Question

Crash Hospital Costs. A study was conducts 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 $9,004 and a standard deviation of $5,629 (based on data from the U.S. Department of Transportation). (a) Construct a 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 set belts

Answer 1

Answer:

a) The 99% confidence interval would be given by (5004.168;12603.832)

b) For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =9004$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=5629 represent the sample standard deviation

n=20 represent the sample size  

2) Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=20-1=19$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,19)".And we see that $t_{\alpha/2}=2.86$

Now we have everything in order to replace into formula (1):

$9004-2.86\frac{5629}{\sqrt{20}}=5004.168$    

$9004+2.86\frac{5629}{\sqrt{20}}=12603.832$

So on this case the 99% confidence interval would be given by (5004.168;12603.832)    

3) Part b

For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I