Suppose that $X_1$ and $X_2$ are independent random variables with

\[X_1\sim\chi^2(n_1)\quad\text{and}\quad X_2\sim\chi^2(n_2)\\\]

Define a new random variable

\[F=\frac{X_1/n_1}{X_2/n_2}\]

$F$ has an “$F$ distribution” with $n_1$ and $n_2$ degrees of freedom.

\[F\sim F(n_1,n_2)\]

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PDF:

\[f(x;n_1,n_2)=\\ \frac{1}{B(n_1/2,n_2/2)}\bigg(\frac{n_1}{n_2}\bigg)^{n_1/2}x^{n_1/2}\bigg(1+\frac{n_1}{n_2}x\bigg)^{-(n_1+n_2)/2}\\ (\text{for }x\gt0)\]

Mean: $\frac{n_2}{n_2-1}\quad(\text{if }n_2\gt2)$

Variance: $\frac{2n_2^2(n_1+n_2-2)}{n_1(n_2-2)^2(n_2-4)}\quad(\text{if }n_2\gt4)$$

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In R:

qf(0.95, 5, 1) # 6.608
pf(6.608,5, 1) # 0.9499824

The Mean:

\[\begin{align} E[F]&=E\bigg[\frac{X_1/n_1}{X_2/n_2}\bigg]=\frac{n_2}{n_1}E\bigg[\frac{X_1}{X_2}\bigg]\\ &\stackrel{indep}{=}\frac{n_2}{n_1}\underbrace{E[X_1]}_{n_1}.E\bigg[\frac{1}{X_2}\bigg]\\ &=n_2E\bigg[\frac{1}{X_2}\bigg]\\ &=n_2\int_{-\infty}^{\infty}\frac{1}{x}f_{x_2}(x)dx\\ &=n_2\int_{0}^{\infty}\frac{1}{x}\frac{1}{\Gamma(n_2/2)}\bigg(\frac{1}{2}\bigg)^{n2/2}x^{n_2/2-1}e^{-x/2}dx\\ &=n_2\int_{0}^{\infty}\frac{1}{\Gamma(n_2/2)}\bigg(\frac{1}{2}\bigg)^{n2/2}\underbrace{x^{n_2/2-2}e^{-x/2}}_{\text{like a }\Gamma(n_2/2-1,1/2)}dx\\ &=n_2\frac{\Gamma(n_2/2-1)}{\Gamma(n_2/2)}\frac{1}{2}.\\ &\qquad\underbrace{\int_{0}^{\infty}\frac{1}{\Gamma(n_2/2-1)}\bigg(\frac{1}{2}\bigg)^{n2/2-1}x^{n_2/2-2}e^{-x/2}dx}_{\text{intergrate to }1}\\ &=n_2\frac{\Gamma(n_2/2-1)}{(n_2/2-1)\Gamma(n_2/2-1)}\frac{1}{2}\\ &=\frac{n_2}{n_2-2} \end{align}\]

  • Suppose that $X_{11},X_{12},…,X_{1,n_1}$ is a random sample of size $n_1$ from the $N(\mu_1,\sigma_1^2)$.

  • Suppose that $X_{21},X_{22},…,X_{2,n_2}$ is a random sample of size $n_2$ from the $N(\mu_2,\sigma_2^2)$.

Derive a test of size $\alpha$ for

\[H_0:\sigma_1^2=\sigma_2^2\quad\text{vs.}\quad H_1:\sigma_1^2\neq\sigma_2^2\\ \huge{\Downarrow}\\ H_0:\sigma_1^2/\sigma_2^2=1\quad\text{vs.}\quad H_1:\sigma_1^2/\sigma_2^2\neq1\]

Let $S_1^2$ and $S_2^2$ be the sample variances for the first and second samples, respectively.

We know that

\[\frac{(n_1-1)S_1^2}{\sigma_1^2}\sim\chi^2(n_1-1)\]

and

\[\frac{(n_2-1)S_2^2}{\sigma_2^2}\sim\chi^2(n_2-1)\]

are independent.

So, define a test statistic $F$ as

\[F:=\frac{[(n_1-1)S_1^2/\sigma_1^2]/(n_1-1)}{[(n_2-1)S_2^2/\sigma_2^2]/(n_2-1)}=\frac{\sigma_2^2}{\sigma_1^2}.\frac{S_1^2}{S_2^2}\]

Then

\[\color{red}{F\sim F(n_1-1,n_2-1)}\\\]

Similarly

\[\frac{\sigma_1^2}{\sigma_2^2}.\frac{S_2^2}{S_1^2}\sim F(n_2-1,n_1-1)\\\]

Under the assumption that $H_0$ is true, we have that

\[\frac{S_1^2}{S_2^2}\sim F(n_1-1,n_2-1)\]

and that

\[\frac{S_2^2}{S_1^2}\sim F(n_2-1,n_1-1)\]

Derive a test of size $\alpha$ for

\[H_0:\sigma_1^2=\sigma_2^2\quad\text{vs}\quad H_1:\sigma_1^2\neq\sigma_2^2\]

  • We will reject $H_0$ if $S_1^2/S_2^2$ is too small or too large.

  • Equivalently, we reject $H_0$ if $S_2^2/S_1^2$ is too large or too small.

Convention: Put the larger sample variance in the numerator and reject $H_0$ is above the appropriate upper $\alpha/2$ critical value.

Example

Fifth grade students from two neighboring counties took a placement exam.

Group 1, from County A, consisted of $18$ students. The sample mean score for these students was $77.2$.

Group 2, from County B, consisted of $15$ students and had a sample mean score of $75.3$.

From previous years of data, it is believed that the scores for both counties are normally distributed.

The sample variances of scores from Counties A and B, respectively, are $15.3$ and $19.7$.

Derive a test of size $\alpha$ for

\[H_0:\sigma_1^2=\sigma_2^2\quad\text{vs}\quad H_1:\sigma_1^2\neq\sigma_2^2\] Step One:

Choose a test statistic.

\[F:=\frac{S_1^2}{S_2^2}\quad\text{or}\quad F:=\frac{S_2^2}{S_1^2}\]

Use the one that is greater than 1.

Step Two:

Give the form of the test.

Reject $H_0$, in favor of the alternative if $F$ is either too large or too small.

Note that this upper tail will have area $\alpha/2$.


Step Three:

Find the cutoff using $\alpha/2$.

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Step Four:

Conclusion.

Reject $H_0$, in favor of $H_1$, if

\[F\gt F_{\alpha/2,n_i-1,n_j-1}\]
\[\begin{align} &n_1=18\quad s_1^2=15.3\quad \alpha=0.05\\ &n_2=15\quad s_2^2=19.7 \end{align}\] \[F:=\frac{S_2^2}{S_1^2}=\frac{19.7}{15.3}\approx1.288\]

Critical value:

\[F_{0.025,14,17}=2.753\]

In R:

qf(1-0.025,14,17) #2.753

The test statistic $F\approx1.288$ does not fall above

\[F_{0.025,14,17}=2.753\]

Thus we fail to reject $H_0$.