I first posted this one MSE, but it was pointed out to me that this site might be more suited for these types of questions (which is probably why this question wasn't answered for about 11 hours) over there.

So we have the following data about a research:

Group 1:

$n_1 = 10$

$\mu \bar{x}_1 = 433$

$ \sigma \bar{x}_1 = 65$

Group 2:

$n_2 = 12$

$ \mu \bar{x}_2 = 367$

$ \sigma \bar{x}_2 = 84$


$H_0: \mu _v = \mu _1 - \mu _2 =0$

$H_1: \mu _v \neq 0$

My teacher then made the following calculations, which I quite frankly don't understand at all:

$ \sigma_v = \sqrt{\sigma_x^2 + \sigma_y^2}$, so:

$ \sigma _v = (\sqrt{\dfrac{ \sigma \bar{x}_1}{\sqrt{n_1}}})^2 + (\sqrt{\dfrac{ \sigma \bar{x}_2}{\sqrt{n_2}}})^2 =... \approx 32 $ , so we can reject $H_0$

My complaints and confusion:

  • In his calculation he uses $\sigma {x} = \dfrac{\sigma \bar{x}}{\sqrt{n}}$, but I've always seen the formula: $\sigma \bar{x} = \dfrac{\sigma_x}{\sqrt{n}}$. Isn't his calculation just plain incorrect?

  • When we have the (I'm assuming wrong) $\sigma _v$, why can we immediately reject $H_0$? I think there are some steps (which I don't understand) missing.


Also, now that I think about it, isn't $\mu \bar{x}$ redundant, since $\bar{x}$ is already the sample mean? Or is there a difference between the two?

My answer:

$ \mu _v = 0$

$ \sigma_v = \sqrt{(\sigma \bar{x_1} \times \sqrt{n_1})^2 + (\sigma \bar{x}_2 \times \sqrt{n_2})^2} \approx 356$

$P (X \geq 66 | \mu = 0 , \sigma = 356) \approx 0.426$

$ 0.426 > \dfrac{1}{2} \alpha $ , so the nullhypothesis holds.

ps - Keep in mind this is rudimentary hypothesis testing, not anything advanced. We've only been taught the binomial distribution and the normal distributions. Quite frankly, I have no idea what z-scores or t-scores (and such) are and mentioning them in your answer is not needed (since we never discussed them and he probably doesn't know what they are himself, since he isn't a statistics teacher. He didn't assume we knew t-tests when we tackled the hypothesis).

  • 1
    $\begingroup$ This question is now redundant. My statistics teacher sent me an answer saying that I was correct and he was wrong. $\endgroup$ Jan 20, 2013 at 12:57
  • $\begingroup$ I meant he said we were both wrong but 'because' of him (he didn't want to make it too complicated so he just divided by n). $\endgroup$ Jan 20, 2013 at 20:46
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    $\begingroup$ You should explain what those variables are to improve your question. $\endgroup$
    – Magpie
    Apr 19, 2014 at 19:32

1 Answer 1


This was originally a comment but my edit of it grew too long; I guess it's an answer now.

It's critical to carefully distinguish between sample quantities (like $\bar{x}$ or $s_x$) and population quantities (like $\mu_x$ or $\sigma_x$). The notation you have there seems to muddle them together with abandon and you have little hope of understanding what you're doing from there. Hypotheses are about population quantities. The inference is based on sample quantities. It looks like your teacher is trying to do a two-sample t-test; there are many teaching resources online that explain how to set them up and do them.

  • $\begingroup$ I don't really understand what you mean by the comment in italics, what I have done from 'there' (whatever there might be) is written under My answer. My teacher does exactly the same and I understand what he does, my question only is why he used a wrong form of the formule he taught me Also, the two-sample t-test is significantly different from what we've both done, so.. (besides, we are doing very basic hypothesis testing, no t-tests etc.) $\endgroup$ Jan 20, 2013 at 12:31
  • $\begingroup$ The notation is so confusing it's very difficult to tell what you're testing; the two-sample-t was my guess based on what I could work out. You're welcome to try to clarify or give more details. That you don't understand what you're doing is clear from the question. My point was that the confusion is unlikely to be properly resolvable without better notation, because the notation will lead you into errors you won't even realize you're making. There's good reason why you are having trouble. $\endgroup$
    – Glen_b
    Jan 21, 2013 at 8:05

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