Question A computer system administrator notices that computers running a partic
ID: 3158159 • Letter: Q
Question
Question A computer system administrator notices that computers running a particular operating system seem to freeze up more often as the installation of the operating system ages. She measures the time (in minutes) before freeze-up for 7 computers one month after installation, and for 9 computers seven months after installation. The results are as follows: One month after install: 207.2 233.3 215.2 234.9 224.6 243.9 245.2 Seven months after install: 83.9 52.9126.8 200.9 173.7 245.8 149.3 155.8 102.4 Find a 95% confidence interval for the mean difference in time to freeze-up between the first month and the seventh. Carry five significant digits throughout your calculations. Lower confidence bound for the mean difference in time = Upper confidence bound for the mean difference in time =Explanation / Answer
One month after install: 207.2 233.3 215.2 234.9 224.6 243.9 245.2
Seven months after install: 83.9 52.9 126.8 200.9 173.7 245.8 149.3 155.8 102.4
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=229.1857
Standard deviation( sd1 )=14.2744
Sample Size(n1)=7
Mean(x2)=143.5
Standard deviation( sd2 )=59.7763
Sample Size(n2)=9
CI = [ ( 229.1857-143.5) ±t a/2 * Sqrt( 203.75849536/7+3573.20604169/9)]
= [ (85.6857) ± t a/2 * Sqrt( 426.1313) ]
= [ (85.6857) ± 2.447 * Sqrt( 426.1313) ]
= [35.1724 , 136.199]