A small computer network consists of ten users and one printer. The average turn
ID: 3380400 • Letter: A
Question
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?Explanation / Answer
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.