z-test, t-test, chi-squared goodness of fit, F-test: Worked examples

This page is a drill. The parent note built the machinery — here we walk through every case class the four tests can throw at you: known vs unknown spread, one-sided vs two-sided, tiny samples, degenerate/limiting inputs, and exam traps.

Before symbols show up, one reminder in plain words so line one is followable:

Definition The three symbols we reuse everywhere below

$H_{0}$ = the null hypothesis, the "boring guess" or status-quo claim we try to knock down — e.g. "the mean is exactly $μ_{0}$ " or "the die is fair." Every test assumes $H_{0}$ is true while computing the luck curve, then asks whether the data still fits. We either reject $H_{0}$ (data too weird for luck) or fail to reject $H_{0}$ (luck is a fine explanation). We never "prove $H_{0}$ true."
$k$ = the number of categories (bins) in a chi-squared goodness-of-fit test — e.g. the $6$ faces of a die give $k = 6$ .
Critical-value subscript convention: we write $c_{p, ν}$ where the subscript $p$ is the tail area to the right (the area cut off in the far tail) and $ν$ is the degrees of freedom. So $z_{0.025} = 1.96$ means "the point with $2.5%$ of a standard Normal above it," and $t_{0.025, 5} = 2.571$ means the same idea for a $t$ with $ν = 5$ . For a two-sided test at level $α$ we split $α$ into two tails and use tail area $α /2$ ; for a one-sided test we keep the whole $α$ in one tail and use tail area $α$ .

The scenario matrix

Each row is a case class you must be able to handle. The right column names the worked example that lands on it.

#	Case class	What makes it tricky	Example
A	Mean, $σ$ known, two-sided	denominator is a fixed constant	Ex 1
B	Mean, $σ$ known, one-sided	tail lives on one side only (sign of cutoff)	Ex 2
C	Mean, $σ$ unknown, small $n$	must use $s$ → fatter-tailed $t$	Ex 3
D	Limiting case: $n \to \infty$	$t$ collapses into $z$	Ex 4
E	GOF, uniform expected	all $E_{i}$ equal, $ν = k - 1$	Ex 5
F	GOF, estimated parameter	lose an extra d.f., $ν = k - 1 - m$	Ex 6
G	Degenerate GOF: $O_{i} = E_{i}$	zero surprise, $χ^{2} = 0$	Ex 7
H	F-test, two variances, two-sided	larger $s^{2}$ on top, $F \geq 1$	Ex 8
I	Word problem end-to-end	choosing the right test yourself	Ex 9
J	Exam twist: same data, wrong test	how the verdict flips	Ex 10

We lean on Central Limit Theorem, Student's t-distribution, Chi-squared Distribution, F-distribution, Degrees of Freedom and Bessel's Correction throughout — the parent note defined them; here we use them.

Figure — z-test, t-test, chi-squared goodness of fit, F-test

How to read the figure (four small panels):

Top-left overlays the standard Normal (blue, the $z$ curve) and a $t$ with $ν = 3$ (orange). The shaded orange slivers past $\pm 1.9$ are the extra tail area the $t$ carries — that shaded region is precisely why the $t$ critical value ( $2.571$ in Ex 3) sits farther out than the $z$ 's $1.96$ .
Top-right shows two $χ^{2}$ curves (used in Ex 5–7); the red dashed line at $7.82$ is the $ν = 3$ upper-tail fence — note $χ^{2}$ lives only on the positive axis, so there is no lower tail.
Bottom-left shows $F_{7, 10}$ (Ex 8) with its upper critical value at $4.03$ .
Bottom-right stacks $t$ curves for $ν = 1$ then $ν = 8$ against the Normal, so you can watch the tails thin as $ν$ grows — the visual meaning of " $t \to z$ " in Ex 4.

Keep this figure in view: every example below points back to one of its four panels.

