# DF for 2 sample t-test with unequal variances

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• #45145

Ropp
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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,

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#146476

Robert Butler
Participant

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;

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