What are the boundaries of a confidence interval at the 99 c

What are the boundaries of a confidence interval at the .99 confidence level if the 14 samples comes from a distribution with a known variance of 6, and the sample mean was 8?

Solution

Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=8
Standard deviation( sd )=2.449
Sample Size(n)=14
Confidence Interval = [ 8 ± Z a/2 ( 2.449/ Sqrt ( 14) ) ]
= [ 8 - 2.58 * (0.655) , 8 + 2.58 * (0.655) ]
= [ 6.311,9.689 ]