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Say, I have two normally distributed random variables: $X \sim \mathcal N(\mu_x, \sigma_x^2), Y \sim \mathcal N(\mu_y, \sigma_y^2)$.

I want to test if $\mu_x = \mu_y$. Further, I assume that these variables are independent of each other. Then, I believe the test statistic would be:

${\mu_x - \mu_y}\over{\sqrt{\sigma_x^2 + \sigma_y^2)}}$

EDIT:

I want to be a bit more specific. In a journal article they test if two regression coefficients from two DIFFERENT regressions are equal.

So, I want to test if two estimated OLS coefficients from DIFFERENT regression models (the sample size is identical) are different from each other. I assume that the coefficients are independent.

Christian
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    There seems to be some confusion, at least about notation. If you know $\mu_x$ and $\mu_y$, you know whether $\mu_x = \mu_y$ without testing anything. On the other hand, if you do not know $\mu_x$ and $\mu_y$, you cannot compute your test statistic. The test statistic should be a function of sample/data, but no sample is described in your question. – Juho Kokkala May 20 '14 at 16:47
  • ... because $\mu$ is taken to mean the population mean and $\sigma$ is taken to be the population variance. – russellpierce May 20 '14 at 16:50
  • Agreed. Please modify your question to reflect (1) your actual question (which I believe is actually already answered on this site, so please try search), and (2) make your test statistic something that can actually be calculated. – Glen_b May 21 '14 at 04:20
  • You might want to provide some more details for the question asker as how to map their question onto the other answer. The conversion is non-trivial for those who are poorly versed in regression and the question asker has not framed their question in a regression framework. – russellpierce May 21 '14 at 13:07

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Close, your denominator could be $\sqrt{(\sigma_x^2 + \sigma_y^2)\over{2}}$ if you had equal sample sizes, you assumed that $\sigma_x^2 \approx \sigma_y^2$, and were trying to get a better estimate of the true variance for the two variables. But as @Juho Kokkala pointed out it is sort of a silly question give that you know the population $\mu$ and $\sigma$. The solution above can be expanded by replacing $\mu$ with the sample mean $M$ and $\sigma$ with the sample estimate of the population variance $s^2$.

russellpierce
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