LSD is used to compare means for signification differences. It is generally done by computing differences between means and then comparing that differences with calculated LSD value at certain p value. If the difference is greater than LSD value, the differences are significant and vice versa. some times the differences between means are negative numbers e.g. suppose the difference is -2.3 and LSD value is 6 at p=0.05, will this value (-2.3) be significant or not?
First, the LSD test is a post hoc test (made after one anova). it is calculated as following :
LSD (AB) = t(0.05/2,DF)* SQRT(MS(1/nA+ 1/nB), where A and B are 2 groups. MS is the within Mean square drawn from the anova table.
Next, if |YA - YB| > LSD(AB) we reject the null hypothesis H0: Mu(A) = Mu(B), where YA and YB are the calculated means obtained from the data.
In brief the answer to your question is "there is sno significant dfference since |-2.3|
First, the LSD test is a post hoc test (made after one anova). it is calculated as following :
LSD (AB) = t(0.05/2,DF)* SQRT(MS(1/nA+ 1/nB), where A and B are 2 groups. MS is the within Mean square drawn from the anova table.
Next, if |YA - YB| > LSD(AB) we reject the null hypothesis H0: Mu(A) = Mu(B), where YA and YB are the calculated means obtained from the data.
In brief the answer to your question is "there is sno significant dfference since |-2.3|
I believe that more important to the answer above, that is strictly correct, is to say that you should look for better post-hoc tests... LSD is the least conservative post-hoc test and practically makes no multiple comparison corrections at all! It means that you use a post-hoc test to control the type 1 error rate and, in practice, have no control at all...
If you pay attention to the formulae describe by Patrick, you can see that it is almost the same as the calculations used to t-tests. The only difference is the use of MS instead of s1 and s2. Since the variance uniformity is one of the ANOVA's assumptions, you will be back to t-test...
LSD test is usually performed by researchers that finds no difference using other post-hoc tests to find some difference to feels that "the research worked"...
To answer the question more directly, the LSD is used as a non-directional "yardstick". So the absolute value of the mean difference is used (as mentioned by Patrick).
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