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:

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:

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:

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:

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:

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:

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:

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: μ_c≤μ_a using /TAIL=1. Here μ_c represents the mean of the control wave while μ_a 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

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: μ_c≥μ_a ):

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

The results are displayed in the table:

5. Multiple Contrasts

Suppose we wanted to test the hypothesis H₀: μ₁+μ₂+μ₃=μ₄+μ₅+μ₆.

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):