Explicación de los grados de libertad no enteros en la prueba t con variaciones desiguales

El procedimiento de prueba t de SPSS informa 2 análisis cuando se comparan 2 medias independientes, un análisis con variaciones iguales asumidas y otro con variaciones iguales no asumidas. Los grados de libertad (df) cuando se asumen variaciones iguales son siempre valores enteros (e iguales n-2). El df cuando no se asumen variaciones iguales no son enteros (por ejemplo, 11.467) y no se acercan a n-2. Estoy buscando una explicación de la lógica y el método utilizado para calcular estos df no enteros.

Se puede demostrar que Welch-Satterthwaite df es una media armónica ponderada a escala de los dos grados de libertad, con pesos proporcionales a las desviaciones estándar correspondientes.

La expresión original dice:

$$\nu_{_W} = \frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)^2}{\frac{s_1^4}{n_1^2\nu_1}+\frac{s_2^4}{n_2^2\nu_2}}$$

Tenga en cuenta que es la varianza estimada de la i ésima media muestral o el cuadrado del i -ésimo error estándar de la media . Sea r = r 1 / r 2 (la razón de las varianzas estimadas de las medias muestrales), entonces$r_i=s_i^2/n_i$$i^\text{th}$$i$$r=r_1/r_2$

$$\begin{align} \nu_{_W} &= \frac{\left(r_1+r_2\right)^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \newline \newline &=\frac{\left(r_1+r_2\right)^2}{r_1^2+r_2^2}\frac{r_1^2+r_2^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \newline \newline &=\frac{\left(r+1\right)^2}{r^2+1}\frac{r_1^2+r_2^2}{\frac{r_1^2}{\nu_1}+\frac{r_2^2}{\nu_2}} \end{align}$$

El primer factor es , que aumenta de 1 en r = 0 a 2 en r = 1 y luego disminuye a 1 en r = ∞ ; es simétrico en log r .$1+\text{sech}(\log(r))$$1$$r=0$$2$$r=1$$1$$r=\infty$$\log r$

El segundo factor es una media armónica ponderada :

$$H(\underline{x})=\frac{\sum_{i=1}^n w_i }{ \sum_{i=1}^n \frac{w_i}{x_i}}\,.$$

$w_i=r_i^2$

$r_1/r_2$$\nu_1$$r_1/r_2$$0$$\nu_2$$r_1=r_2$$s_1^2=s_2^2$ you get the usual equal-variance t-test d.f., which is also the maximum possible value for $\nu_{_W}$.

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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.

What you are referring to is the Welch-Satterthwaite correction to the degrees of freedom. The $t$-test when the WS correction is applied is often called Welch's $t$-test. (Incidentally, this has nothing to do with SPSS, all statistical software will be able to conduct Welch's $t$-test, they just don't usually report both side by side by default, so you wouldn't necessarily be prompted to think about the issue.) The equation for the correction is very ugly, but can be seen on the Wikipedia page; unless you are very math savvy or a glutton for punishment, I don't recommend trying to work through it to understand the idea. From a loose conceptual standpoint however, the idea is relatively straightforward: the regular $t$-test assumes the variances are equal in the two groups. If they're not, then the test should not benefit from that assumption. Since the power of the $t$-test can be seen as a function of the residual degrees of freedom, one way to adjust for this is to 'shrink' the df somewhat. The appropriate df must be somewhere between the full df and the df of the smaller group. (As @Glen_b notes below, it depends on the relative sizes of $s^2_1/n_1$ vs $s_2^2/n_2$; if the larger n is associated with a sufficiently smaller variance, the combined df can be lower than the larger of the two df.) The WS correction finds the right proportion of way from the former to the latter to adjust the df. Then the test statistic is assessed against a $t$-distribution with that df.