Cell A — Mean, $σ$ known, two-sided

Worked example Ex 1 · Cereal box weight

A machine should fill boxes to $μ_{0} = 500$ g. Long factory history gives a known $σ = 6$ g. A sample of $n = 36$ boxes averages $\overset{x}{ˉ} = 502.4$ g. At $α = 0.05$ , is the machine off-target (either way)?

Forecast: guess now — will $∣ Z ∣$ beat $1.96$ , or not? Jot a yes/no.

Standard error $SE = \frac{σ}{n} = \frac{6}{36} = \frac{6}{6} = 1$ . Why this step? Averaging $36$ boxes shrinks the wobble by $36 = 6$ ; the mean is $6 \times$ steadier than one box.
Z-statistic $Z = \frac{x ˉ - μ _{0}}{SE} = \frac{502.4 - 500}{1} = 2.4$ . Why this step? This counts how many standard errors the sample sits above the target — a pure, unit-free distance.
Critical-value rule: two-sided tail area $α /2 = 0.025$ gives $z_{0.025} = 1.96$ ; $∣2.4∣ > 1.96$ . Why this step? "Either way" off-target means both tails count, so we split $α$ and use tail area $0.025$ per side (top-left panel of the figure).
p-value rule: for a two-sided $z = 2.4$ , the tail area beyond $\pm 2.4$ is $2 \times P (Z > 2.4) \approx 2 \times 0.0082 = 0.0164$ . Since $0.0164 < 0.05$ , same verdict. Why this step? The p-value quantifies how rare — under $H_{0}$ we'd see a deviation this big only about $1.6%$ of the time.
Both rules $\Rightarrow$ reject $H_{0}$ . The machine is significantly off-target. Why this step? Statistic beyond the fence and p-value below $α$ are the same fact stated two ways.

Verify: $σ$ was known, so the denominator is a constant — only $\overset{x}{ˉ}$ is random, and by the Central Limit Theorem it is Normal. Using $z$ (not $t$ ) is correct. Units: g $/$ g $=$ dimensionless ✓.

Cell B — Mean, $σ$ known, one-sided (mind the sign of the cutoff)

Worked example Ex 2 · Is the drug

faster? A standard painkiller acts in $μ_{0} = 30$ min, known $σ = 8$ min. A new pill on $n = 64$ patients averages $\overset{x}{ˉ} = 28$ min. Does it act faster ( $H_{1} : μ < 30$ )? Use $α = 0.05$ .

Forecast: one-sided cutoff is closer to zero than $1.96$ — will that change the verdict?

$SE = \frac{8}{64} = \frac{8}{8} = 1$ . Why this step? Same shrink-by- $n$ logic; $n = 64 \Rightarrow$ factor $8$ .
$Z = \frac{28 - 30}{1} = - 2.0$ . Why this step? Negative because the sample is below the claim — exactly the direction "faster" predicts.
Sign of the cutoff. Our subscript rule gives $z_{0.05} = 1.645$ , which is the point with $5%$ of the Normal above it. But $H_{1} : μ < 30$ pushes the rejection region into the lower tail, so the actual fence is the mirror-image point $- z_{0.05} = - 1.645$ . Critical-value rule: reject if $Z < - 1.645$ . Why this step? The subscript $p$ always names an upper tail area, so a lower-tail test flips the sign: the fence is $- z_{0.05}$ , not $+ z_{0.05}$ . Forgetting the minus is a classic error.
$- 2.0 < - 1.645 \Rightarrow$ statistic passes the lower fence. Why this step? The result sits farther into the lower tail than the cutoff allows under luck alone.
p-value rule: one-sided, lower tail, $p = P (Z < - 2.0) \approx 0.0228$ . Since $0.0228 < 0.05$ , same verdict. Why this step? We only count the predicted (lower) direction, so we do not double it — a one-sided p-value is half of the two-sided one for the same $∣ Z ∣$ .
Both rules $\Rightarrow$ reject $H_{0}$ . Evidence the pill is faster.

