A random sample of 45 life insurance policy holders showed that the average premiums paid on...

A random sample of 45 life insurance policy holders showed that the average premiums paid on their life insurance policies was $345 per year with a sample standard deviation of $65.

1. Construct a 90% confidence interval for the population mean. Make a statement about this in context of the problem.

Solutions

Expert Solution

Solution :

Given that,

= $345

s = $65

n = 45

Degrees of freedom = df = n - 1 = 45 - 1 = 44

)At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t_/2,df = t_0.05,44=1.680

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 1.680 * (65/ $\sqrt$ 45) = 16.2786

The 90% confidence interval estimate of the population mean is,

- E < < + E

345 - 16.2786 < < 345+ 16.2786

328.7214 < < 361.2786

(328.7214, 361.2786)

orchestra answered 2 years ago

Next > < Previous

Related Solutions

Joan paid $10,000 in premiums for her whole life insurance policy that now has an $11,000...

Joan paid $10,000 in premiums for her whole life insurance policy that now has an $11,000 cash value. How much must she include in her income, if any, if she takes a $10,000 cash value loan from the policy?

Over the years, Luke paid $65,000 in premiums on a whole life policy with a surrender...

Over the years, Luke paid $65,000 in premiums on a whole life policy with a surrender value of $200,000 and $500,000 death benefit. How much is taxable if upon reaching 65, Show and Label ALL work a) Luke passes away and $200,000 is paid to his sister b) Luke cashes the policy in while in good health c) Luke cashes in the policy after being diagnosed with a terminal illness with a life expectancy of 20 months

A random sample of 100 patients at RWJ Hospital showed that 45 of them are middle...

A random sample of 100 patients at RWJ Hospital showed that 45 of them are middle income. Using 99% level of confidence, estimate the percentage of middle income patients of the hospital.

A disability income insurance policy on which premiums are paid weekly must have a grace period...

A disability income insurance policy on which premiums are paid weekly must have a grace period of at least seven days. ten days. thirty-one days. there is no grace period on weekly premium policies.

A random sample of 89 tourist in the Chattanooga showed that they spent an average of...

A random sample of 89 tourist in the Chattanooga showed that they spent an average of $2,860 (in a week) with a standard deviation of $126; and a sample of 64 tourists in Orlando showed that they spent an average of $2,935 (in a week) with a standard deviation of $138. We are interested in determining if there is any significant difference between the average expenditures of those who visited the two cities? a) Determine the degrees of freedom for...

For the same amount of premiums in comparison to whole life insurance, term life insurance generally...

For the same amount of premiums in comparison to whole life insurance, term life insurance generally purchases a.a greater amount of coverage. b.less amount of coverage. c.the same amount of coverage. d.a higher deductible. e.a lower deductible. Annual Cost-of-Living-Adjustment (COLA) changes of coverage levels within long-term care insurance, disability insurance, or homeowner's insurance policies is generally deigned to protect policyholders against a.the adverse event occurring. b.lower rates of return. c.lack of insurability. d.the company defaulting on its payments. e.inflation. Over...

1. For the same amount of premiums in comparison to whole life insurance, term life insurance...

1. For the same amount of premiums in comparison to whole life insurance, term life insurance generally purchases a.a greater amount of coverage. b.less amount of coverage. c.the same amount of coverage. d.a higher deductible. e.a lower deductible. 2. Annual Cost-of-Living-Adjustment (COLA) changes of coverage levels within long-term care insurance, disability insurance, or homeowner's insurance policies is generally designed to protect policyholders against a.the adverse event occurring. b.lower rates of return. c.lack of insurability. d.the company defaulting on its payments....

A random sample of 25 workers showed that the average workweek in the United States is...

A random sample of 25 workers showed that the average workweek in the United States is down to only 35 hours, largely because of a rise in part-time workers. The variance of the sample was 18.49 hours. Assuming that the population is normally distributed, calculate a 98% confidence interval for the population standard deviation.

A random sample of 81 cars traveling on a section of an interstate showed an average...

A random sample of 81 cars traveling on a section of an interstate showed an average speed of 68 mph. The distribution of speeds of all cars on this section of highway is normally distributed with a standard deviation of 10.5 mph. The 94.52 confidence interval for μ is from _____ to ____?

An insurance policy costs $110 and will pay policy holders $12,000 if they suffer a major...

An insurance policy costs $110 and will pay policy holders $12,000 if they suffer a major injury (resulting in hospitalization) or $3500 if they suffer a minor injury (resulting in lost time from work). The company estimates that 1 in every 1985 policy holders will suffer a major injury and that 1 in 468 will suffer a minor injury. a) Create a probability model for the profit on a policy. b) What's the company's expected profit on the policy? c)...

Subjects