F distribution

Definition taught in Statistical inference, part of Statistical inference

F distribution

Type	Definition
Taught in	Statistical inference
Part	Statistical inference
Properties	2
Worked examples	2

If U is \chi^2_{m_1}, V is \chi^2_{m_2}, and U and V are independent, then F=(U/m_1)/(V/m_2) has the F distribution with (m_1,m_2) degrees of freedom, written F_{m_1,m_2}.

Properties

Variance-ratio F pivot

Main article: Variance-ratio F pivot

For two independent normal samples,

\frac{s_1^2/\sigma_1^2}{s_2^2/\sigma_2^2}

has the F distribution with (n_1-1,\,n_2-1) degrees of freedom.

ANOVA F distribution

Main article: ANOVA F distribution

In the one-way model, \mathrm{SS}_{\mathrm{error}}/\sigma^2 is \chi^2_{n-k}, and it is independent of \mathrm{SS}_{\mathrm{treatment}}. Under H_0:\mu_1=\cdots=\mu_k, \mathrm{SS}_{\mathrm{treatment}}/\sigma^2 is \chi^2_{k-1}, and therefore

F=\frac{\mathrm{SS}_{\mathrm{treatment}}/(k-1)}{\mathrm{SS}_{\mathrm{error}}/(n-k)}=\frac{\mathrm{MS}_{\mathrm{treatment}}}{\mathrm{MS}_{\mathrm{error}}}

has the F_{k-1,\,n-k} distribution.

Examples

A squared t variable

Question page: A squared t variable

Using only the definitions of the t and F distributions, show that if T\sim t_m then T^2\sim F_{1,m}. Hence explain why a two-sided t test at level \alpha and an F test with one numerator degree of freedom reject on the same samples.

Worked answer (click to show)

Write T=Z/\sqrt{V/m} with Z standard normal, V\sim\chi^2_m and the two independent. Then

T^2=\frac{Z^2/1}{V/m}.

Z^2 is \chi^2_1 by definition, and it is independent of V, so T^2 is F_{1,m}.

The two-sided t test rejects when |T|\ge t_{m;1-\alpha/2}, that is when T^2\ge t_{m;1-\alpha/2}^2. Because \mathbb{P}(T^2\le c^2)=\mathbb{P}(-c\le T\le c), the quantile F_{1,m;1-\alpha} equals t_{m;1-\alpha/2}^2. For example, t_{4;0.975}^2=2.776^2=7.71=F_{1,4;0.95}. The two tests reject on the same samples.

Checking equal variances

Question page: Checking equal variances

Lost working hours averaged 91 per month, with standard deviation 14.1, over 50 months after a safety programme, and 108 per month, with standard deviation 14.3, over the 50 months before. Assuming two independent normal samples, give a 95% confidence interval for \sigma_1^2/\sigma_2^2, using F_{49,49;0.975}=1.762. Does it support pooling?

Worked answer (click to show)

s_1^2/s_2^2=14.1^2/14.3^2=0.972. Both samples have 49 degrees of freedom, so the two F quantiles in the interval are the same, 1.762, and the interval is

\Bigl(\frac{0.972}{1.762},\ 0.972\times1.762\Bigr)=(0.552,\ 1.713).

It contains 1. At the 95% level the data are consistent with equal variances, so pooling is reasonable.

Used in

Taught in Statistical inference, where the derivation and examples are.
All 4 questions on Statistical inference