Let a random sample be taken of size n 36 from a population

Let a random sample be taken of size n = 36 from a population with a known standard deviation of sigma = 3. Suppose that the mean of the sample is X- = 37. Find the 95% confidence interval for the mean, mu, of the population from which the sample was drawn. (Round the values to whole numbers for your confidence interval.)

Solution

Note that              
              
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 =    0.025          
X = sample mean =    37          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    3          
n = sample size =    36          
              
Thus,              
              
Lower bound =    36.02001801          
Upper bound =    37.97998199          
              
Thus, the confidence interval is              
              
(   36.02001801   ,   37.97998199   )

or

(36, 38) [ANSWER]