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?
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...
