Statistics Practice Problem, please help me out here?

Question

The average years of employee experience at a company with 10,000 employees is normally distributed, with a standard deviation of 12 years. If a sample of 50 employees indicates a mean age of 38, calculate 90 percent and 99 percent confidence intervals for the population mean.

Answer 1

For this kind of problem, the formula is a little different that for proportions.

You'll need
mean +/- critical value (standard deviation/square root of sample size)

The mean is 38, standard deviation is 12, sample size is 50.

The critical value is 1.645 for 90% and 2.575 for 99%

I can't work it out for you here because I need new batteries in my calculator, but just plug the numbers in the formula and you'll be OK

Answer 2

In statistics, a confidence interval (CI) for a population parameter is an interval between two numbers with an associated probability p which is generated from a random sample of an underlying population, such that if the sampling was repeated numerous times and the confidence interval recalculated from each sample according to the same method, a proportion p of the confidence intervals would contain the population parameter in question.




In a Normal Distribution, 90% of the data lies approximately between the Z scores of -1.65 and 1.65 (reverse lookup in the Z-score table).

Rearranging the z score formula to calculate the mean:

µ = x - z*σ (where x is our sample mean)

So the interval of 90% confidence in predicting our actual mean µ is:

( (38 - 1.65*12), (38 + 1.65*12) )
( (38 - 19.8), (38 + 19.8) )
(18.2, 57.8)


Now for 99% the z score is 2.58, so the interval of confidence is:

( (38 - 2.58*12), (38 + 2.58*12) )
( (38 - 30.96), (38 + 30.96) )
(7.04, 68.96)