Verify: the same $∣ Z ∣ = 2.0$ gives two-sided p $\approx 0.0455$ and one-sided p $\approx 0.0228$ — exactly half. The one-sided test is easier to reject in the predicted direction. Consistent ✓.

Cell C — Mean, $σ$ unknown, small $n$

Worked example Ex 3 · Handmade nails length

Target $μ_{0} = 50$ mm. We measure $n = 6$ nails: $51, 49, 52, 50, 48, 50$ . We have no known $σ$ . Is the length off-target at $α = 0.05$ (two-sided)?

Forecast: with only $6$ nails and unknown spread, do you expect the critical value to be near $1.96$ or noticeably bigger?

Mean $\overset{x}{ˉ} = \frac{51 + 49 + 52 + 50 + 48 + 50}{6} = \frac{300}{6} = 50$ . Why this step? The centre of the data — our best guess of $μ$ .
Deviations $1, - 1, 2, 0, - 2, 0$ ; squares $1, 1, 4, 0, 4, 0$ ; sum $= 10$ . Why this step? Squaring removes signs so pluses and minuses don't cancel; big misses count more.
Sample variance with Bessel's Correction: $s^{2} = \frac{10}{n - 1} = \frac{10}{5} = 2$ , so $s = 2 \approx 1.414$ . Why divide by $n - 1$ ? Deviations are from $\overset{x}{ˉ}$ , not the true $μ$ ; the data hugs its own mean, so we'd underestimate spread using $n$ . One d.f. is spent on the constraint $\sum (x_{i} - \overset{x}{ˉ}) = 0$ .
$SE = \frac{s}{n} = \frac{1.414}{6} \approx 0.577$ . Why this step? Convert the spread of single nails into the spread of their average by dividing by $n$ .
$t = \frac{x ˉ - μ _{0}}{SE} = \frac{50 - 50}{0.577} = 0$ , with $ν = n - 1 = 5$ . Why this step? Same "distance in SE" idea, but with the estimated $s$ → we read it against Student's t-distribution (top-left/bottom-right panels).
Critical-value rule: two-sided tail area $α /2 = 0.025$ gives $t_{0.025, 5} = 2.571$ ; $∣0∣ < 2.571$ . p-value rule: $p = 2 \times P (t_{5} > 0) = 2 \times 0.5 = 1.0 > 0.05$ . Both $\Rightarrow$ fail to reject. Why this step? A statistic of exactly $0$ is the least surprising outcome possible; its p-value is the maximum, $1.0$ .

Verify: $\overset{x}{ˉ}$ landed exactly on $μ_{0}$ , so the numerator is $0$ and $t = 0$ regardless of $s$ — the maximally boring outcome. No evidence of off-target ✓. Note $2.571 > 1.96$ : the $t$ is more demanding, as its fatter tails require.

Cell D — Limiting case $n \to \infty$

Worked example Ex 4 · When

$t$ becomes $z$ Same style as Ex 3 but with a huge sample: $n = 2001$ , $\overset{x}{ˉ} = 50.2$ , $s = 1.414$ , target $μ_{0} = 50$ . Two-sided, $α = 0.05$ .

Forecast: with $ν = 2000$ , will the critical value be closer to $1.96$ or to $2.571$ ?

$SE = \frac{1.414}{2001} \approx \frac{1.414}{44.73} \approx 0.03162$ . Why this step? Enormous $n$ makes the mean extremely steady.
$t = \frac{50.2 - 50}{0.03162} \approx 6.32$ , $ν = 2000$ . Why this step? Same standardized-distance formula as Ex 3; the tiny $SE$ turns a modest $0.2$ mm gap into a huge number of standard errors.
Critical-value rule: two-sided tail area $0.025$ gives $t_{0.025, 2000} \approx 1.961$ — essentially the $z$ value $1.96$ ; $6.32 > 1.961$ . p-value rule: $p = 2 \times P (t_{2000} > 6.32) \approx 3 \times 1 0^{- 10}$ , far below $0.05$ . Both $\Rightarrow$ reject $H_{0}$ . Why this step? As $n \to \infty$ , $s \to σ$ , the extra denominator-noise vanishes, and $t \to N (0, 1)$ ; the fat tails thin out (bottom-right panel).

