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Home Q & A Board Science-Related Homework Help Mathematics Statistics and Probability
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Biraj Biraj
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4 years ago
Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. The insurance company knows that, last year, the life expectancy of its policyholders was 77 years. They want to know if their clients this year have a longer life expectancy, on average, so the company randomly samples some of the recently paid policies to see if the mean life expectancy of policyholders has increased. The insurance company will only change their premium structure if there is evidence that people who buy their policies are living longer than before.



For more accurate cost determination, the insurance companies want to estimate the life expectancy to within one year with 95% confidence. How many randomly selected records would they need to have?
Textbook 
Stats: Modeling the World

Stats: Modeling the World


Edition: 4th
Authors:
Read 66 times
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#1
4 years ago
We wish to find the sample size, n, that would allow a 95% confidence level for the mean life expectancy of a policy holder from the insurance company to have a margin of error of only one year.

First estimate:
ME = z* × SE()
1 = 1.96 ×
n = 77.1 ≈ 78
Although not necessary, since 78 is quite large, we could find a better estimate using t*75 = 1.992, from Table T.
ME = t*75 × SE()
1 = 1.992 ×
n = 79.6 ≈ 80
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