Probability of Variances of normal populations using F Distribution If $S_1^2$ and $S_2^2$ are the variances of independent random samples of size $n_1=10$ and $n_2=15$, from two different normal populations with equal variances $\sigma ^2 $ , then find $P(\frac{s_1^2}{s_2^2} <4.03)$
Using the property that $\frac{(n-1)s^2}{\sigma^2}$ has $\chi^2$ distribution with $(n-1)$ degrees of freedom, I come up with the following:
$P(\frac{s_1^2}{s_2^2}<4.03)$
$=P(\frac{9s_1^2}{\sigma^2}*\frac{\sigma^2}{14s_2^2}<4.03*\frac{9}{14})$
$=P(\large{\frac{\chi^2_9}{\chi^2_{14}}}\normalsize <2.59071)$
This becomes an F distribution with numerator $\nu=9$ and denominator $\nu=14$. Using a TI-89, I come up with $F-PDF(2.59071,9,14)=.069128$
Is this correct? I'm very new to using F-distributions and I'm not sure I necessarily trust the calculator, but calculating by hand seems rather cumbersome.
 A: You are not right. It is
$\large{\frac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}} \normalsize\sim F(m,n)$,
where $F(m,n)$ is the cdf of the F-distribution with $m$ and $n$ are the degrees of freedom: $m=n_1-1=9$ and $n=n_2-1=14$.
Since $\sigma_1=\sigma_2$ we get $\large{\frac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}}=\large{\frac{s_1^2}{s_2^2}} \normalsize\sim F(m,n)$
Using a calculator with the given values we get $P\left(\frac{s_1^2}{s_2^2} <4.03\right)=0.99$
How to use the TI 98
Step 1: Press APPS.
Step 2:Press ENTER twice to get to the list entry screen.
Step 3: Press F5 for “F5-Distr.”
Step 4: Scroll down to “A:F Cdf” and press ENTER.
Step 5: Leave the box for “Lower Value” blank, then press the down arrow key.
Step 6: Enter $4.03$ in the box for “Upper Value”, then press the down arrow key.
Step 7: Enter $9$ in the “Num df” box, then press the down arrow key.
Step 8: Enter $14$ in the “Den df” box.
Step 9: Press ENTER. The calculator will return $.990$ as the answer
Remark
$\frac{\chi_m^2/m}{\chi_n^2/n}\sim F(m,n)$
$\frac{9\cdot s_1^2}{\sigma_1^2}\sim \chi_9^2$
$\frac{14\cdot s_2^2}{\sigma_2^2}\sim \chi_{14}^2$
$\Large{\frac{\frac{9\cdot s_1^2}{9\cdot \sigma_1^2}}{\frac{14\cdot s_2^2}{14\cdot \sigma_2^2}}} \normalsize=\frac{s_1^2}{s_2^2} \sim F(9,14)$
