A random sample of 16 pharmacy customers showed the waiting

A random sample of 16 pharmacy customers showed the waiting times below (in minutes). 17 18 16 20 20 12 19 19 15 27 14 23 15 25 18 20 Find a 90 percent confidence interval for , assuming that the sample is from a normal population. (Round your standard deviation answer to 4 decimal places and t-value to 3 decimal places. Round your answers to 3 decimal places.)

The 90% confidence interval from, to?

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 =    18.625          
t(alpha/2) = critical t for the confidence interval =    1.753          
s = sample standard deviation =    3.981205847          
n = sample size =    16          
df = n - 1 =    15          
Thus,              
              
Lower bound =    16.88023654          
Upper bound =    20.36976346          
              
Thus, the confidence interval is              
              
(   16.880   ,   20.370   ) [ANSWER]