And how does it depend on degrees of freedom? And why isn't the answer 0?

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    $\begingroup$ Re "why isn't the answer zero" - why do you expect it to be? It is difficult to explain the flaw in your reasoning unless you say why you thought it would be zero. By chi squared test, do you mean the chi squared test for variance or the co squared test for goodness of fit? (Not that this changes the expected value but it might change the intuition for why it isn't zero.) $\endgroup$ – Silverfish Mar 3 '16 at 7:59
  • $\begingroup$ This question seems clear enough to me, & has a good answer to boot. $\endgroup$ – gung Mar 3 '16 at 11:14
  • $\begingroup$ i'm talking about a chi squared test for the variance. sorry if missing many obvious details, as just briefly exposed to this, but by the null, wouldn't there be no explained or observed variance in the numerator? $\endgroup$ – cfye14 Mar 4 '16 at 7:59

If all the other assumptions hold (and, depending on which particular chi-square test you're doing), also assuming that the chi-square approximation to the distribution of the test statistic holds*, then under the null hypothesis, the statistic has the same distribution as the sum of squares of $\nu$ independent standard normal random variables.

* chi-squared tests of counts have a discrete distribution that is approximated by a chi-square.

This has expectation $\nu$ and variance $2\nu$.

why isn't the answer 0?

Assuming you're doing some kind of "observed minus expected" calculation -- because under the null hypothesis the observed values are random quantities whose mean happens to be the expected values, but they vary around that mean. Hence their squared-standardized-differences from those expected values won't be 0; $\mathbb{E}(O_i-E_i)= 0$ (under the null) but $\mathbb{E}[(O_i-E_i)^2]$ - essentially a variance - is not.

  • $\begingroup$ sorry, hastily written and mistitled. What is the expected chi square stat assuming the null is true? Using chi-square to test multiple regression parameters $\endgroup$ – cfye14 Mar 3 '16 at 5:18
  • $\begingroup$ How are you using the chi-square to do that? Are you doing an asymptotic Wald chi-square? If so, why not just do an F test? (and if that's what you are doing then the small sample distribution doesn't have an expectation of $\nu$ under the null; under iid normal errors the expectation would be $\nu\cdot\frac{\nu_e}{\nu_e-2}$) $\endgroup$ – Glen_b Mar 3 '16 at 5:21

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