A random sample of 16 pharmacy customers showed the waiting

A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Find a 90 percent confidence interval for mu, assuming that the sample is from a normal population.

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 =    0.05          
X = sample mean =    17.75          
t(alpha/2) = critical t for the confidence interval =    1.753050356          
s = sample standard deviation =    5.105552533          
n = sample size =    16          
df = n - 1 =    15          
Thus,              
              
Lower bound =    15.51242733          
Upper bound =    19.98757267          
              
Thus, the confidence interval is              
              
(   15.51242733   ,   19.98757267   ) [ANSWER]