Chi-squared, t, F distributions — definition, degrees of freedom: Worked examples

This page is the case-crunching drill room for the parent topic. The definitions live there. Here we hunt down every scenario a problem can throw at you — small and large degrees of freedom, degenerate inputs, limiting behaviour, a real-world word problem, and an exam-style trap — and work each one from zero.

Before we start, some plain-word groundwork so no symbol is unearned:

Recall What the three distributions mean (one line each)

$χ_{k}^{2}$ ::: add up $k$ independent squared standard normals $Z_{i}^{2}$ ; measures total squared wobble. $t_{k}$ ::: a standard normal $Z$ divided by an estimated standard deviation built from $k$ df; a bell with fatter tails. $F_{d_{1}, d_{2}}$ ::: a ratio of two chi-squareds, each first divided by its own df; asks "how many times bigger is one wobble than the other?"

The scenario matrix

Every question this topic can ask falls into one of these case classes. Read the whole table first — the examples below are labelled with the cell they hit, and together they fill every cell.

Cell	Case class	What makes it tricky	Example
A	$χ^{2}$ mean/variance, small $k$	just apply $E = k$ , $Var = 2 k$	Ex 1
B	df from a constraint ( $n$ vs $n - 1$ )	remembering estimating $\overset{x}{ˉ}$ costs one df	Ex 2
C	building a $t$ statistic (unknown $σ$ )	why $t$ not $Z$	Ex 3
D	$t^{2} = F_{1, k}$ identity	recognising $Z^{2}$ is $χ_{1}^{2}$	Ex 4
E	$F$ for two variances + reciprocal swap	which df is numerator; $1/ F$ symmetry	Ex 5
F	Degenerate / boundary inputs ( $k = 1$ , $k = 2$ , $s^{2} = 0$ )	formulas that give $\infty$ or undefined	Ex 6
G	Limiting behaviour ( $k \to \infty$ )	$t_{k} \to Z$ , $F \to$ narrow, why	Ex 7
H	Real-world word problem (must extract $n$ , $s$ , df)	translating English into $\overset{x}{ˉ}, s, n$	Ex 8
I	Additivity of independent $χ^{2}$	$χ_{a}^{2} + χ_{b}^{2} = χ_{a + b}^{2}$	Ex 9
J	Exam twist: variance of $t$ , $F$ mean, when they don't exist	the $(k - 2)$ and $(d_{2} - 2)$ traps	Ex 10

Notice there are no "quadrant/sign" cells here as there are for angles — these distributions live on $[0, \infty)$ ( $χ^{2}$ , $F$ ) or all of $R$ ( $t$ ). The "sign" analogue for us is the boundary values of the df ( $k = 1, 2$ ; $d_{2} \leq 2$ ), where means and variances stop existing. Cell F and J guard those.

The figure below draws that geography: the black curves are $χ_{5}^{2}$ and $F_{5, 8}$ , both pinned to the left edge $0$ (they can never go negative — they are built from sums of squares), while the red curve is $t_{5}$ , symmetric about $0$ and spreading to both sides. Keep this picture in mind: $χ^{2}$ and $F$ are one-sided; $t$ is two-sided. Every example below sits somewhere on this map.

Figure — Chi-squared, t, F distributions — definition, degrees of freedom

Worked examples

Worked example Example 2 — Cell B: sample variance to

$χ^{2}$ , and why df $= n - 1$ A sample of $n = 12$ from $N (μ, σ^{2})$ gives $s^{2} = 5$ ; the true $σ^{2} = 4$ . Give the test statistic's value and distribution.

Forecast: will the df be $12$ or $11$ ?

Step 1. Statistic $\frac{( n - 1 ) s ^{2}}{σ ^{2}} = \frac{11 \cdot 5}{4} = \frac{55}{4} = 13.75$ . Why this step? This is the standard result $\frac{( n - 1 ) s ^{2}}{σ ^{2}} \sim χ_{n - 1}^{2}$ from Sample Variance and Bessel's Correction — the only combination of $s^{2}$ and $σ^{2}$ that is a clean chi-squared.

