounded to TENTHS of a percent. Do not type in the % sign.
How do I find the confidence interval? Once I get this do I just subtract the intervals to get the margin of error?
n = 1000
x = 112
p = x/n
p = 112/1000 = 0.112
z critical value = 1.645
E = za/2 *sqrt((p (1-p)/)/n))
E = 1.645 *sqrt((.112*.888/1000)
E = 0.016
p ± ('z critical value') * SQRT[p * (1 - p)/n] = 0.112 ± 1.645 * SQRT[0.112 * (1 - 0.112)/1000] = [0.096, 0.128]
The 90% confidence interval is 12.8% to 9.6%
Wow! Thank you for the help! Also, wouldn't the 90% confidence interval be 9.6% to 12.8%? Also, would the margin of error be 3.2 or 1.6?
The 90% confidence interval is 9.6 % to 12.8%.
The margin of error is 1.6.