Worked examples — Hypothesis testing — null - alternative, test statistic, p-value, errors (Type I & II)

This page is a shooting range: every kind of hypothesis-test question the topic can fire at you, one after another, fully worked. We start by drawing the map of all cases, then walk each cell with a picture-first, guess-first method. Everything here builds on the parent Hypothesis testing — null - alternative, test statistic, p-value, errors (Type I & II) and leans on Central Limit Theorem, Standard Error, Normal Distribution and the Student t-distribution.

Before a single symbol appears, here is the plain-words dictionary we will keep re-using, so nothing is ever "assumed":

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The scenario matrix

Every question in this topic is one row of this table. The examples below tick off every cell.

#	Case class	What changes	Example that hits it
A	Two-tailed, known $\sigma$ → z	direction $\ne$	Ex 1
B	Right-tailed, known $\sigma$ → z	direction $>$, statistic positive	Ex 2
C	Left-tailed, unknown $\sigma$ → t	direction $<$, statistic negative	Ex 3
D	Boundary / degenerate: $p=\alpha$ exactly, and $\bar{X}={\mu }_0$ so $z=0$	zero input	Ex 4
E	Non-rejection read correctly (absence of evidence)	large p-value	Ex 5
F	Limiting behaviour: same effect, growing $n$ → power rises	$n\to \infty$	Ex 6
G	Type-II / power computation	compute $\beta$	Ex 7
H	Real-world word problem (proportions)	different data type	Ex 8
I	Exam twist: they hand you the p-value and ask a trap question	conceptual	Ex 9

The three signs of the test statistic — positive, negative, zero — are covered by B, C, D. The three tail directions — two-tailed, right, left — are A, B, C. Good coverage means: no student hits a case we skipped.

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Ex 1 — Case A: two-tailed z-test (both signs matter)

Forecast: guess first — is $3$ ml below target across 64 bottles surprising, or normal wobble? Jot a yes/no before reading on.

Step 1. Write $H_0:\mu =500$, $H_1:\mu \ne 500$. Why this step? "Miscalibrated" says nothing about direction — too full is as bad as too empty — so both tails count. That is what $\ne$ means.

Step 2. Standard error $\text{SE}=\frac{\sigma }{\sqrt{n}}=\frac{8}{\sqrt{64}}=\frac{8}{8}=1$ ml. Why this step? The sample mean wobbles less than a single bottle because averaging cancels noise; the Central Limit Theorem says that wobble equals $\sigma /\sqrt{n}$. See Standard Error.

Step 3. $z=\frac{\bar{X}-{\mu }_0}{\text{SE}}=\frac{497-500}{1}=-3$. Why this step? We convert "3 ml below" into "3 wobbles below," a unit-free count. Look at figure s01: the observed mean sits 3 tick-marks left of centre.

Step 4. Two-tailed p-value $=2P(Z\ge |-3|)=2P(Z\ge 3)=2(0.001350)=0.0027$. Why this step? "As extreme or more" for $\ne$ means either far-left or far-right, so we take one tail's area and double it (the Normal Distribution is symmetric).

Step 5. $0.0027<0.05\Rightarrow$ reject $H_0$. The machine is off.

Verify: SE units are ml ($8\text{ml}/8$), $z$ is unitless (ml/ml) ✓. A $z$ of $-3$ is famously rare (the "0.13%" tail), and doubling gives $0.27\%$ — matches intuition that a 3-wobble miss is a red flag.

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Ex 2 — Case B: right-tailed z-test (positive statistic)

Forecast: 2 minutes over, 100 people. Reject or not? Guess.

Step 1. $H_0:\mu =30$, $H_1:\mu >30$. Why this step? The suspicion has a direction ("longer") → right tail only. We ignore the left tail entirely.

Step 2. $\text{SE}=\frac{10}{\sqrt{100}}=\frac{10}{10}=1$ min.

Step 3. $z=\frac{32-30}{1}=+2$. Why this step? Positive because $\bar{X}$ sits above the claim — the sign of $z$ literally records "which side."

Step 4. Right-tailed p-value $=P(Z\ge 2)=0.0228$. Why this step? Only the right tail is "as extreme or more" when $H_1$ points right. No doubling here (contrast Ex 1).

Step 5. $0.0228<0.05\Rightarrow$ reject $H_0$: evidence commutes are longer.

Verify: Same $|z|=2$ would give a two-tailed p of $0.0456$ — still significant, but larger, showing the one-tailed test is more powerful in the pre-chosen direction. Units: min/min = unitless ✓.

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Ex 3 — Case C: left-tailed t-test (negative statistic, σ unknown)

Forecast: small sample, 1.5 kg under. Reject?

