Fisher’s Least Significant Difference ( Fisher’s LSD )

Why not using t test to compare each two population means? Overall error rate 1 – ( 1- a )^c .

1) Perform an analysis of variance to test Ho: m1 = m2 =…= mt

2) If there is insufficient evidence to reject Ho – proceed no further

3) If Ho is rejected, calculate LSD

LSD = ta/2 √ S²w ( 1/ni + 1/nj )

If ni = nj = n

LSD = ta/2 √ 2S²w/n

Note: df used in t is for S²w

4) If absolute value |Yi – Yj| ³ LSD , the population means are different

e. g.

Treatment ( n= 5)	Sample mean
1	505
2	528
3	564
4	498
5	600
6	470

The ANOVA table is then

Source	Df	SS	MS	F
Between	5	56360	11272	4.60
Within	24	58824	2451
Total	29

From tables F _5,24 (0.05) = 2.62

LSD = t a/2 √ 2S2w / n

1) Rank the sample means from lowest to highest

Pop	6	4	1	2	3	5
Mean	470	498	505	528	564	600

2) Compute sample differences

Y5 - Y6 , Y3 - Y6 , Y2 - Y6 ….

6 4 1 2 3 5

Y5 - Y4 , Y3 - Y4 , Y2 - Y4

6 4 1 2 3 5

Tukey’s W procedure

1) rank the t sample means

2) two populations are different if

| Yi - Yj | ³ W

W = q a (t,v) √ S²w / n

Note: v is df for mean square within samples

Tukey’s W is more conservative compared to LSD

If different number of samples per treatment

W ij = q a (t,v) √ S2w/2 ( 1/ni + 1/nj )

Student-Newman-Keuls Procedure ( SNK )

The SNK procedure make use of a critical value

Wr = qa (r,v) √ S²w/n

For means that are r steps apart when the sample means are ranked from lowest to highest

Pop	6	4	1	2	3	5
Mean	470	498	505	528	564	600

e. g. W6 = qa (6,v) √ S²w /n

r	2	3	4	5	6
qa(r,v)	2.92	3.53	3.90	4.17	4.37
Wr	64.65	78.16	86.35	92.33	96.75

1) rank the t sample means from lowest to highest

2) for two means Yi and Yj that are r steps apart, mI and mj are different if

| Yi - Yj | ³ Wr

Wr as defined above, n is the number of observations per sample , v is the df for mean square for within samples

Note: replace t by r when finding the q values from the table

SNK is less conservative than Tukey’s W procedure – will declare more significant differences than W