Stats confidence interval - please help!

I am getting really frustrated with this problem.... I am already late on the submission so it doesn't matter but I want to know how to do the ****ing thing.

a = .01, n = 38, x
bar = 34, s2 = 12

what is the 95% CI for mu?

I need to know how to do it, and the answer please.. so I know if I am doing it right.

95% confidence interval for population mean (Mu) is
sample mean (Xbar) +/- Margin of error (ME)
Margin of error = critical value for the level of confidence*standard error of sample mean
critical value depends on the level of significance
if alpha is 0.01, level of confidence will be 99% and critical value z-score = 2.575
if alpha is 0.05, level of confidence will be 95% and critical value z-score = 1.96
WHY a = 0.01 and 95% CI are given ??? These are inconsistent.
standard deviation = s = sqrt variance = sqrt s^2 = sqrt 12
standard error of sample mean = s/sqrt n = sqrt 12 / sqrt 38 = 0.3158

Margin of error for 95% CI is 1.96*0.3158 = 0.62
( if CI is 99%, ME = 2.575*0.3158 = 0.81)

95% confidence interval for Mu is
34 +/- 0.62
lower limit is 34 - 0.62 = 33.38
upper limit is 34 + 0.62 = 34.62
The 95% CI for Mu is (33.38, 34.62)
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<,
Remember that standard deviation (s) is different from standard error (s/sqrt n)

This is a T Interval test.
At C = 0.95, my TI-83 gives the mean lying between 30.056 and 37.944.
The formula is T = (X - µ)/SE where SE is the standard error of the mean
SE = sample standard deviation/√n

Hope this is helpful. Good luck!