#### Analyze data from a two-factor experiment that has no replication

The sample data are contained in a (6 rows by 4 columns) 2D wave dataWave1 where Factor A corresponds to rows and Factor B corresponds to columns:

 4.04 5.08 2.83 3.06 4.91 4.68 3.17 2.71 2.94 1.9 1.61 2.02 5.95 5.26 4.85 4.45 3.9 4.85 2.69 2.93 5.72 5.14 4.39 3.29

To run the test execute the command:

`StatsANOVA2NRTest/T=1 dataWave1`

The results of the operation are displayed in the table ANOVA2(NR):

 SS DF MS F Fc Accept Total 36.5039 23 Factor B 11.2881 3 3.7627 14.427 3.28738 0 Factor A 21.3036 5 4.26072 16.3364 2.90129 0 Remainder 3.91216 15 0.260811

The assumption in this test is that there is no interaction between Factor A and Factor B so the operation computes the results for both factors. H0 provides that there is no difference in the means of the populations. SS are the sum of the squares, DF are the degrees of freedom, MS are the mean square, F are the computed statistics, Fc are the critical F values for the corresponding degrees of freedom. Evidently in this case H0 is rejected for both rows and columns.

#### Analysis of data of the same dimensions but having less variation as shown in dataWave2

 4.6 4.92 4.39 4.26 4.52 5 5.51 4.63 5.06 5.7 5.72 5.43 4.51 5.35 4.43 4.86 5.74 4.85 5.38 5.17 4.89 4.74 4.66 5.4

To repeat the test for these data use the command:

`StatsANOVA2NRTest/T=1 dataWave2`

The resulting table is:

 SS DF MS F Fc Accept Total 4.71293 23 Factor B 0.137833 3 0.0459444 0.303508 3.28738 1 Factor A 2.30443 5 0.460887 3.04461 2.90129 0 Remainder 2.27067 15 0.151378

In this case we can't reject H0 for Factor B (columns), i.e., the the mean reading is the same for all columns. The situation is different for rows where F>Fc at least for the default 0.05 significance level. To test with relaxed significance (0.01) execute the command:

`StatsANOVA2NRTest/T=1/ALPH=0.01 dataWave2`

This results in:

 SS DF MS F Fc Accept Total 4.71293 23 Factor B 0.137833 3 0.0459444 0.303508 5.41696 1 Factor A 2.30443 5 0.460887 3.04461 4.55561 1 Remainder 2.27067 15 0.151378

As expected, the change in significance was enough to drive F,sub>c above the value of F for Factor A and so at this significance level we can't reject H0 for Factor A.

#### Handling One Missing Value

In the following examples we use the same data as dataWave1 but we introduce one missing value. To do so we set (row 3, column 2) to NaN.

`StatsANOVA2NRTest/T=1/FOMD dataWave3`

The operation prints in the history:

`Single missing value computed to be: 4.34667`
`SS bias estimate: 0.572033`

The result table contains the following values:

 SS DF MS F Fc Accept Total 35.7387 22 Factor B 11.3438 3 3.78126 12.2375 3.34389 0 Factor A 20.0691 5 4.01382 12.9901 2.95825 0 Remainder 4.32585 14 0.30899

The computed datum 4.34667 is used as a substitute for 4.85 which we erased from the original data. The test results are essentially unchanged. Forum Support Gallery