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Confidence interval, error of estimate, sample size

In the publication Employment and Earnings, information on the ages of people in the civilian labor force was published. Seventy-five people in the civilian labor force are randomly selected. The mean age is 37.4 years with a standard deviation of 10.3 years.

a. Find the 99% confidence interval for the mean age.
b. What is the error of estimate for the 90% confidence interval?
c. What is the minimum sample size that is needed for the 95% confidence interval, if the maximum error of estimate is 2.1 years?

Solution Preview

a. Find the 99% confidence interval for the mean age.

Mean=M =37.4years

Standard deviation =s=10.3years

sample size=n=75

sx=standard error of mean=s/square root of n=1.1893= ( 10.3 /square root of 75)

Confidence level=99%

a (alpha) =1%=100% -99%

or Significance level=a (alpha) =0.01(expressed as a number )

No of tails=2

Since sample size=75>=30use normal distribution

Z at the ...

Solution Summary

The solution calculates confidence interval of the mean age, error of estimate for for the confidence interval and sample size for the confidence interval.

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