A random sample of 49 lunch customers was taken at a restaur

A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed in the restaurant was 33 minutes. From past experience, it is known that the standard deviation equals 10 minutes. Construct a 99% confidence interval for the true average amount of time customers spent in the restaurant.

Solution

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 =    0.005          
X = sample mean =    33          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    10          
n = sample size =    49          
              
Thus,              
Margin of Error E =    3.679756148          
Lower bound =    29.32024385          
Upper bound =    36.67975615          
              
Thus, the confidence interval is              
              
(   29.32024385   ,   36.67975615   ) MINUTES [ANSWER]