A small computer network consists of ten users and one print

A small computer network consists of ten users and one printer. The average turn around time for the system is 15 minutes. Ten new users and a second printer are added to the system. A sample of 30 turnaround times for the new network yields an average turnaround time of X = 14 and a sample variance of s^2 = 0. Calculate the 95% confidence interval for the mean turnaround time. Is there evidence to support the assertion that the modifications to the system have had no effect on turnaround times?

Solution

a)

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 =    0.025          
X = sample mean =    14          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    3          
n = sample size =    30          
              
Thus,              
Margin of Error E =    1.073516486          
Lower bound =    12.92648351          
Upper bound =    15.07351649          
              
Thus, the confidence interval is              
              
(   12.92648351   ,   15.07351649   ) [ANSWER]

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b)

Yes, there is evidence, as the old turnaround time, 15 minutes, is still inside the confidence inetrval.