UL 1342 1443 1535 14. Crash Hospital Costs A study was conducted to estimate hos

Question

UL 1342 1443 1535 14. Crash Hospital Costs A study was conducted to estimate hospital costs for accident victims who wore seat belts. Twenty randomly selected cases have a distribution that appears to be bell-shaped with a mean of $9004 and a standard deviation of $5629 (based on data from the U.S. Department of Transportation). a. Construct the 99% confidence interval for the mean of all such costs. b. If you are a manager for an insurance company that provides lower rates for drivers who wear seat belts, and you want a conservative estimate for a worst case scenario, what amount should you use as the possible hospital cost for an accident victim who wears seat belts?

Explanation / Answer

here, sample size=n=20, sample mean=9004 and population standard deviation=5629

since population standard deviation is known/given so we use z-value for calcuation of confidence interval

(a)(1-alpha)*100% confidence interval for population mean=sample mean±z(alpha/2)*sd/sqrt(n)

99% confidence interval for population mean=9004±z(0.05/2)*5629/sqrt(20)

=9004±2.5758*5629/sqrt(20)=9004±3242=(5762,12246)

(b) in worst case for insuarnce company possible hospital cost will be upper limit of confidence interval calcuated in the part(a) i.e. $12246

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