# Need help calculating margin of sampling error and confidence interval!

• Apr 1st 2009, 11:51 PM
thecleric
Need help calculating margin of sampling error and confidence interval!
Problem:
Assume that the following random samples are taken from populations with normal distributions. In each case, find the margin of sampling error and a confidence interval for $u$ for the indicated level.

(a). n=12, X bar = 122, standard deviation = 25, 90% level of confidence.

What I know:
the confidence interval (CI) = point estimate +- margin of sampling error

margin of sampling error = (quantile value) x (standard error of point estimator)

standard error of point estimator = standard deviation divided by square root of n
-or-
CI = Point estimate +- margin of sampling error

What I need help with:
I have the answer, but cannot seem to arrive to it. I have tried to work this over and over again with no luck.

According to what i know, the quantile value should .95 because (1-(.1/2)) and the standard error of point estimator should be 25 divided by square root of 12. But that is not the answer....Help!
• Apr 2nd 2009, 12:05 AM
matheagle
Since n=12 is small you can't use the central limit theorem.
But, the problem here is in your standard deviation.
I don't know what you mean by that.
Is that the population $\sigma$ st deviation or the sample
st deviation, S?

If it's $\sigma$ ,then the interval is

$(\bar X-z_{\alpha /2}\sigma/\sqrt{n}, \bar X+z_{\alpha /2}\sigma/\sqrt{n})$

and if you mean S, then the interval is

$(\bar X-t_{n-1,\alpha /2}S/\sqrt{n},\bar X+t_{n-1,\alpha /2}S/\sqrt{n})$.
• Apr 2nd 2009, 12:40 AM
thecleric
here is what i am supposed to work with ... left side of the page
• Apr 2nd 2009, 07:45 AM
matheagle
that clearly says, when $\sigma$ is known
so use the first interval in my last post