Step 2. Distribution is $χ_{n - 1}^{2} = χ_{11}^{2}$ . Why this step? Computing $\overset{x}{ˉ}$ forces $\sum (x_{i} - \overset{x}{ˉ}) = 0$ : one deviation is determined by the other eleven, so only $11$ are free. Df $=$ data $-$ estimated parameters $= 12 - 1$ .

Verify: the observed $13.75$ sits near the mean of $χ_{11}^{2}$ (which is $11$ ) — plausible, not extreme, so $s^{2} = 5$ against $σ^{2} = 4$ is unremarkable. ✓

Worked example Example 3 — Cell C: build a

$t$ statistic with unknown $σ$ $n = 25$ , sample mean $\overset{x}{ˉ} = 103$ , sample sd $s = 10$ . Test whether $μ = 100$ .

Forecast: $Z$ or $t$ ? And what df?

Step 1. Standard error $= s / n = 10/ 25 = 10/5 = 2$ . Why this step? $\overset{x}{ˉ}$ has spread $σ / n$ ; since $σ$ is unknown we use its estimate $s / n$ .

Step 2. $t = \frac{x ˉ - μ}{s / n} = \frac{103 - 100}{2} = 1.5$ . Why this step? We standardise the mean, but the denominator is now random (built from $s$ ), which is exactly what makes this a $t$ , not a $Z$ .

Step 3. Reference distribution: $t_{24}$ , since $k = n - 1 = 24$ . Why this step? The chi-squared hidden in the denominator, $\frac{( n - 1 ) s ^{2}}{σ ^{2}}$ , carries $n - 1$ df — see Hypothesis Testing for how this $t$ becomes a $p$ -value.

Verify: with $24$ df, $t_{24}$ is almost indistinguishable from $Z$ ; a value of $1.5$ is well inside the usual $\pm 2$ band, so we would not reject $μ = 100$ . ✓

Worked example Example 5 — Cell E:

$F$ for two variances + reciprocal swap Sample 1: $s_{1}^{2} = 12$ , $n_{1} = 8$ . Sample 2: $s_{2}^{2} = 3$ , $n_{2} = 6$ . Assume equal true variances. Give $F$ and its df, then the value if you had put sample 2 on top.

Forecast: guess $F$ and guess whether swapping gives $1/ F$ .

Step 1. $F = \frac{s _{1}^{2}}{s _{2}^{2}} = \frac{12}{3} = 4$ . Why this step? Under equal true variances, $\frac{s _{1}^{2} / σ ^{2}}{s _{2}^{2} / σ ^{2}} = \frac{s _{1}^{2}}{s _{2}^{2}}$ , and each $s^{2} / σ^{2}$ is a $χ^{2}$ /df — exactly the $F$ recipe.

Step 2. Df: numerator $d_{1} = n_{1} - 1 = 7$ , denominator $d_{2} = n_{2} - 1 = 5$ . So $F \sim F_{7, 5}$ . Why this step? Each sample variance carries its own $n - 1$ ; the top variance's df is the numerator df.

Step 3. Swap: $\frac{s _{2}^{2}}{s _{1}^{2}} = \frac{3}{12} = 0.25 = \frac{1}{4}$ , distributed as $F_{5, 7}$ . Why this step? Reciprocal symmetry: $1/ F_{d_{1}, d_{2}} \sim F_{d_{2}, d_{1}}$ — inverting the ratio swaps the df.

Verify: $0.25 \times 4 = 1$ , confirming the two statistics are exact reciprocals, and the df have indeed swapped from $(7, 5)$ to $(5, 7)$ . This $F$ machinery is what ANOVA runs internally. ✓

Worked example Example 6 — Cell F: degenerate & boundary inputs

(a) A sample gives $s^{2} = 0$ . What is the $χ^{2}$ statistic $\frac{( n - 1 ) s ^{2}}{σ ^{2}}$ ? (b) Does $Var (t_{2})$ exist? (c) Does $E [F_{4, 2}]$ exist?

Forecast: which of these blow up or vanish?

Step 1 (a). $s^{2} = 0$ means every data point equals $\overset{x}{ˉ}$ ; the statistic is $\frac{( n - 1 ) \cdot 0}{σ ^{2}} = 0$ . Why this step? Zero spread means zero total squared deviation — the smallest a $χ^{2}$ can be, sitting exactly at the left edge $0$ of its support.

