SOLUTION: A sample of 62 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 25.1 years. The population variance is 26. (a) Give a poin

Click here to see ALL problems on Probability-and-statistics

Question 965061: A sample of 62 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 25.1 years. The population variance is 26.
(a) Give a point estimate for μ. (Give your answer correct to one decimal place.)

Correct: Your answer is correct.

(b) Find the 95% confidence interval for μ. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
(c) Find the 99% confidence interval for μ. (Give your answer correct to two decimal places.)
Lower Limit
Upper Limit
I got part a) correct. And I know I have to deduct in either direction from the mean to find the limits (I think). But I am not sure how to figure out what to deduct. I have looked online and can't seem to find anything similar with clear cut, step by step instructions.
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A sample of 62 night-school students' ages is obtained in order to estimate the mean age of night-school students. x-bar = 25.1 years. The population variance is 26.
(a) Give a point estimate for μ. (Give your answer correct to one decimal place.)
Ans: 25.1
Correct: Your answer is correct.
(b) Find the 95% confidence interval for μ. (Give your answer correct to two decimal places.)
Margin of Error = z*s/sqrt(n) = 1.96*26/sqrt(62) = 6.47
---------
Lower Limit = x-bar - ME = 25.1-6.47 = 18.63
---------------
Upper Limit = x-bar + ME = 25.1+6.47 = 31.57
==============================
(c) Find the 99% confidence interval for μ. (Give your answer correct to two decimal places.)
ME = 2.5758*26/sqrt(62) = 8.51
Lower Limit = 25.1-8.51
Upper Limit = 25.1+8.51
==============
Cheers,
Stan H.