Verify: the ratio $t_{0.025, 2000} /1.96 \approx 1.0005$ — within $0.1%$ of the Normal. This is the concrete meaning of " $t \to z$ ." Using $z$ here would give the same verdict ✓.

Cell E — GOF, uniform expected counts

Worked example Ex 5 · Is the spinner fair?

A $4$ -colour spinner is spun $n = 100$ times: $O = (30, 20, 22, 28)$ . A fair spinner predicts each colour equally. Here the number of categories is $k = 4$ . Test fairness at $α = 0.05$ .

Forecast: the counts wobble around $25$ — do you think that wobble is "just luck"?

Expected each cell $E_{i} = \frac{100}{4} = 25$ . Why this step? "Fair" spreads the total evenly across $k = 4$ categories.
Check the $E_{i} \geq 5$ rule. All $E_{i} = 25 \geq 5$ , so the $χ^{2}$ approximation is valid — proceed. Why this step? The chi-squared curve only approximates the true count-distribution when expected cells aren't tiny; we confirm before trusting the fence.
Cell contributions $\frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$ : deviations $5, - 5, - 3, 3$ ; squares $25, 25, 9, 9$ ; each over $25$ : $1, 1, 0.36, 0.36$ . Why divide by $E_{i}$ ? A count in a cell behaves roughly Poisson with variance $\approx E_{i}$ , so $\frac{O _{i} - E _{i}}{E _{i}}$ is a standardized $z$ -like quantity; squaring and summing gives a Chi-squared Distribution.
$χ^{2} = 1 + 1 + 0.36 + 0.36 = 2.72$ . Why this step? Summing the per-cell surprises gives one total "how badly does the model fit" number.
Degrees of freedom $ν = k - 1 = 4 - 1 = 3$ (the total is fixed at $100$ , costing one d.f.). Why this step? The counts must sum to $n$ , so only $k - 1$ of them wander freely. See Degrees of Freedom.
Critical-value rule: tail area $α = 0.05$ gives $χ_{0.05, 3}^{2} = 7.815$ ; $2.72 < 7.815$ . p-value rule: $p = P (χ_{3}^{2} > 2.72) \approx 0.437 > 0.05$ . Both $\Rightarrow$ spinner looks fair. Why this step? Chi-squared has only an upper tail (surprise can't be negative), so "extreme" means "large" and the whole $α$ sits on the right.

Verify: all $E_{i} = 25 \geq 5$ ✓. Our surprise ( $2.72$ ) is well under what luck routinely produces (up to $\sim 7.8$ ); p-value $\approx 0.44$ says such a mismatch happens $44%$ of the time by chance ✓.

Cell F — GOF with an estimated parameter

Worked example Ex 6 · Counting typos per page (Poisson fit)

We count typos on $n = 100$ pages and bin them into $k = 4$ categories (0, 1, 2, 3+). We estimated the Poisson rate $λ$ from the same data (that's $m = 1$ fitted parameter). The expected counts come out $E = (37, 37, 18, 8)$ and observed $O = (40, 34, 20, 6)$ . Does the Poisson model fit at $α = 0.05$ ?

Forecast: estimating $λ$ costs a d.f. — will $ν$ be $3$ or $2$ ?

