Thread: 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 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!

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 st deviation or the sample
st deviation, S?

If it's ,then the interval is



and if you mean S, then the interval is

.

here is what i am supposed to work with ... left side of the page

that clearly says, when is known
so use the first interval in my last post