From a population of 820 student a sample of 100 had a mean

From a population of 820 student, a sample of 100 had a mean test score of 550 and a standard deviation of 120. Find a 97% confidence interval for the true mean. (finite population corrector ?).

Solution

Here,

fpc = sqrt((N-n)/(N-1)) = sqrt((820-100)/(820-1)) = 0.937614462

Thus, the effective standard deviation is

s = sigma*fpc = 120*0.937614462 = 112.5137354

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.015          
X = sample mean =    550          
z(alpha/2) = critical z for the confidence interval =    2.17          
s = sample standard deviation =    112.5137354          
n = sample size =    100          
              
Thus,              
Margin of Error E =    24.41548058          
Lower bound =    525.5845194          
Upper bound =    574.4154806          
              
Thus, the confidence interval is              
              
(   525.5845194   ,   574.4154806   ) [ANSWER]