The **Scheirer Ray Hare Test **is the two factor version of the Kruskal-Wallis test. The assumptions are the same as the Kruskal-Wallis test; in particular the interaction groups must be equal-sized and contain at least 5 sample members.

**Real Statistics Data Analysis Tool**:** **The Real Statistics Resource Pack provides the **Scheirer Ray Hare **data analysis tool, as demonstrated in the next example.

**Example 1**: Repeat Example 1 of using the **Scheirer Ray Hare **data analysis tool (the input data is repeated on the left side of Figure 2).

To perform the analysis, press **Ctrl-m** and double click on the **Analysis of Variance** option. On the dialog box that appears choose the **Anova – two factor** option and fill in the dialog box that appears as shown in Figure 1 and press the **OK** button.

**Figure 1 – Two Factor ANOVA dialog box**

We first note that if we use the Shapiro-Wilks test on each of the interaction samples we would see that there is no significant departure from normality (although these samples only contain 5 elements) and so we really don’t need to use the Scheirer Ray Hare test, since the usual ANOVA is a more accurate and powerful test. Despite this, we will perform the Scheirer Ray Hare test to show how it is done and to compare the results with ANOVA.

The first thing the Scheirer Ray Hare data analysis tool does is to calculate the ranks for all the input data elements, as shown on the right side of Figure 2. E.g. cell G5 contains the Excel formula =RANK.AVG(B5,$B$5:$E$19,1), or for Excel 2007 users the equivalent Real Statistics formula =RANK_AVG(B5,$B$5:$E$19,1).

**Figure 2 – Ranking of the input data elements**

A two factor ANOVA is then performed using the ranks (i.e. range G4:K19) as input, which is then modified as shown in Figure 3.

**Figure 3 – Scheirer Ray Hare Test**

To explain how the output shown on the right side of Figure 3 is obtained, we first show in Figure 4 the usual output for the data in G4:K19 from the Real Statistics two factor ANOVA.

**Figure 4 – Two Factor ANOVA on the ranked data**

As you can see, the values of *SS _{Row}, SS_{Col}, SS_{Int}, df_{Row}, df_{Col}, df_{Int} *and

*df*are the same.

_{Tot}The *H* values in range X6:X8 are calculated by divided the corresponding *SS* value by *MS _{Tot}* – e.g.

*H*(in cell X6) is calculated by the formula =U6/W10. The Scheirer Ray Hayes test calculates the significance of the Rows, Column and Interaction factors using the fact that the corresponding

_{Row}*H*statistic has approximately a chi-square distribution with the corresponding df value.

We see, for example, that the null hypothesis for the Rows factor, namely that the three fertilizers are equally effective, is rejected since p-value = CHISQ.DIST(X6,V6) = .001407 < .05 = α. In fact, we see from column Z that only the null hypothesis for the Row factor is rejected.

The main difference from the test using ANOVA is that the Interaction factor (Crops Blend) is not significant (p-value = .14839), while it is significant using ANOVA (p-value = .04347).