Check the $E_{i} \geq 5$ rule. The expecteds are $37, 37, 18, 8$ — the smallest is $8 \geq 5$ , so the $χ^{2}$ approximation holds and no pooling is needed. Why this step? If any $E_{i}$ had fallen below $5$ (say the "3+" bin came out $E = 2$ ), we would merge it with a neighbouring bin (e.g. combine "2" and "3+"), recompute, and reduce $k$ by one — tiny expecteds inflate the statistic and break the approximation.
Cell contributions $\frac{( O _{i} - E _{i} ) ^{2}}{E _{i}}$ :
- $\frac{( 40 - 37 ) ^{2}}{37} = \frac{9}{37} \approx 0.2432$
- $\frac{( 34 - 37 ) ^{2}}{37} = \frac{9}{37} \approx 0.2432$
- $\frac{( 20 - 18 ) ^{2}}{18} = \frac{4}{18} \approx 0.2222$
- $\frac{( 6 - 8 ) ^{2}}{8} = \frac{4}{8} = 0.5$ Why this step? Each cell's squared miss is scaled by the count we expected there, turning raw errors into standardized, comparable surprises.
$χ^{2} \approx 0.2432 + 0.2432 + 0.2222 + 0.5 = 1.2087$ . Why this step? Adding the four standardized surprises gives the single total we test against the Chi-squared Distribution.
Degrees of freedom $ν = k - 1 - m = 4 - 1 - 1 = 2$ . Why this step? Two constraints now: the total is fixed (−1) and we forced $λ$ to match the data (−1 more). Each fitted parameter buys one extra constraint.
Critical-value rule: tail area $α = 0.05$ gives $χ_{0.05, 2}^{2} = 5.991$ ; $1.21 < 5.991$ . p-value rule: $p = P (χ_{2}^{2} > 1.2087) \approx 0.546 > 0.05$ . Both $\Rightarrow$ Poisson model fits. Why this step? The total surprise is tiny next to the upper-tail fence, and a mismatch this small happens over half the time by chance.

Verify: compare to the trap of using $ν = 3$ : critical would be $7.815$ and $p = P (χ_{3}^{2} > 1.2087) \approx 0.751$ — the verdict survives either way here, but with a borderline statistic the wrong $ν$ flips it. Always subtract fitted parameters $m$ ✓.

Cell G — Degenerate GOF: perfect match

Worked example Ex 7 · The suspiciously perfect die

A die is rolled $60$ times and lands $O = (10, 10, 10, 10, 10, 10)$ — exactly the fair expectation $E_{i} = 10$ . Here the number of categories is $k = 6$ .

Forecast: what is the smallest a $χ^{2}$ can ever be?

Check the $E_{i} \geq 5$ rule. Every $E_{i} = 10 \geq 5$ ✓ — the approximation is valid. Why this step? Same guard as Ex 5–6; even a degenerate case needs the expecteds large enough for the fence to mean anything.
Every cell $\frac{( O _{i} - E _{i} ) ^{2}}{E _{i}} = \frac{( 10 - 10 ) ^{2}}{10} = 0$ . Why this step? Zero deviation everywhere means zero squared surprise.
$χ^{2} = 0$ . Why this step? A sum of zeros is zero — the model matched every category exactly.
With $ν = k - 1 = 5$ : critical-value rule — any fence $> 0$ , so we never reject. p-value rule — $p = P (χ_{5}^{2} > 0) = 1.0$ , the maximum. Both $\Rightarrow$ never reject; the fit is as good as arithmetically possible. Why this step? $χ^{2} \geq 0$ always (a sum of squares over positives); $0$ is the floor, so nothing pushes it into the rejection tail, and its p-value is $1$ .

Verify: $χ^{2}$ can never be negative and equals $0$ iff every $O_{i} = E_{i}$ ; then p-value $= 1$ . In real life a perfect match is itself odd, but the GOF test only flags bad fits in its upper tail, not "too good" fits — so a suspiciously tidy dataset needs a different check, not this one. ✓

Cell H — F-test, two variances, two-sided

How to read this figure: the blue curve is the $F_{7, 10}$ luck curve for the ratio of two sample variances. The red dashed line at $4.03$ is the upper-tail fence; the red-shaded region beyond it is the rejection zone. The orange line marks our observed $F = 3.0$ — sitting left of the fence, inside the "could be luck" region.