Step 2 (b). $Var (t_{k}) = \frac{k}{k - 2}$ requires $k > 2$ . At $k = 2$ the denominator is $0$ : the variance is infinite / undefined. Why this step? With so few df the fat tails are so heavy that the squared values don't have a finite average — the mean $E [t_{2}] = 0$ still exists (needs only $k > 1$ ), but the spread does not.

Step 3 (c). $E [F_{d_{1}, d_{2}}] = \frac{d _{2}}{d _{2} - 2}$ needs $d_{2} > 2$ . Here $d_{2} = 2$ gives $\frac{2}{0}$ : the mean is undefined. Why this step? Only the denominator df controls whether the mean exists (a small $χ^{2}$ on the bottom can be near zero, sending the ratio to $\infty$ ).

Verify: the three edges line up with the parenthetical conditions in the parent's formula boxes: $E [t]$ needs $k > 1$ , $Var (t)$ needs $k > 2$ , $E [F]$ needs $d_{2} > 2$ . ✓

Worked example Example 7 — Cell G: limiting behaviour as df

$\to \infty$ As the df grows without bound, what happens to (a) $t_{k}$ , (b) $Var (t_{k})$ , (c) $\frac{χ _{k}^{2}}{k}$ , and (d) $F_{d_{1}, d_{2}}$ ?

Forecast: which of these converge to a bell, which to the number $1$ , and which collapse to a spike?

Step 1 (a). $t_{k} = \frac{Z}{V / k}$ with $V \sim χ_{k}^{2}$ . As $k \to \infty$ , $V / k \to 1$ , so $t_{k} \to Z \sim N (0, 1)$ . Why this step? $V / k$ is an average of $k$ independent $Z_{i}^{2}$ 's, each mean $1$ ; by the law of large numbers this average locks onto $1$ , so the random denominator stops being random. (This is the "average settling to its mean" law, distinct from the Central Limit Theorem which describes the shape of that average's fluctuations.)

Step 2 (b). $Var (t_{k}) = \frac{k}{k - 2} \to 1$ as $k \to \infty$ . Why this step? $\frac{k}{k - 2} = \frac{1}{1 - 2/ k} \to \frac{1}{1 - 0} = 1$ , matching $Var (Z) = 1$ . The fat tails shrink to normal tails.

Step 3 (c). $\frac{χ _{k}^{2}}{k} \to 1$ (in probability). Why this step? Same law of large numbers — the mean of $k$ terms each averaging $1$ .

Step 4 (d). $F_{d_{1}, d_{2}} = \frac{U / d _{1}}{V / d _{2}} \to 1$ as both $d_{1}, d_{2} \to \infty$ : numerator $U / d_{1} \to 1$ and denominator $V / d_{2} \to 1$ , so the ratio collapses onto the single value $1$ (its spread $\to 0$ ). Why this step? Each $χ^{2}$ divided by its own df is exactly the object from Step 3, which pins to $1$ ; a ratio of two things both pinned to $1$ is pinned to $1$ . Its mean $E [F_{d_{1}, d_{2}}] = \frac{d _{2}}{d _{2} - 2} \to 1$ confirms the centre, and the density narrows to a spike at $1$ — this is why an $F$ near $1$ means "equal variances."

Verify: check the trend numerically. $Var (t_{10}) = \frac{10}{8} = 1.25$ , $Var (t_{100}) = \frac{100}{98} \approx 1.0204$ — marching toward $1$ . And $E [F_{10, 10}] = \frac{10}{8} = 1.25$ while $E [F_{100, 100}] = \frac{100}{98} \approx 1.0204$ — also marching to $1$ . ✓

The figure shows all three limits at once: the black $t$ curves lose their heavy tails and sink onto the red normal curve as $k$ climbs ( $t_{k} \to N (0, 1)$ ), while the inset dashed curves for $χ_{k}^{2} / k$ and $F_{k, k}$ narrow onto the value $1$ — the visual meaning of "everything settles to $1$ ."

