The SPSS t-Test procedure reports 2 analyses when comparing 2 independent means, one analysis with equal variances assumed and one with equal variances not assumed. The degrees of freedom (df) when equal variances are assumed are always integer values (and equal n-2). The df when equal variances are not assumed are non-integer (e.g., 11.467) and nowhere near n-2. I am seeking an explanation of the logic and method used to calculate these non-integer df’s.

**Answer**

The Welch-Satterthwaite d.f. can be shown to be a scaled weighted harmonic mean of the two degrees of freedom, with weights in proportion to the corresponding standard deviations.

The original expression reads:

νW=(s21n1+s22n2)2s41n21ν1+s42n22ν2

Note that ri=s2i/ni is the estimated variance of the ith sample mean or the square of the i-th standard error of the mean. Let r=r1/r2 (the ratio of the estimated variances of the sample means), so

νW=(r1+r2)2r21ν1+r22ν2=(r1+r2)2r21+r22r21+r22r21ν1+r22ν2=(r+1)2r2+1r21+r22r21ν1+r22ν2

The first factor is 1+sech(log(r)), which increases from 1 at r=0 to 2 at r=1 and then decreases to 1 at r=∞; it’s symmetric in logr.

The second factor is a weighted harmonic mean:

H(x_)=∑ni=1wi∑ni=1wixi.

of the d.f., where wi=r2i are the relative weights to the two d.f.

Which is to say, when r1/r2 is very large, it converges to ν1. When r1/r2 is very close to 0 it converges to ν2. When r1=r2 you get twice the harmonic mean of the d.f., and when s21=s22 you get the usual equal-variance t-test d.f., which is also the maximum possible value for νW (given the sample sizes).

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With an equal-variance t-test, if the assumptions hold, the square of the denominator is a constant times a chi-square random variate.

The square of the denominator of the Welch t-test isn’t (a constant times) a chi-square; however, it’s often not too bad an approximation. A relevant discussion can be found here.

A more textbook-style derivation can be found here.

**Attribution***Source : Link , Question Author : Joel W. , Answer Author : Glen_b*