The StatsNPMCTest operation supports a number of comparison tests. The following examples illustrate the tests and the optional flags.

1. Tukey type test analysing the data in the following 5 waves:

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

To run the test execute the command:

`StatsNPMCTest/T=1/Q/TUK data1,data2,data3,data4,data5`

The results are displayed in the NP Multiple Comparison (Tukey) table:

 Pair Difference SE q qc Conclusion data4_vs_data3 163 33.0656 4.9296 3.85766 0 data4_vs_data1 116 33.0656 3.50818 3.85766 1 data4_vs_data5 61 33.0656 1.84482 3.85766 1 data4_vs_data2 40 33.0656 1.20972 3.85766 1 data2_vs_data3 123 33.0656 3.71988 3.85766 1 data2_vs_data1 76 33.0656 2.29846 3.85766 1 data2_vs_data5 21 33.0656 0.635101 3.85766 1 data5_vs_data3 102 33.0656 3.08478 3.85766 1 data5_vs_data1 55 33.0656 1.66336 3.85766 1 data1_vs_data3 47 33.0656 1.42142 3.85766 1

The Tukey test conclusions indicate that at the 0.05 significance level only they hypothesis that the mean of data4 equals the mean of data3 is rejected. All other means are taken to be the same. If we increase the significance level to 0.1 we have:

`StatsNPMCTest/T=1/Q/TUK/ALPH=0.1 data1,data2,data3,data4,data5`

The results are displayed in the table:

 Pair Diff SE q qc Conclusion data4_vs_data3 163 33.0656 4.9296 3.47828 0 data4_vs_data1 116 33.0656 3.50818 3.47828 0 data4_vs_data5 61 33.0656 1.84482 3.47828 1 data4_vs_data2 40 33.0656 1.20972 3.47828 1 data2_vs_data3 123 33.0656 3.71988 3.47828 0 data2_vs_data1 76 33.0656 2.29846 3.47828 1 data2_vs_data5 21 33.0656 0.635101 3.47828 1 data5_vs_data3 102 33.0656 3.08478 3.47828 1 data5_vs_data1 55 33.0656 1.66336 3.47828 1 data1_vs_data3 47 33.0656 1.42142 3.47828 1

In this case H0 has to be rejected also when comparing data4 with data1 and when comparing data2 with data1.

2. Student-Newman-Keuls test example.

This is a non-parametric variation on the Student-Newman-Keuls test where the standard error is a function of the parameter p (the rank difference). The test requires that all input waves have the same number of points. To run the test execute:

`StatsNPMCTest/T=1/Q/SNK data1,data2,data3,data4,data5`

The results are displayed in the NP Multiple Comparison (SNK) table:

 Pair Diff SE p qp qc Conclusion data4_vs_data3 163 33.0656 5 4.9296 3.85766 0 data4_vs_data1 116 26.533 4 4.37191 3.63316 0 data4_vs_data5 61 20 3 3.05 3.31449 1 data4_vs_data2 40 13.466 2 2.97044 2.77181 0 data2_vs_data3 123 26.533 4 4.63574 3.63316 0 data2_vs_data1 76 20 3 3.8 3.31449 0 data2_vs_data5 21 13.466 2 1.55948 2.77181 1 data5_vs_data3 102 20 3 5.1 3.31449 0 data5_vs_data1 55 13.466 2 4.08436 2.77181 0 data1_vs_data3 47 13.466 2 3.49027 2.77181 0

In this test the statistic q depends on the value of p and therefore differences between waves tend to be amplified relative to the Tukey test above.

3. Multi-Comparison when waves do not have the same number of points

The Student-Newman-Keuls test above requires that all waves have the same number of points. The Dunn-Holland-Wolfe test supports unequal number of points as well as ties in ranks. In this example we analyze the following waves:

 data2 data3 data4 data5 data6 data7 58.55 50.42 41.41 47.81 38.17 50.55 57 40.04 64.21 46.28 36.96 49.63 44.9 41.77 58.33 59 38.28 49.57 52.66 42.61 60.37 55.73 33.13 53.92 67.99 37.75 55.32 54.58 39.97 54.4 52.98 41.42 55.75 43.74 47.99 39.62 51.1 39.48 63.74 59.02 49.66 47.96 43.79 67.86 46.75 42.51

