I have read enough papers on the subject of human iron level

I have read enough papers on the subject of human iron levels that I believe the distribution is normal with standard deviation of iron level of 1.42. If a sample of 12 men had an average iron level of 1.783, what would the 99% confidence interval be?

Solution

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    1.783          
t(alpha/2) = critical t for the confidence interval =    3.105806516          
s = sample standard deviation =    1.42          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
              
Lower bound =    0.509871858          
Upper bound =    3.056128142          
              
Thus, the confidence interval is              
              
(   0.509871858   ,   3.056128142   ) [ANSWER]