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$
Hypotheses:
$H_0: \mu _v = \mu _1 - \mu _2 =0$
$H_1: \mu _v \neq 0$
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.
EDIT:
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).