
alternative hypotheses
Grades for 2 classes, X and Y
X:78,89,81,94,60,67,51,73,90,93,81,68,45,100,60,10 2,95,87,85,96,77,79,55,89,86,75,85,76,95,84
I calculated $\displaystyle \bar{x}=79.866667$, $\displaystyle s^{2}_{x}=216.46$ and n=30
Y:65,95,94,72,56,87,89,28,71,78,54,93,91,68,88,90, 69,78,69,91
$\displaystyle \bar{Y}=76.3$, $\displaystyle s^{2}_{Y}=293.27$ and m=30
a. test the hypothesis $\displaystyle H_{0}: \mu_{X}\mu_{Y}$ against the alternative $\displaystyle H_{1}: \mu_{X}>\mu_{Y}$ at the $\displaystyle \alpha =0.10$ significance level assuming X and Y are $\displaystyle N(\mu_{X},\sigma^{2}_{X})$ and $\displaystyle N(\mu_{Y},\sigma^{2}_{Y})$ where $\displaystyle \sigma^{2}_{X}=\sigma^{2}_{Y}$
I started by using $\displaystyle t=\frac{\bar{X}\bar{Y}}{S_{p}\sqrt{1/n+1/m}}$ and critical region $\displaystyle t \ge t_{0.10}(48)=1.299$
b. test the hypothesis $\displaystyle H_{0}: \sigma^{2}_{X}=\sigma^{2}_{Y}$ against the alternative hypothesis $\displaystyle H_{1}: \sigma^{2}_{X} doesn't = \sigma^{2}_{Y}$ at the $\displaystyle \alpha$=10% significance level.
I started this part with $\displaystyle F=\frac{s^{2}_{x}}{s^{2}_{Y}}$ and $\displaystyle F \ge F_{0.05,29,19}=2.04$ I don't know how to do this because 29 and 19 are not in the table. If this equality is true though then is H0 rejected?

Here's a Ftable that should help.
http://www.danielsoper.com/statcalc/calc04.aspx
Also, remember that
$\displaystyle F_{a,b,\alpha}={1\over F_{b,a,1\alpha}}$
I get (1/0.481414)=2.0772.... using 19,29,.95 in that calculator.

Thats a great tool! if i put in 19, 29 .05, I believe it gives me the F for the 1/F test $\displaystyle (\frac{s^{2}_{y}}{s^{2}_{x}})$ Does everthing else seem ok to use? All my equations?

Do I have to also include $\displaystyle \frac{1}{F}=\frac{s^{2}_{y}}{s^{2}_{x}}=\frac{293. 46}{216.46}=1.3548$?
$\displaystyle 1.3548 \ge F_{0.05,19,29}=1.93$
Or is only one test sufficient