0
$\begingroup$

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?

Improve this question
$\endgroup$
9
  • $\begingroup$ 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. $\endgroup$
    – Charlie Chang
    Commented Dec 3, 2020 at 16:47
  • 1
    $\begingroup$ 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 ♦
    Commented 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
    Commented Dec 3, 2020 at 17:15
  • 2
    $\begingroup$ 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
    Commented Dec 3, 2020 at 17:16
  • $\begingroup$ In the last line do you mean deviation instead of variance? $\endgroup$
    – Massimo Ortolano
    Commented Dec 3, 2020 at 17:48

1 Answer 1

Reset to default
1
$\begingroup$

$\bar X$ 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 $\sqrt{\frac{\sigma_x^2+\sigma_y^2}{N}}=\frac{\sqrt{\sigma_x^2+\sigma_y^2}}{\sqrt{N}}$

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Charlie Chang
    Commented 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. $\endgroup$
    – Charlie Chang
    Commented Dec 3, 2020 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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