DF for 2 sample t-test with unequal variances
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Robert Butler.
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November 3, 2006 at 2:18 pm #45145
With a 2-sample t-test assuming equal variances the total degrees of freedom is simply N-2. If you can’t assume that the variances are equal the df of course will be lower, presumably because it’s eaten up by having to account for the multiple std dev’s. I understand why this would be the case but would like to know the formula/calculations of how to arrive at the correct df.
Thanks,
0November 3, 2006 at 2:28 pm #146476
Robert ButlerParticipant@rbutlerInclude @rbutler in your post and this person will
be notified via email.The equations are from pp.299-303 Brownlee Statistical Theory and Methodology in Science and Engineering – 2nd edition 1965
We are assuming you have the two means of interest, their respective standard deviations, and the associated number of samples/mean
data values1; input xbar1 std1 n1 xbar2 std2 n2; rat1 = (std1*std1)/n1; /*computation of fprime the synthetic
degrees of freedom*/ rat2 = (std2*std2)/n2; nval1 = (n1 – 1); nval2 = (n2 – 1);
fnumerator = (rat1 + rat2)*(rat1 + rat2);
fdenom1 = rat1*rat1/nval1; fdenom2 = rat2*rat2/nval2;
fprime = fnumerator/(fdenom1 + fdenom2);
meandiff = xbar1 – xbar2; /*computation of the t statistic*/ tdenom = sqrt(rat1+rat2);
tvalue = meandiff/tdenom;0 -
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