This is one of the following seven articles on Two-Factor ANOVA With Replication in Excel
Two-Factor ANOVA With Replication in 5 Steps in Excel 2010 and Excel 2013
Variance Tests: Levene’s and Brown-Forsythe For 2-Factor ANOVA in Excel 2010 and Excel 2013
Shapiro-Wilk Normality Test in Excel For 2-Factor ANOVA With Replication
2-Factor ANOVA With Replication Effect Size in Excel 2010 and Excel 2013
Excel Post Hoc Tukey’s HSD Test For 2-Factor ANOVA With Replication
2-Factor ANOVA With Replication – Test Power With G-Power Utility
Scheirer-Ray-Hare Test Alternative For 2-Factor ANOVA With Replication
Scheirer-Ray-Hare Test in
Excel As a Nonparametric
Alternative To Two-Factor
ANOVA With Replication
A relatively unknown but very useful nonparametric substitute for two-way ANOVA with replication (must be balanced ANOVA) is the Scheirer-Ray-Hare test. It is an extension of the Kruskal-Wallis test. It is done in this way
Scheirer-Ray-Hare Test
Procedures
1) Replace each data observation with its overall rank (lowest number is ranked 1 and tied observations are all given the average rank)
2) Run the two-way ANOVA as usual with the ranks instead of the actual data values.
3) Discard the MS, F, and p value terms in the ANOVA output.
4) Sum SS for SS factors, SS interaction, and SS error. Divide this sum by df total. The result is MS total.
5) The test statistic, H, for each factor and interaction equals its SS / MS total
6) The Excel formula for the p value for each is: CHISQ.DIST.RT(H, df). The df is the usual df for each factor and interaction. The Excel output provides these df figures.
Just like the Kruskal-Wallis test, the Scheirer-Ray-Hare test requires that the data groups be symmetrical about an axis. The normality of all data groups has already been confirmed with the Shapiro-Wilks test. A data group that is normally distributed will be symmetrical about its mean. The Scheirer-Ray-Hare test symmetry requirement is therefore validated.
Just like the Kruskal-Wallis test, the Scheirer-Ray-Hare test statistic H for each F Test is very nearly distributed as the Chi-Square distribution with k – 1 degrees of freedom as long as the number of samples in each group is at least 5. The Factor 1 Main Effects F Test has three level groups each with 8 data observations. The Factor 2 Main Effects F Test has two level groups each with 12 data observations. The test statistics for each of these F Tests will distributed nearly as the Chi-Square distribution with k -1 degrees of freedom.
There are, however, only four replicates in each treatment cell. The requirement of at least five samples for each group is not met for the Interaction Effects F Test. The Scheirer-Ray-Hare test statistic for this F Test is not confirmed to be distributed similar to the Chi-Square distribution.
(Click On Image To See a Larger Version)
(Click On Image To See a Larger Version)
(Click On Image To See a Larger Version)
The p Value formula used here is for Excel versions prior to 2010. Excel 2010 and later would use the following formula:
p Value = CHISQ.DIST.RT(H,df)
The p Values generated by the Scheirer-Ray-Hare test are compared here with the p Values generated by the two-factor ANOVA with replication test perform on the data set.
The Interaction p Value from the Scheirer-Ray-Hare test from this data set is not considered valid because there is less than five samples in each sample group of this F Test.
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Thank you Sir for your post. But I have a question. How to determine/calculate/analyze post hoc test after we use Scheirer-Ray-Hare Test?
ReplyDeleteThank you (I'm sorry if my english is bad)