Direct mail advertisers send solicitations aka junk mail to

Direct mail advertisers send solicitations (a.k.a. \"junk mail\") to thousands of potential customers in the hope that some will buy the company\'s product. The acceptance rate is usually quite low. Suppose a company wants to test the response to a new flyer, and sends it to 1000 people randomly selected from their mailing list of over 200,000 people. They get orders from 123 of the recipients. Create a 90% confidence interval for the percentage of people the company contacts who may buy something. Explain what this interval means. Explain what \"90% confidence\" means. The company must decide whether to now do a mass mailing. The mailing won\'t be cost-effective unless it produces at least a 5% return. What does your confidence interval suggest? Explain.

Solution

a)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.123          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.010386096          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.017083608          
lower bound = p^ - z(alpha/2) * sp =   0.105916392          
upper bound = p^ + z(alpha/2) * sp =    0.140083608          
              
Thus, the confidence interval is              
              
(   0.105916392   ,   0.140083608   ) [ANSWER]

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b)

We are 90% confident that the true proportion of people the company contacts who may buy something is between 0.1059 and 0.1401.

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c)

It means that the probability that the true population proportion is within this interval is 0.90.

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d)

As the whole interval is greater than 0.05, then it suggests that they do a mass mailing, as it we are 90% confident that it will be cost effective.