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Qualitative Analysis: Confidence Intervals

Annual family dental expenses for the families of a random sample of ten employees of a company are given below. You can assume the family dental expenses for the employees of the company are normally distributed.
$450, $390, $550, $140, $690, $600, $250, $330, $490, $810

a) Compute the mean and standard deviation of the sample.
b) Determine the point estimate and 95% confidence interval for the mean family dental expenses of all employees of the company.
c) Interpret the confidence interval.
d) How would you communicate this information to the Vice-President of Human Resources in non-statistical language.
e) How could this information be used in negotiations with the dental insurer of the company in setting the premiums for the next year.

Solution Preview

a. Mean=(450+390+550+140+690+600+250+330+490+810)/10=470
Standard deviation=sqrt(((450-470)^2+(390-470)^2+(550-470)^2+(140-470)^2+(690-470)^2+(600-470)^2+(250-470)^2+(330-470)^2+(490-470)^2+(810-470)^2)/(10-1))= ...

Solution Summary

This solution provides a detailed, step by step explanation which demonstrates how to compute basic statistical measures, such as the mean of a sample, and determine the 95% confidence interval for the mean. Written responses for parts c-e are briefly discussed in 60 words.

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