Worked example Ex 8 · Which lathe is more consistent?

Lathe A: $s_{1}^{2} = 45$ , $n_{1} = 8$ . Lathe B: $s_{2}^{2} = 15$ , $n_{2} = 11$ . Are their variances significantly different at $α = 0.05$ (two-sided)?

Forecast: the ratio is $3$ — big enough to beat the critical $F$ ?

Put the larger $s^{2}$ on top: $F = \frac{s _{1}^{2}}{s _{2}^{2}} = \frac{45}{15} = 3.0$ . Why this step? Forcing $F \geq 1$ lets us use a single upper-tail critical value instead of chasing the lower tail. See F-distribution.
Degrees of freedom $ν_{1} = n_{1} - 1 = 7$ (numerator), $ν_{2} = n_{2} - 1 = 10$ (denominator). Why this step? Each variance estimate is a $χ^{2} / ν$ ; the $F$ is a ratio of two such, one d.f. per estimate.
Critical-value rule: two-sided at $α = 0.05$ puts $2.5%$ in each tail; with the bigger variance on top we read the upper value with tail area $α /2 = 0.025$ : $F_{0.025, 7, 10} \approx 4.03$ ; $3.0 < 4.03$ . Why this step? "Different" allows either lathe to be noisier, so we split $5%$ ; placing the bigger on top means only the upper $2.5%$ tail is ever tested.
p-value rule: $p = 2 \times P (F_{7, 10} > 3.0) \approx 2 \times 0.0503 \approx 0.101 > 0.05$ . Same verdict. Why this step? We double the one-tail area because a two-sided variance test allows an extreme ratio either way; $0.101$ says a ratio this far from $1$ happens about $10%$ of the time under equal variances.
Both rules $\Rightarrow$ fail to reject. Variances not significantly different.

Verify: the figure shows $F = 3.0$ (orange) left of $4.03$ (red dashed) — inside the "could be luck" region, matching p $\approx 0.10 > 0.05$ . Order-of-magnitude $F \approx 3$ with $\sim 10$ d.f. is common under $H_{0}$ ✓.

Cell I — Word problem, choose the test yourself

Worked example Ex 9 · The café owner

"My espresso machine is supposed to dispense $30$ ml per shot. I don't trust the factory's spread, so I measured $10$ shots myself: $\overset{x}{ˉ} = 31.2$ ml, $s = 2.4$ ml. Am I over-dispensing at $α = 0.05$ ?"

Forecast: which of the four tests, and why? Decide before reading step 1.

Pick the test. We compare one mean to a target, $σ$ is unknown (owner doesn't trust the factory value) and $n = 10$ is small $\Rightarrow$ one-sample $t$ -test, one-sided ( $H_{1} : μ > 30$ ). Why this step? Unknown $σ$ + estimated $s$ = the exact signature that forces $t$ over $z$ (Hypothesis Testing).
$SE = \frac{s}{n} = \frac{2.4}{10} \approx 0.7589$ . Why this step? Turn the spread of single shots into the spread of their average by dividing by $n$ — the wobble of the quantity we actually test.
$t = \frac{x ˉ - μ _{0}}{SE} = \frac{31.2 - 30}{0.7589} \approx 1.581$ , $ν = 9$ . Why this step? The standardized distance: how many standard errors the measured average sits above the target, read against $t$ with $ν = n - 1 = 9$ .
Critical-value rule. "Over-dispensing" is a one-direction upper claim, so the whole $α = 0.05$ sits in the upper tail: $t_{0.05, 9} = 1.833$ ; reject if $t > 1.833$ . Here $1.581 < 1.833$ — no sign flip needed because the claim is upper-tail, so the fence is $+ t_{0.05, 9}$ . Why this step? Contrast with Ex 2: there the claim was lower-tail so we used $- z_{0.05}$ ; here it is upper-tail so we keep the plus sign. The direction of $H_{1}$ dictates the sign.
p-value rule. $p = P (t_{9} > 1.581) \approx 0.074 > 0.05$ . Same verdict. Why this step? One-sided, upper direction only, so no doubling; $0.074$ narrowly exceeds $α$ .
Both rules $\Rightarrow$ fail to reject. Not enough evidence of over-dispensing.

