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For the mean of sample $\bar X$, $\frac {\bar X -\mu}{\sigma_X/\sqrt{N}}$ has normal distribution. According to CLT, $\bar X$ has a variance of $\sigma_x/\sqrt{N}$.

For two means $\bar X, \bar Y$ of two samples, according to $\bar X-\bar Y=\frac{(X_1+\dots+X_M)-(Y_1+\dots+Y_N)}{M+N}=\frac{M\bar X-N \bar Y}{M+N}$ would have a deviation of $\frac { {M \sigma_x^2+N\sigma_y^2}}{(M+N)^2}$, and so $\frac {\bar X-\bar Y}{\frac {\sqrt {M \sigma_x^2+N\sigma_y^2}}{M+N}}$ has a normal distribution, right?

But Data Anal of Life Sci (Michael Love) says that$\frac {\bar X-\bar Y} {\sqrt {\sigma_x^2/M+\sigma_y^2/N}}$ has a normal distribution; and that $\bar X-\bar Y$ has a variance of ${\sqrt {\sigma_x^2+\sigma_y^2}}/\sqrt N$. Where does it go wrong?

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  • $\begingroup$ It seems that the author sometimes assumes the two sample sizes are N, as here stats.stackexchange.com/q/401214/301417 (but even if so it should be 2$\sqrt N$ in the denominator, right?); sometimes assume that the two sample sizes are M and N separately. $\endgroup$ Dec 3, 2020 at 16:47
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    $\begingroup$ It's important to pay attention to the context. It's highly likely the second formula is proposed in the context of a hypothesis test; and the specific null and alternative hypotheses are important determinants of a good test statistic. $\endgroup$
    – whuber
    Dec 3, 2020 at 17:15
  • $\begingroup$ Are you assuming both samples have the same mean? When using "normal", do you mean "standard normal"? $\endgroup$
    – Xi'an
    Dec 3, 2020 at 17:15
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    $\begingroup$ The equality$$\bar X-\bar Y=\frac{X_1+\dots+X_M+Y_1+\dots+Y_N}{M+N}$$is definitely wrong. $\endgroup$
    – Xi'an
    Dec 3, 2020 at 17:16
  • $\begingroup$ In the last line do you mean deviation instead of variance? $\endgroup$ Dec 3, 2020 at 17:48

1 Answer 1

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$\bar X$ has variance $\frac{\sigma_x^2}{M}$ and $\bar Y$ has variance $\frac{\sigma_y^2}{N}$

If the two samples are independent then $\bar X - \bar Y$ has variance $\frac{\sigma_x^2}{M}+\frac{\sigma_y^2}{N}$ and standard deviation $\sqrt{\frac{\sigma_x^2}{M}+\frac{\sigma_y^2}{N}}$

If then $M=N$ then $\bar X - \bar Y$ has variance $\frac{\sigma_x^2+\sigma_y^2}{N}$ and standard deviation $\sqrt{\frac{\sigma_x^2+\sigma_y^2}{N}}=\frac{\sqrt{\sigma_x^2+\sigma_y^2}}{\sqrt{N}}$

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  • $\begingroup$ I see it’s $\bar X-\bar Y$, there is no difference in weights (according to sizes of the two samples) of the two variables...or say it’s not weighted sum. $\endgroup$ Dec 3, 2020 at 17:56
  • $\begingroup$ That is, $\bar X-\bar Y=\frac{(X_1+\dots+X_M)}M-\frac{(Y_1+\dots+Y_N)}N$. I made a mistake here. What I calculated in the post was mean of the two samples put together, i.e. weighted sum of means gotten from two means. $\endgroup$ Dec 3, 2020 at 17:59

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