11-2 Chapter 11 Confidence Intervals and Hypothesis Tests for Means
11.5.6 Create and interpret confidence intervals for the mean.
3. Insurance companies track life expectancy information to assist in determining the
cost of life insurance policies. Last year the average life expectancy of all policyholders
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% confidence interval and interpret.
b. For more accurate cost determination, ABI Insurance wants to estimate the average life
expectancy to within one year with 95% confidence. 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% confidence interval and interpret.
11.5.6 Create and interpret confidence intervals for the mean.
4. A sample of students from an introductory business course were polled regarding the
number of hours they spent studying for the last exam. All students anonymously
submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one
section of the course polled. The data was used to make inferences regarding the other
students taking the course. The data are shown below:
4.5 22 7 14.5 9 9 3.5 8 11 7.5 18 20
7.5 9 10.5 15 19 2.5 5 9 8.5 14 20 8
a. Based on the sample results, find the 95% confidence interval.
b. Interpret the results.
c. Do you expect a 90% confidence interval to be wider or narrower and why?
d. A previous study of a large cross-section of students in this course showed that
students studies 12 hours per week. Are the results of this current study statistically
different than the assumption of 12 hours per week studying? Define the hypotheses and
compare the t-value to the critical t value, evaluating at α = 0.05.