A I and II B III only C I only D II and III E II only Insura

A) I and II
B) III only
C) I only
D) II and III
E) II only

Insurance companies track life expectancy information to assist in determining the cost of life insurance policies. Five years ago the average life expectancy of all policyholders was 77 years. ABI Insurance wants to determine if its clients now have a longer life expectancy compared to five years ago, on average, so it randomly samples some of its recently paid policies. The insurance company will only change its premium structure if there is evidence that people who buy their policies now are living longer than they were five years ago. The sample has a mean of 78.6 years and a standard deviation of 4.48 years. At a =0.05, I. we fail to reject the null hypothesis. II. ABI Insurance shouldn\'t need to increase its premiums because there is little evidence to indicate an increase in average life expectancy. III. we reject the null hypothesis. A) I and II B) III only C) I only D) II and III E) II only

Solution

a. b. d.ANSWER: Conclusion: H0 is true
Why????
SINGLE SAMPLE TEST, ONE-TAILED, 6 - Step Procedure for t Distributions, \"one-tailed test\"

Step 1: Determine the hypothesis to be tested.
Lower-Tail
H0: %u03BC %u2265 %u03BC0 H1: %u03BC < %u03BC0
or
Upper-Tail
H0: %u03BC %u2264 %u03BC0 H1: %u03BC > %u03BC0

hypothesis test (lower or upper) = upper


Step 2: Determine a planning value for %u03B1 [level of significance] = 0.05

Step 3: From the sample data determine x-bar, s and n; then compute Standardized Test Statistic: t = (x-bar - %u03BC0)/(s/SQRT(n))

x-bar: Estimate of the Population Mean (statistical mean of the sample) = 78.6
n: number of individuals in the sample = 20
s: sample standard deviation = 4.48
%u03BC0: Population Mean = 77
significant digits = 3

Standardized Test Statistic t = ( 78.6 - 77 )/( 4.48 / SQRT( 20 )) = 1.597


Step 4: Use Students t distribution, \'lookup\' the area to the left of t (if lower-tail test) or to the right of t (if upper-tail test) using Students t distribution Table or Excel TDIST(x, n-1 degrees_freedom, 1 tail) =TDIST( 1.597 , 19 , 1 )


Step 5: Area in Step 4 is equal to P value [based on n -1 = 19 df (degrees of freedom)] = 0.063

Table look-up value shows area under the 19 df curve to the right of t = 1.597 is (approx) probability = 0.063

Step 6: For P %u2265 %u03B1, fail to reject H0; and for P < %u03B1, reject H0 with 95% confidence.
Conclusion: H0 is true