## Scheffe's Test Demo

Scheffe's test is a parametric multi-comparison procedure which tests the hypotheses that the means of each pair of waves are equal. In this example we consider the following six waves:

 data1 data2 data3 data4 data5 data6 36.34 58.55 50.42 41.41 47.81 38.17 52.08 57 40.04 64.21 46.28 36.96 50.09 44.9 41.77 58.33 59 38.28 37.81 52.66 42.61 60.37 55.73 33.13 58.98 67.99 37.75 55.32 54.58 39.97 63.93 52.98 41.42 55.75 43.74 47.99 37.55 51.1 39.48 63.74 59.02 49.66 40.64 47.96 43.79 67.86 46.75 42.51

1. Basic Scheffe's test.

To run the test execute the command:

`StatsScheffeTest/T=1/Q data1,data2,data3,data4,data5,data6`

The results are displayed in the Scheffe Test table:

 Pair Diff SE S Sa Conclusion data6_vs_data1 -6.34375 3.61625 1.75424 3.4912 1 data6_vs_data2 -13.3087 3.61625 3.68027 3.4912 0 data6_vs_data3 -1.32625 3.61625 0.366748 3.4912 1 data6_vs_data4 -17.54 3.61625 4.85033 3.4912 0 data6_vs_data5 -10.78 3.61625 2.98099 3.4912 1 data5_vs_data1 4.43625 3.61625 1.22676 3.4912 1 data5_vs_data2 -2.52875 3.61625 0.699275 3.4912 1 data5_vs_data3 9.45375 3.61625 2.61424 3.4912 1 data5_vs_data4 -6.76 3.61625 1.86934 3.4912 1 data4_vs_data1 11.1962 3.61625 3.0961 3.4912 1 data4_vs_data2 4.23125 3.61625 1.17007 3.4912 1 data4_vs_data3 16.2138 3.61625 4.48359 3.4912 0 data3_vs_data1 -5.0175 3.61625 1.38749 3.4912 1 data3_vs_data2 -11.9825 3.61625 3.31352 3.4912 1 data2_vs_data1 6.965 3.61625 1.92603 3.4912 1

The table indicates that hypotheses μ62, μ64 and μ43 must be rejected while the remaining hypotheses for the equality of the means are accepted.

2. Testing multiple contrasts.

The contrast hypothesis is defined by a contrast wave. For example, if

denotes the mean of the wave datai then to test the hypothesis

We start by creating the contrast wave:

`Make/N=6/O cWave={1,2,1,2,2,1}`

Next we execute the test:

`StatsScheffeTest/T=1/Q/CONW=cWave data1,data2,data3,data4,data5,data6`

The results are displayed in the Scheffe test table (shown here transposed):

 Difference 11.3196 SE 2.08784 S 5.42167 Salpha 3.4912 Conclusion 0

As the statistic S exceeds the critical value Salpha we reject the contrast H0.