To run the test execute the command:

`StatsNPMCTest/T=1/Q/DHW data2,data3,data4,data5,data6,data7`

The results are displayed in the NP Multiple Comparison (DHW) table:

 Pair Diff SE Q Qc Conclusion data4_vs_data6 26.5 6.71131 3.94856 2.9352 0 data4_vs_data3 24.375 6.71131 3.63193 2.9352 0 data4_vs_data7 12.5 7.24904 1.72437 2.9352 1 data4_vs_data5 8.875 6.71131 1.32239 2.9352 1 data4_vs_data2 5.625 6.71131 0.838138 2.9352 1 data2_vs_data6 20.875 6.71131 3.11042 2.9352 0 data2_vs_data3 18.75 6.71131 2.79379 2.9352 1 data2_vs_data7 6.875 7.24904 0.948401 2.9352 1 data2_vs_data5 3.25 6.71131 0.484257 2.9352 1 data5_vs_data6 17.625 6.71131 2.62616 2.9352 1 data5_vs_data3 15.5 6.71131 2.30953 2.9352 1 data5_vs_data7 3.625 7.24904 0.500066 2.9352 1 data7_vs_data6 14 7.24904 1.93129 2.9352 1 data7_vs_data3 11.875 7.24904 1.63815 2.9352 1 data3_vs_data6 2.125 6.71131 0.31663 2.9352 1

In this case, the test indicates that we should reject H0 (equality of means) for the pairs data4 and data6, data4 and data3, data2 and data6.

4. Multi-Comparison to a control.

Using the waves data1-data6 above, we can test the hypothesis of equal means (default /TAIL) of each wave compared with the control wave designated as data4. To execute the test, select the blue line below and type Ctrl-Enter:

`StatsNPMCTest/T=1/Q/CIDX=3 data1,data2,data3,data4,data5,data6`

The results are displayed in NP Multiple Comparison to Control table:

 Pair Diff SE q q' conclusion data4_vs_data6 205 56 3.66071 2.51146 0 data4_vs_data3 183 56 3.26786 2.51146 0 data4_vs_data1 135 56 2.41071 2.51146 1 data4_vs_data5 66 56 1.17857 2.51146 1 data4_vs_data2 41 56 0.732143 2.51146 1

The test concludes that H0 (equal means) has to be rejected for the wave data3 and the wave data6 when compared with the control wave data4.

We can modify the test for H0: μc≤μa using /TAIL=1. Here μc represents the mean of the control wave while μa is the mean of any other wave in the input group.

`StatsNPMCTest/T=1/Q/CIDX=3/TAIL=1 data1,data2,data3,data4,data5,data6`
 Pair Diff SE q q' Conclusion data4_vs_data6 205 56 3.66071 2.23382 0 data4_vs_data3 183 56 3.26786 2.23382 0 data4_vs_data1 135 56 2.41071 2.23382 0 data4_vs_data5 66 56 1.17857 2.23382 1 data4_vs_data2 41 56 0.732143 2.23382 1

Clearly the mean of the control wave is greater than the means of the waves data1, data3 and data6.

Testing for the other possible tail (H0: μc≥μa ):

`StatsNPMCTest/T=1/Q/CIDX=3/TAIL=2 data1,data2,data3,data4,data5,data6`

The results are displayed in the table:

 Pair Diff SE q q' Conclusion data4_vs_data6 205 56 3.66071 2.23382 1 data4_vs_data3 183 56 3.26786 2.23382 1 data4_vs_data1 135 56 2.41071 2.23382 1 data4_vs_data5 66 56 1.17857 2.23382 1 data4_vs_data2 41 56 0.732143 2.23382 1

5. Multiple Contrasts

Suppose we wanted to test the hypothesis H0: μ123456.

You can create the contrast wave using the command:

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

To run the test execute:

`StatsNPMCTest/T=1/Q/CONW=contrastW data1,data2,data3,data4,data5,data6`

The results are displayed in the NP Multiple Contrasts table (shown here transposed):

 Contrast 3.66667 SE 4.04145 S 0.907265 Critical_Chai 3.32724 Conclusion 1

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