Computing Tests of Comparisons (3 of 6)

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As can be seen from the analysis of variance summary table shown below, MSE = 21.722.

Source  df    Ssq       Ms     F    p
 Groups  4  274.913   68.728  3.16 .039
 Error  18  391.000   21.722
Total   22  665.913   30.269

= (1)²/5 + (1)²/4 + (-1)²/5 + (-1)²/5 + (0)²/4 = .85

Therefore,

= 4.297. (Click here for formula for s_L)

Finally,

= 10.3/4.297 = 2.397.

The degrees of freedom error = N-a = 23 - 5 =18 (see the summary table). From a t table it can be determined that the probability value associate with a t of 2.397 with 18 df is 0.028.