Verify: the p-value $\approx 0.074 > 0.05$ is consistent with "fail to reject," and it is close to $α$ — a slightly larger $\overset{x}{ˉ}$ would tip it over. Units ml $/$ ml cancel ✓.

Cell J — Exam twist: same data, wrong test flips the verdict

Worked example Ex 10 · The tempting shortcut

Take Ex 9's numbers ( $\overset{x}{ˉ} = 31.2$ , $s = 2.4$ , $n = 10$ , target $30$ ). A student says: "The CLT makes everything Normal, so I'll just use a $z$ -test with $1.645$ ." Does the verdict change?

Forecast: same statistic value $1.581$ — but a different reference curve. Reject or not?

The student computes the same ratio $z = \frac{31.2 - 30}{2.4/ 10} \approx 1.581$ (they plug $s$ where $σ$ should be). Why this step? Numerically identical to Ex 9's $t$ ; only the reference curve differs, so the arithmetic can't reveal the mistake.
Wrong critical value. They compare to the one-sided $z_{0.05} = 1.645$ instead of $t_{0.05, 9} = 1.833$ . Why this step? A $z$ -test reads the Normal, whose one-sided $5%$ point is $1.645$ ; the correct $t$ curve for $ν = 9$ has a farther fence at $1.833$ .
Wrong p-value too. The Normal p-value $P (Z > 1.581) \approx 0.057$ is smaller than the correct $t$ p-value $P (t_{9} > 1.581) \approx 0.074$ , because the $t$ 's fatter tail piles more probability past $1.581$ . Why this step? Both the fence and the p-value get distorted; the shortcut makes the result look more significant than it is.
$1.581 < 1.645 \Rightarrow$ still fail to reject here — but only barely. Had $\overset{x}{ˉ}$ been $31.3$ , the wrong-test $z$ ( $z$ p $\approx 0.041 < 0.05$ ) would reject while the correct $t$ would not. Why this step? Since $1.645 < 1.833$ , the $z$ rejection region is wider: the shortcut rejects too easily, inflating the Type I error rate above the promised $5%$ .

Verify: with unknown $σ$ and $n = 10$ , the correct critical value is $1.833$ , not $1.645$ . The gap $1.833 - 1.645 = 0.188$ is the exact cost of pretending $s$ is $σ$ ✓.

Recall Fast self-check

Which cell has $χ^{2}$ exactly $0$ , and when? ::: Cell G — degenerate GOF, whenever every $O_{i} = E_{i}$ ; $χ^{2}$ is a sum of squares so $0$ is its floor and its p-value is $1$ . In Ex 2 why is the fence $- 1.645$ not $+ 1.645$ ? ::: The subscript $p$ names an upper tail area, but $H_{1} : μ < 30$ puts the rejection region in the lower tail, so we mirror to $- z_{0.05}$ . In Ex 8 why is the larger variance on top? ::: So $F \geq 1$ and we only ever read the single upper-tail critical value. In Ex 6 why is $ν = 2$ not $3$ ? ::: We estimated one parameter ( $λ$ ) from the data, costing an extra degree of freedom: $ν = k - 1 - m = 4 - 1 - 1$ . State both verdict rules. ::: Critical-value rule: reject if statistic beyond $c_{p, ν}$ . p-value rule: reject if p-value $< α$ . They always agree.