Worked example Example 8 — Cell H: real-world word problem

A coffee machine is supposed to pour $250$ mL. An inspector pulls $n = 16$ cups, finds an average of $\overset{x}{ˉ} = 246$ mL and a sample standard deviation $s = 8$ mL. Set up and evaluate the appropriate test statistic and name its reference distribution.

Forecast: which quantities map to $n$ , $\overset{x}{ˉ}$ , $s$ , $μ$ ? Is $σ$ known?

Step 1. Extract: $μ = 250$ (the claimed mean), $\overset{x}{ˉ} = 246$ , $s = 8$ , $n = 16$ . True $σ$ is not given. Why this step? "Supposed to pour $250$ " is the hypothesised mean; the sd we have is a sample sd, so $σ$ is unknown → this is a $t$ situation, not $Z$ .

Step 2. Standard error $= s / n = 8/ 16 = 8/4 = 2$ mL. Why this step? The mean of $16$ cups wobbles $16 = 4$ times less than a single cup.

Step 3. $t = \frac{x ˉ - μ}{s / n} = \frac{246 - 250}{2} = \frac{- 4}{2} = - 2$ . Why this step? Negative because the machine underfills; magnitude $2$ measures how many standard errors below target.

Step 4. Reference distribution $t_{15}$ ( $k = n - 1 = 15$ ). Why this step? The estimated standard deviation $s$ hides a $χ_{n - 1}^{2}$ (we computed $\overset{x}{ˉ}$ from the same data, spending one df on the constraint $\sum (x_{i} - \overset{x}{ˉ}) = 0$ ), so with $n = 16$ the $t$ carries $16 - 1 = 15$ df — never $n$ .

Verify: units cancel correctly — $mL / mL$ leaves $t$ dimensionless, as any test statistic must be. A value of $- 2$ against $t_{15}$ is near the $5%$ two-sided boundary, so the underfill is borderline-significant. ✓

Worked example Example 10 — Cell J: the exam twist (

$t$ variance, $F$ mean, existence) (a) Give $Var (t_{5})$ . (b) Give $E [F_{6, 10}]$ . (c) For $t_{30}$ , by how much (percent) does its variance exceed the normal's?

Forecast: all three plug into the $(k - 2)$ / $(d_{2} - 2)$ formulas — watch the denominators.

Step 1 (a). $Var (t_{5}) = \frac{k}{k - 2} = \frac{5}{3} \approx 1.6667$ . Why this step? Here $k = 5 > 2$ so the variance exists; the answer exceeds $1$ , confirming fatter-than-normal tails.

Step 2 (b). $E [F_{6, 10}] = \frac{d _{2}}{d _{2} - 2} = \frac{10}{8} = 1.25$ . Why this step? Only the denominator df $d_{2} = 10$ enters the mean; $10 > 2$ so it exists. The numerator df $d_{1} = 6$ never appears in the mean — a classic trap.

Step 3 (c). $Var (t_{30}) = \frac{30}{28} \approx 1.0714$ , so it exceeds the normal's variance of $1$ by about $7.14%$ . Why this step? $\frac{30}{28} - 1 = \frac{2}{28} = \frac{1}{14} \approx 0.0714$ — a $7%$ excess, small because $30$ df is already close to normal.

Verify: all denominators satisfy their existence conditions ( $k > 2$ , $d_{2} > 2$ ), so none of these is undefined — contrast with Ex 6 where they were. And $Var (t_{5}) = 1.6667 > Var (t_{30}) = 1.0714 > 1$ , correctly showing more df ⇒ closer to normal. ✓

Recall Rapid self-test

Which cell: " $n = 20$ , unknown $σ$ , test the mean"? ::: Cell C — build a $t_{19}$ . $t \sim t_{9}$ ; distribution of $t^{2}$ ? ::: $F_{1, 9}$ (Cell D). Does $Var (t_{2})$ exist? ::: No — needs $k > 2$ (Cell F). $χ_{3}^{2} + χ_{5}^{2}$ independent equals? ::: $χ_{8}^{2}$ (Cell I). $E [F_{5, 10}]$ ? ::: $10/8 = 1.25$ (Cell J).