0
$\begingroup$

I don't understand why the test statistic of a two sided two sample t-test actually t distributed is. The test statistic is definded as:

$T=\sqrt{\frac{n_1n_2}{n_1+n_2}}(\frac{\bar{Y}-\bar{X}}{\sqrt{\frac{1}{n_1+n_2-2}(\sum_{i=1}^{n_1}(X_i-\bar{X})^2+\sum_{j=1}^{n_2}(Y_j-\bar{Y})^2})})$ with $X_i$ and $Y_j$ normally distributed with mean m and variance v (under $H_0$) for all $i=1...n_1$ and $j=1...n_2$ .

To show that T is t distributed it has to be shown that $A:= \bar{Y}-\bar{X}$ and $B:=X_i-\bar{X}$ and $C:=Y_j-\bar{Y}$ are independent and standard normally distributed. Is that right?

I was able to calculate that the mean of $A$ is indeed zero and but the variance of $\sqrt{\frac{n_1n_2}{n_1+n_2}}\cdot A$ is equal to v (not 1).

I don't understand how to show the independence. Can someone help me?

$\endgroup$
  • $\begingroup$ Showing A,B and C to be independent and normal would establish that the numerator is normal (i.e. $\bar{Y}-\bar{X}$) but that alone won't establish that you have a $t$ distribution for the statistic. $\endgroup$ – Glen_b Sep 10 '17 at 7:00
  • $\begingroup$ In the lecture we were given this defiition for a t-distribution: If $Y, X_1, ... , X_n$ are independent and standard normally distributed then $\frac{Y}{\sqrt{\frac{1}{n}\sum\limits_{i=1}^n X_i^2}}$ is t distributed with n degrees of freedom $\endgroup$ – Lucy Sep 10 '17 at 8:32
  • $\begingroup$ YepThat's right. So if you have numerator and denominator independent you can get somewhere. $\bar{Y}$ and $s^2_X$ will be independent for the obvious reason (and vice versa), so you need independence of $\bar{X}$ and $s^2_X$ (and similarly for $Y$). You should have something about this; if you haven't had this proved or at least stated for you, you can get it as a consequence of Basu's theorem for example. en.wikipedia.org/wiki/Basu%27s_theorem $\endgroup$ – Glen_b Sep 10 '17 at 8:59
  • $\begingroup$ So the numerator would be $\frac{1}{n_2}\sum\limits_{j=1}^{n_2} Y_j-\frac{1}{n_1}\sum\limits_{i=1}^{n_1} X_i$ and the denominator $\sqrt{\frac{1}{n_1+n_2-2}(\sum\limits_{i=1}^{n_1} X_i^2-\frac{1}{n_1}(\sum\limits_{i=1}^{n_1} X_i)^2+\sum\limits_{j=1}^{n_2} Y_j^2-\frac{1}{n_2}(\sum\limits_{j=1}^{n_2} Y_j)^2)}$ How do I show that they are independent? $\endgroup$ – Lucy Sep 10 '17 at 9:11
  • $\begingroup$ Cincerning your last comment: I thought I have to show that $\bar{X}$ is indpendent of $X_i-\bar{X}$ and the same for $Y$. Or does this follow from your comment? $\endgroup$ – Lucy Sep 10 '17 at 9:23
1
$\begingroup$

I can see why you might struggle with such a definition. (I can't make any sense of it.) I suggest it is better to start with a straightforward definition that uses the names of things rather than the functions for calculation of those things. Like this:

$$t=\frac{\bar{x}-\mu_{H_0}}{SE_{\bar{x}}}$$

where $\bar{x}$ is the observed mean difference, $\mu_{H_0}$ is the null hypothesised value of the population mean difference, and $SE_{\bar{x}}$ is the standard error of $\bar{x}$.

If you move from that towards your complicated formula in a stepwise manner you may be able to make sense of it.

$\endgroup$
  • $\begingroup$ If I do this with $\bar{x}=\bar{Y}-\bar{X}$ , $\mu_{H_0}=E_{H_0}(\bar{Y}-\bar{X})=0$ and $SE_{\bar{x}}=\frac{1}{n_1+n_2} V(\bar{Y}-\bar{X})=\frac{v(n_1-n_2)}{(n_1+n_2)n_1 n_2}$ (V is the variance) This is standard normally distributed right? How do I go from there to get the t distribution? $\endgroup$ – Lucy Sep 10 '17 at 8:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.