1
$\begingroup$

I have $n_X$ observations of variable X, $n_Y$ of variable Y, and $n_Z$ of variable Z. I'd like to test the hypothesis that the true mean of $X$ is equal to the sum of the means of Y and Z. $$H_0: \mu_X - (\mu_Y+\mu_Z) = 0$$

Initial thoughts

I can use the sample means to define an estimator $\hat{\gamma} = \bar{y}_X - (\bar{y}_Y+\bar{y}_Z)$. Can I estimate the variance using $$\hat{\sigma}^2(\hat{\gamma}) = \sigma^2 (1/n_1+1/n_2+1/n_3)?$$ If so, is the test statistic $\hat{\gamma}/\sqrt{\hat{\sigma}^2(\hat{\gamma})}$ t-distributed?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Your calculations suggest all $n_X+n_Y+n_Z$ observations are independent. Are you sure this is the case? (BTW, $\hat\gamma$ needs two, not one, minus signs.) What you are testing might be termed non-additivity; linearity is a much stronger condition. $\endgroup$ – whuber♦ May 7 '14 at 18:51
  • 1
    $\begingroup$ I'm sure the observations are independent. $\endgroup$ – KDM May 7 '14 at 20:27
  • $\begingroup$ How does your question relate to your title? Is the aim to test that the means $\hat{\mu}_X$, $\hat{\mu}_Y$ and $\hat{\mu}_Z$ are equispaced? $\endgroup$ – Glen_b Mar 21 '15 at 11:54
0
$\begingroup$

Do you know $\sigma^2$? If not then you will need to estimate it from the data (and assume that all 3 variances are equal). So your final formula will probably be a little more complicated than you show, but it can be reduced to something that is (at least approximately) t distributed (though probably done using $t^2=F$ if the 3 variables are normal or the sample sizes are large enough. One way to test this is with the General Linear Hypothesis (google for it).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thanks! I don't know $\sigma^2$ and was going to estimate it from the pooled sample variances (assuming the underlying variances are equal, which I think is a pretty good assumption). $\endgroup$ – KDM May 7 '14 at 20:31

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.