# Sample Standard Deviation - Did my teacher make a mistake?

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 statistics 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?

$\mu _v = 0$
$\sigma_v = \sqrt{(\sigma \bar{x_1} \times \sqrt{n_1})^2 + (\sigma \bar{x}_2 \times \sqrt{n_2})^2} \approx 356$
$P (X \geq 66 | \mu = 0 , \sigma = 356) \approx 0.426$
$0.426 > \dfrac{1}{2} \alpha$ , so the nullhypothesis holds.