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A multiple range test for comparing means in an analysis of variance This procedure is broadly similar to that for calculating the LSD but it gives us more confidence in comparing the means within a table. In calculating LSD, we first found s_{d}^{2} (= 2 x residual mean square / n) and from this we found s_{d} (the standard deviation of the difference between any two means) and multiplied it by a t value (for the degrees of freedom of the residual mean square). For a multiple range test, we use essentially the same procedure but instead of a t value we use a Q value obtained from a table "The Studentized Range". We list our means in order of magnitude, from highest to lowest, then we test for significant difference between the highest and lowest  it must be greater than Qxs_{d}). If this is significant, we test the highest against the second lowest mean, and continue in this way until all means have been tested against one another. However the Q value changes each time. For the first test (highest against lowest mean) we look up the Q value for the number of treatments (i.e. for the number of means in our table of results). For the next test (second highest against lowest, we use the Q value for the number of means minus 1 (because we are excluding the highest mean now), and so on. However the degrees of freedom does not change  it is always the df of the residual (error) mean square. Although each step in this procedure is simple, you need to be organised for testing each mean against all others  see a worked example for the best way to do this. Having done an analysis like this, most people construct a table as follows, using letters to show which treatments differ from others. They would say that treatments that are not followed by the same letter differ significantly from one another (P, 0.05). For example, in the fictitious table below, the means for pH 4 and 5 do not differ from one another but differ from the means for all other pH values. The mean for pH 6 differs from the means at all other pH values. The means for pH 7 and 8 do not differ from one another but differ from all other means, and the mean for pH 9 differs from all others.
We will now apply the Multiple Range Test to the data on bacterial biomass that we analysed by ANOVA Summary table of data:
s_{d}^{2} = 2 x residual mean square / n = 17.56 /3 = 5.85. s_{d} = 2.42 (obtained as the square root of s_{d}^{2}). First: Rank the treatments from highest to lowest mean. Then compare the highest (39) with the lowest (12) mean:
Second: compare the second highest (20.7) with the lowest (12) mean:
Third: continue in this way, down the table of ranked means, until you get a nonsignificant result. [In our case, we have reached the end because there are only 3 treatments] Fourth: [This is not necessary in our case because we ran out of means!] Compare the second lowest mean with the highest, then continue with the second lowest and second highest, etc. Comparison of the Multiple Range Test with the Least Significant Difference It is interesting to compare the findings of these two types of test. Using the LSD method, we found that the least significant difference between any two means would need to be t(s_{d} ) = 2.45 x 2.42 = 5.92. Using the Multiple Range Test, we had to meet stricter criteria: the highest and lowest means had to differ by 9.72, and the secondhighest and lowest means had to differ by 8.37. If we had had more means to compare (e.g. the thirdhighest and lowest) then the critical value would have been reduced again. The Multiple Range Test is much more discriminating that the LSD  it greatly reduces the chance of error when making multiple comparisons between treatments. Suggested procedure for comparing all means Suppose that we did an ANOVA on 5 treatments, with 4 replicates per treatment, and found means of 36, 42, 74, 10, 80. We calculated s_{d }as 2.50. With 5 treatments and 4 replicates we have 15 degrees of freedom for the residual (error) mean square. We will need the Q values for comparing 5, then 4, then 3, then 2 treatments (always with 16 df). It is sensible to make a small table:
Now we can rank our means from highest to lowest and see if they differ by Qxs_{d}. Again make a table (see below). Start in box 1, comparing the means 80 and 10. These differ by more than 10.925 (Qxs_{d} for 5 treatments) so we insert * in box 1. Repeat this for box 2 (comparing means 80 and 36, using Q for 4 treatments). Again insert *. We would also get a significant difference in box 3 (means 80 and 42) but not in box 4, so we insert ns (not significant). Now start in the next column (boxes 5, 6 and 7) then the third column (boxes 8, 9) and the fourth column.
Finally, we put letters against the means to show significant differences. We state that: means followed by the same letter do not differ significantly from one another (p = 0.05). Do this in a series of steps as shown below, to reach the letters (shown in blue) in the final column. These are the letters that we retain from the early steps if they are not contradicted by a later step. [Of course, having done this, we can present our means in any order, with their letters; we do not need to keep them in ranked order]
During ANOVA we do an F_{max} test to check for homogeneity of variance, i.e. to check that it is safe to pool all the treatment variances  an essential condition for performing an Analysis of Variance. What should we do if the F_{max} test shows a major discrepancy in the variances, thereby invalidating ANOVA? The answer is to use some mathematical transformation of the original data, then perform ANOVA with the transformed data. There are several types of transformation, each most appropriate for particular circumstances. 1. When our data consist of small, wholenumbered counts the variance is often proportional to the mean. This is overcome by converting each value (X) to Ö X and analysing the Ö X data. If the counts are low and contain zeros then use Ö (X + 0.5). 2. More generally, it is appropriate to use log_{10} X, or log_{10} (X+1) if there are zero values. 3. Percentages and proportions (after multiplying by 100) can be converted to arcsin values. In all these cases the transformed data are analysed in exactly the same way as in a normal ANOVA, and we can use LSD or a multiple range test, as we did before, to test for significant differences between the treatment means. BUT remember that these tests tell us the difference between the transformed values, and it is not valid to detransform an LSD and show it as a significant difference between the ‘true’ means. This problem does not arise with a multiple range test, where we use letters to show significant differences. The way to overcome this is to present the data in a table as follows, showing both the true means and the transformed means, and the LSD that applies to the transformed means:

This site is no longer maintained and has been left for archival purposes
Text and links may be out of date