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Now that we know the S values, we can finally calculate the confidence interval. After all the necessary manipulations have been done, we got three performance values for each hard disk drive: average performance according to the experiments, the minimum and maximum values of the confidence interval.

It turned out that the picture is not quite clear only in one subtest.

Note that the confidence intervals of the two drives coincide. To be more exact, the confidence interval of WD2500JB “includes” the confidence interval of Hitachi 180GXP. It means that despite the higher average performance of WD2500JB than that of Hitachi 180GXP (the average obtained as a result of 10 experiments), we can’t claim with 0.95 probability that the situation will not change in case of a different sample considered.

Let’s use the Student’s t-criterion to check the importance of the performance difference in Application Loading test for Hitachi 180GXP and WD2500JB HDDs.

According to this criterion, the sample average values Xa and Xb vary significantly if their difference exceeds the standard deviation of  by more than  times, where  is the Student’s coefficient for the confidence probability  and there are freedom degrees for the selection .

In practice, the following ratios are usually calculated:

where

And then these values are compared with the Student’s coefficient. Let’s do the same thing now :)

The obtained value (0.530484) is lower than Student’s coefficient (t=2.10), so the average results difference is not so meaningful.

If the performance difference between these two HDDs is not statistically visible, then why don’t we call them “equally fast” (in this test) and pass over to the conclusions? :)

 
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