Question

# Insurance Companies track life expectancy information to assist in determing the cost of life insurance policies....

Insurance Companies track life expectancy information to assist in determing the cost of life insurance policies. Last year the average life expectancy of all policy holders was 77 years. ABI Insurance wants to determine if their clients now have a longer life expectancy, on average, so they randomly sample some of their recently paid policies. The ages of the clients in the sample are shown below.

86 75 83 84 81 77 78 79 79 81

76 85 70 76 79 81 73 74 72 83

a) Based on the sample results, find the 90% cofidence interval and interpret

b) For more accurate cost determination, ABI Insurance wants to estimate the average life expectancy to within one year with 95% cofidence. How many randomly selected recently paid policies would they need to sample?

c) Suppose ABI samples 100 recently paid policies. This sample yields a mean of 77.7 years and a standard deviation of 3.6 years. Find a 90% cofidence interval and interpret.

 A confidence interval for a parameter is an interval of numbers within which we expect the true value of the population parameter to be contained. The endpoints of the interval are computed based on sample information. a.With smaller samples (n< 30) distribution called the t distribution must be used. The t distribution is similar to the standard normal distribution but takes a slightly different shape depending on the sample size. In a sense, one could think of the t distribution as a family of distributions for smaller samples. Instead of "Z" values, there are "t" values for confidence intervals which are larger for smaller samples, producing larger margins of error, because small samples are less precise. t values are listed by degrees of freedom (df). Just as with large samples, the t distribution assumes that the outcome of interest is approximately normally distributed. For n < 30 Use the t table with df=n-1 Based on the above sample the mean calculated is 78.6 and Standard Deviation is : 4.48 Putting values in formula we get 90% CI =78.6+-1.729*(4.48/sqrt(20)) 76.87,80.33 b.95% CI = Mean +- t(distribution)*Standard Deviation/sqrt(n) =Based on the above sample the mean calculated is 78.6 and Standard Deviation is : 4.48 =78.6+-2.093*(4.48/sqrt(20)) =(76.50,80.70) Note value of t is calculated from t Table taking df as n-1

c. If sample size is greater than 30 below formula is used to 90% cofidence interval :

90% CI = Mean+- 1.645*(Standard Deviation/sqrt(N)

=77.7+-1.645*(3.6/sqrt(100)

=(77.11, 78.29)

It means that with 90% confident that sample mean is between 77.11 and 78.29

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