I'm just taking a guess - but are you sure your Bartlett's test is right? Also, the F and t-tests are only comparable when variances are equal (F = t_squared).
My understanding is that the t-test is a special case of a oneway ANOVA.
I am confused by highly discrepant results when comparing the means of 2 variables. I am using Stata 10.
We have a control group and a treated group each with n = 20
control mean = 0.08 sd = .58
treated mean = 0.9 sd = .56
We first note that Bartlett's test for equal variances: chi2(3) = 12.5823 Prob>chi2 = 0.006
Hence, using t-test with unequal population variance, we have
ttest control == treated, unpaired unequal
gives a p value < 0.0001
The command:
ttesti 20 0.08 .58 20 .9 .56, unequal welch
also gives p < 0.0001
But
oneway control treated
gives an F = 2.46
p = .1206
Manual calculation with a log-likelihood ratio test matches the ANOVA p value almost exactly.
Non parameteric Kruskal-Wallis gives p = .3121
My question is:
What assumptions are being violated such that the t-test gives a wildly different answer to the ANOVA?
Cheers
Rob
I'm just taking a guess - but are you sure your Bartlett's test is right? Also, the F and t-tests are only comparable when variances are equal (F = t_squared).
lumhearts is right. F and t-tests are only comparable when variance are equal.
and t test is a special case of ANOVA( when groups are 2 and equal variance). In terms of statistic..
square of student's t become F distribution.
students t = std normal/ sqrt( chi-squre /df)
square of student's t = std normal ^2 / (chi squre /df)
= chisqure(1) /( Chisqure(df) /df)
= F(1,df)
In the long run, we're all dead.
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