Homework SourseA computer scientist is studying the tendency for computers
A computer scientist is studying the tendency for computers running a certain operating system to run more slowly as the computer ages. They measure the opening time of a particular computer program when the computer is one month old versus the computer at six months. Here are the results of this study, times are in seconds. Use this data to test if the opening times are different at six months vs. one month.
One month after install 84.3 53.2 127.3 201.3 174.2 246.2 149.4 156.4 103.3
Six months after install
207.4 233.1 235.1 225.6 245.3
215.9 244.4
Solution
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 143.9555556
X2 = 229.5428571
Calculating the standard deviations of each group,
s1 = 59.75698936
s2 = 14.16884577
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 9
n2 = sample size of group 2 = 7
Thus, df = n1 + n2 - 2 = 14
Also, sD = 20.62633936
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -4.149417891
where uD = hypothesized difference = 0
Now, the critical value for t is
tcrit = +/- 2.144786688
Thus, comparing t and tcrit, we decide to WE REJECT THE NULL HYPOTHESIS.
Also, using p values,
p = 0.000982681
Comparing this to the significance level, WE REJECT THE NULL HYPOTHESIS.
Thus, there is significant evidence that the opening times are different at six months vs. one month. [CONCLUSION]
