In: Statistics and Probability
Along with interest rates, life expectancy is a component in pricing financial annuities. Suppose that you know that last year life expectancy was 77 years for your annuity holders. Now you want to know if your clients this year have a longer life expectancy, on average, so you randomly sample n=20 of recently deceased annuity holders to see actual age at death. Using a 5% level of significance, test whether or not the new data shows evidence of your annuity holders now live longer than 77 years.
Here are the sample data (in years of life): 86 75 83 84 81 77 78 79 79 81 76 85 70 76 79 81 73 74 72 83
a) Does this sample indicate that life expectancy has increased? Test an appropriate hypothesis and state your conclusion (use a 5% level of significance).
b) For more accurate cost determination, suppose you want to estimate the life expectancy to within one year with 95% confidence. How many randomly selected records would you now need to sample?
a)
as test statistic is not in critical region we can not reject null hypothesis
we do nt have evidence to conclude that t life expectancy has increased
b)
for 95 % CI value of z= | 1.960 |
standard deviation σ= | 4.48 |
margin of error E = | 1 |
required sample size n=(zσ/E)2 = | 78.0 |
( please try 77 if this comes wrong due to rounding error)