Mean nonoperated time is 2.54 seconds, so we take A = 0.25 second. The mean of differences between leg times is d = 0.1600, and the standard deviations of differences is s = 0.2868. From the second row of Table 21.1, Sd = s/~/~ = 0.2868/~/g = 0.1014. The standard deviation is estimated from the sample, so we use the t distribution. For 7 df, t0.95 = 1.895.

If the test statistic of Eq. (21.2) > 1.895, there is evidence to reject H0.

The calculated t = (A -- d)/sd = (0.25 -- 0.16)/0.1014 = 0.8876 does not exceed 1.895; therefore, we have no evidence to reject the null hypothesis of difference. We have not shown nonsuperiority.