Question

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 policy holders 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 policy holders 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. The attached excel spreadsheet contains age of death of a sample of policy holders. 86 75 83 84 81 77 78 79 79 81 76 85 70 76 79 81 73 74 72 83

a) To begin to answer this question, we will use a confidence interval. Find and interpret a 90% confidence interval for the mean life expectancy of these policy holders. Be sure to check the conditions required for this interval.

b) Using your confidence interval, what can you conclude about the average life expectancy? Do we have evidence that it is increasing?

c) Your work on part (a) depends on a sampling distribution model. Describe the center, shape, and spread of this model.

Answer 1

(a) Here sample size n = 20

Sample mean x? = 78.60

sample standard deviation s = 4.477

standard error of sample mean se₀= s/sqrt(n) = 4.477/sqrt(20) = 1.00

Here 90% confidence interval = x? +- t_19,0.10se₀ = 78.60 +- 1.00 * 1.729 = (76.87 years , 80.33 years)

Here the conditions are

(i) the sample is a random sample.

(ii) The data are approximately normally distributed.

(b) Here we can conclude about average life expectancy that the life expectancy of its policy holders was 77 years as confidence interval consists the value of 77 years. we failed to reject the company's claim that life expectancy has increase from 77 years.

(c)Here the sampling distribution model is approximately normally distributed where center is sample mean = 78.6 years, shape is bell shaped and normally distributed and spread is 1 years , which is standard error.