Determining level of confidence when provided with intervals

[ShannonL]

Hello to the forum...

Just waiting for the results of my mid-term and would like to thank some members of this forum for their helpful direction! Stuck on a problem though...thought I had a reasonable grasp of how to calculate intervals when the level of confidence is provided but this problem asks for the level of confidence:

A random check of 81 records of inmates from all federal prisons found the confidence interval for mean prison terms per prisoner to be 2.13 years to 2.287 years. Assuming a standard deviation of 0.4 year, what is the level of confidence provided by this interval?

If I'm correct, the mean would be 2.2 and I think I am to use the formula xbar +/- (confidence interval) x s/square root of n; any direction would be greatly appreciated as that is as far as I've been able to get...

Thanks very much...Shannon

[JohnM]

Convert 2.287 and 2.13 into z scores and find the area under the normal curve in between those two z scores:

z = (x-mu) / (s/sqrt(n))

If you check our Examples section, there is a post entitled "The Vaunted Normal Distribution" - this describes the process to be followed....

[ShannonL]

Hi John...thanks for your help! I've calculated the z-scores to be 2.175, determined the areas to be .4852 and the confidence interval to be 97%. Hope that sounds right and again, many thanks for your help!

Shannon :wave:

[JohnM]

Actually, it should be:

z = (2.287-2.2085) / (0.4 / sqrt(81)) = 1.766

and the area between z = 1.766 and z = -1.766 is 0.923, so it's 92.3% confidence

[ShannonL]

My apologies, John...I just realized that I typed in 2.13 rather than 2.113, which is what the question should have read. Based on that, I had

z = (2.287-2.2)/(0.4/sqrt (81) = (rounded off to 3 decimal places) 1.96 which is a 95% confidence interval.

Sorry for the error and thanks for you help yet again!

Have a good evening...Shannon