alternative hypotheses

**chromium** · June 30th 2009, 04:57 PM

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 $\bar{x}=79.866667$ , $s^{2}_{x}=216.46$ and n=30

Y:65,95,94,72,56,87,89,28,71,78,54,93,91,68,88,90, 69,78,69,91

$\bar{Y}=76.3$ , $s^{2}_{Y}=293.27$ and m=30

a. test the hypothesis $H_{0}: \mu_{X}-\mu_{Y}$ against the alternative $H_{1}: \mu_{X}>\mu_{Y}$ at the $\alpha =0.10$ significance level assuming X and Y are $N(\mu_{X},\sigma^{2}_{X})$ and $N(\mu_{Y},\sigma^{2}_{Y})$ where $\sigma^{2}_{X}=\sigma^{2}_{Y}$

I started by using $t=\frac{\bar{X}-\bar{Y}}{S_{p}\sqrt{1/n+1/m}}$ and critical region $t \ge t_{0.10}(48)=1.299$

b. test the hypothesis $H_{0}: \sigma^{2}_{X}=\sigma^{2}_{Y}$ against the alternative hypothesis $H_{1}: \sigma^{2}_{X} doesn't = \sigma^{2}_{Y}$ at the $\alpha$ =10% significance level.

I started this part with $F=\frac{s^{2}_{x}}{s^{2}_{Y}}$ and $F \ge F_{0.05,29,19}=2.04$ 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?

**matheagle** · June 30th 2009, 05:45 PM

Here's a F-table that should help.
http://www.danielsoper.com/statcalc/calc04.aspx
Also, remember that

$F_{a,b,\alpha}={1\over F_{b,a,1-\alpha}}$

I get (1/0.481414)=2.0772.... using 19,29,.95 in that calculator.

**chromium** · June 30th 2009, 06:27 PM

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

**chromium** · July 1st 2009, 11:57 AM

Do I have to also include $\frac{1}{F}=\frac{s^{2}_{y}}{s^{2}_{x}}=\frac{293. 46}{216.46}=1.3548$ ?

$1.3548 \ge F_{0.05,19,29}=1.93$

Or is only one test sufficient

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