Question

12. (The average U.S. household has $235,600 in life insurance. A local insurance agent would like to see how households in his city compare to the national average, and selects a simple random sample of 30 households from the city. For households in the sample, the average amount of life insurance is x¯ = $245, 800, with s = $25, 500.

(a) Using a nondirectional hypothesis test and a level of significance of 95%, what conclusion would be reached?

(b) For the same significance level in part (a), construct the appropriate confidence interval and verify the conclusion reached in part (a).

(c) Suppose we change our confidence level from 95% to 99%. How would this impact the width of the confidence interval?

Answer 1

=235600,  s= 25500, n=30, = 245800, =0.05

Ho: = 235600

H1: 235600

Calculate test statistics

t = 2.191

test statistics = 2.191

Calculate P-Value using calculator with df = n-1 = 29

we get,

P-Value = 0.0366

since (P-Value = 0.0366) < ( =0.05)

Hence Reject the null hypothesis.

Therefore, there is enough evidence to claim that the population mean is different than 235600

b)

formula for confidence interval is

Where tc is the t critical value for c= 95% with df = n-1 = 30-1 = 29

tc =2.045

236278.144 < < 255321.856

confidence interval is ( 236278.144 ,  255321.856)

235600 lies inside the confidence interval hence population mean is = 235600

c)  

we change our confidence level from 95% to 99% then the width of confidence interval gets increased,

The new confidence interval with c=99% is wider than the interval with c=95%.