thedude said:

Thanks for the reply but isn't a one way anova used when you have 3 or more means to compare? I have only 2?

I dont know from where you got the idea that ANOVA cannot be applied for 2 treatments. For more than 2 treatments ANOVA is the way, but for 2 it can also be applied. The result will be same a for a usual t test. I am giving the details in the following. But before that let me say, a problem may be solved in more than one way, but choose the way which is naturally indicated by proposition of the question.

ANOVA has the advantage over Fisher's t test that one does not have to assume the variances of the two groups to be equal.

Let T1 & T2 be sum of 4 treatments in the respective two groups.

Then the test statistic is F= MST/MSE, for testing against the alt hyp: the means are not equal.

SST=Sum(Ti^2/ni)-CF, where CF=G^2/n,

G= Sum(Ti), n=Sum(ni), ni = the ith group size.

SS(T)= Sum of sq of all obsn-CF

SSE=SS(T)-SST

MST=SST/(k-1), MSE=SSE/(n-k), k = the no. of treatments=2 here.

w:F>F(0.05,k-1,n-k)