Question

Policy Holder #	Life Expectancy at 65
1	20.4
2	22.2
3	17.6
4	27.2
5	24.5
6	20.3
7	21.3
8	22.5
9	26.7
10	18.3
11	23.5
12	25.6
13	22.1
14	24.2
15	15.4
16	23.4
17	25.3
18	18.5
19	24.2
20	20.3
21	26.8
22	28.1
23	19.9
24	25.5
25	22.3
26	23.9
27	31.7
28	26.0
29	22.8
30	23.3
31	25.9
32	17.7
33	19.6
34	21.8
35	23.3
36	21.9
37	21.9
38	28.7
39	19.9
40	27.8
41	26.6
42	21.1
43	23.3
44	25.5
45	23.8
46	21.4
47	23.3
48	23.6
49	23.1
50	23.9

1. Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Life expectancy is a statistical measure of average time a person is expected to live, based on a number of demographic factors. Mathematically, life expectancy is the mean number of years of life remaining at a given age, assuming age-specific mortality rates remain at their most recently measured levels. Last year the average life expectancy of all the Life Insurance policyholders in Ontario at age 65 was 22.3 years (meaning that a person reaching 65 last year was expected to live, on average, until 87.3). The insurance company wants to determine if their clients now have a longer average life expectancy, so they randomly sample some of their recently paid policies. 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 sample data is provided in the excel file. Answer the following questions. Results should be support by excel output.

a. Construct a 95% and 99% confidence intervals for the true average life expectancy. Use t-distribution and Descriptive Statistics function from Data Analysis. Interpret each Confidence interval and comment on the difference between the 95% and 99% interval.

b. Write the null and alternative hypotheses for this test:

c. In this context, describe a Type I error possible. How might such an error impact Life Insurance company’s decision regarding the premium structure?

d. What is the value of the t-test statistic?

e. What is the associated P-value?

f. State the conclusion using α = 0.05. Do it using both P-value and critical value.

Answer 1

Answer:

n = 50

  = 23.158

s = 3.162

A). Construct a 95% and 99% confidence intervals for the true average life expectancy.Use t distribution and Descriptive Statistics function from Data Analysis. Interpret each confidence interval .

95% confidence interval:

The critical value for = 0.05 and

= n-1

= 50 - 1

= 49 degrees of freedom is = = 2.01.

The corresponding confidence interval is computed as shown below:

CI =

=

= (22.259 , 24.057)

99% confidence interval:

The critical value for = 0.01 and

= n-1

= 50 - 1

= 49 degrees of freedom is to = = 2.68.

The corresponding confidence interval is computed as shown below:

   CI =

=  

= (21.96 , 24.356 )

A 95% confidence interval would mean there are 95% chances of the interval containing the true population mean.

As we can 99% confidence interval is wider than the 95% confidence interval because of the more confidence needed for the true population mean to lie within the interval.

B). Write the null and alternative hypotheses for this test:

= 22.3

> 22.3

C). In this context, describe a Type I error possible. How might such an error impact life insurance company's decision regarding the premium structure.

Type 1 error means rejecting a null hypothesis when actually it is true.

This mean that the company would have to change its premium structure when they need not do that.

D). What is the value of the t test statistic

T-statistics:

The t-statistic is computed as follows:

=

= 1.919

E). What is the associated P value

= n - 1

= 50 - 1

= 49

p-value = 0.0304

F). State the conclusion using = 0.05 , Doing both P value and critical value

Critical t-value at = 49 and = 0.05 is: 1.677

Since p-value = 0.0304 < 0.05 and t = 1.919 > 1.677

i.e. we can reject and

Hence we can say that there exists enough evidence to conclude that average life expectancy has increased.