Question
In: Statistics and Probability
Direct mail advertisers send solicitations (junk mail) to thousands of potential customers hoping that some will...
Direct mail advertisers send solicitations (junk mail) to thousands of potential customers hoping 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 1000 randomly selected people. The company gets orders from 139 of the recipients and decides to do a mass mailing list to everyone on its mailing list of over 200,000. Create a 95% confidence interval for the percentage of those people who will order something. (Use 1.96 for the z statistic)
Solutions
Expert Solution
| x = | 117 | |||
| sample size | n = | 1500 | ||
| sample proportion p̂ | x/n= | 0.0780 | ||
| std error =Se | =√(p*(1-p)/n) = | 0.0069 | ||
| for 95 % CI value of z= | 1.9600 | |||
| margin of error E=z*std error = | 0.0136 | |||
| lower confidence bound=sample proportion-margin of error | 0.064 | |||
| Upper confidence bound=sample proportion+margin of error | 0.092 | |||
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