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 $\displaystyle u$ 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!