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 $5629.a) Construct the 99% confidence interval for the mean of all such costs and write a sentence that interprets the interval.b) Based on the confidence interval, if you are a manager for an insurance company that provides lower rates for drivers who wear seat belts, and you want a conservative estimate for a worst case scenario, what amount should you use as the possible hospital cost for an accident victim who wears seat belts?
Accepted Solution
A:
Answer:12700$Step-by-step explanation:Given that sample size = 20, mean = 9004 and s= 5629a) For 99% confidence interval, we use t distribution as population std dev is not known.t critical value = 2.861Margin of error = 2.861 *s/sqrt n= [tex]2.869(\frac{5629}{\sqrt{19} } \\=3694.64[/tex]Confidence interval lower bound= [tex]9004-3694.64 = 5309.36[/tex]Upper bound = [tex]9004+3694.64 = 12698.64[/tex]b) Conservative estimate is the upper bound i.e. approxy 12700 dollars.