Step 1. $H_0:\mu =50$, $H_1:\mu <50$ (left tail — "underweight" points left).

Step 2. Why $t$ and not $z$? We do not know the true spread $\sigma$; we only have the sample estimate $s$. Estimating the spread adds extra uncertainty, so we use the fatter-tailed Student t-distribution with $df=n-1=8$.

Step 3. $\text{SE}=\frac{s}{\sqrt{n}}=\frac{2}{\sqrt{9}}=\frac{2}{3}\approx 0.6667$ kg.

Step 4. $t=\frac{48.5-50}{0.6667}=\frac{-1.5}{0.6667}=-2.25$. Why this step? Negative = below claim, matching the left-tail suspicion.

Step 5. Critical value $t_{0.05,8}\approx 1.860$ (right side); for a left tail we compare to $-1.860$. Since $-2.25<-1.860$, we are farther out than the cutoff $\Rightarrow$ reject $H_0$.

Verify: With $df=8$, the one-sided p-value of $t=-2.25$ is about $0.0273<0.05$ ✓ — agrees with the critical-value decision. Two ways, same verdict.

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Ex 4 — Case D: the degenerate boundary ($z=0$ and $p=\alpha$)

Forecast: what should the test say when your data lands dead on the claim?

Step 1 (a). $z=\frac{500-500}{1}=0$. Why this step? Zero distance → zero wobbles. The data is the least surprising it could possibly be.

Step 2 (a). Two-tailed $p=2P(Z\ge 0)=2(0.5)=1$. Why this step? Every possible dataset is "as extreme or more" than the exact centre, so the whole probability mass qualifies → $p=1$. We fail to reject — as we must.

Step 3 (b). For $\alpha =0.05$ two-tailed we need $2P(Z\ge |z|)=0.05$, i.e. $P(Z\ge |z|)=0.025$, giving the famous $|z|=1.96$. Why this step? This is the razor's edge. The convention is: at exactly $p=\alpha$ we reject (the rule is $p\le \alpha$). So $|z|=1.96$ is the smallest $|z|$ that still rejects.

Verify: (a) $2\times 0.5=1$ ✓. (b) $P(Z\ge 1.96)\approx 0.025$ so doubled $=0.05=\alpha$ ✓ — the critical value $1.96$ is recovered exactly.

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Ex 5 — Case E: a correct non-rejection (absence of evidence)

Forecast: does 1 point above prove the method works?

Step 1. $H_0:\mu =70$, $H_1:\mu >70$ (right tail).

Step 2. $\text{SE}=\frac{6}{\sqrt{16}}=\frac{6}{4}=1.5$.

Step 3. $t=\frac{71-70}{1.5}=\frac{1}{1.5}=0.6667$, $df=15$.

Step 4. Critical $t_{0.05,15}\approx 1.753$. Since $0.6667<1.753\Rightarrow$ fail to reject $H_0$.

Step 5 — the wording that scores marks. We did not prove the method useless. A 1-point rise on only 16 students is well inside luck's reach; a bigger $n$ might still reveal a real effect. This is absence of evidence, not evidence of absence.

Verify: The one-sided p-value for $t=0.667$, $df=15$ is about $0.258$ — huge, nowhere near $0.05$ ✓. Consistent with "shrug and keep believing $H_0$."

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Ex 6 — Case F: limiting behaviour, grow $n$ to infinity

Forecast: does a fixed 2-minute gap become more or less significant as data piles up?

Step 1. $z=\frac{2}{10/\sqrt{n}}=\frac{2\sqrt{n}}{10}=0.2\sqrt{n}$. Why this step? Fix the numerator; only SE shrinks. Writing $z$ as a function of $n$ exposes the trend.

Step 2. Plug in: $z(25)=0.2\cdot 5=1.0,z(100)=0.2\cdot 10=2.0,z(400)=0.2\cdot 20=\mathrm{4.0.}$

Step 3 — the limit. As $n\to \infty$, $z=0.2\sqrt{n}\to \infty$, so $p\to 0$. Why this step? A real gap of any size becomes detectable once you gather enough data — SE $\to 0$ narrows the null curve to a spike. Look at figure s03: the null bell tightens as $n$ climbs.

Verify: $z(25)=1<1.96$ (not significant, two-tailed), $z(100)=2>1.96$ (significant), $z(400)=4$ (wildly significant) ✓. Confirms: statistical significance for a fixed effect is bought with sample size — see Statistical Power & Sample Size.

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Ex 7 — Case G: computing a Type II error $\beta$ and power

Forecast: if the true mean is 6 above, how likely are we to miss it?

Step 1 — the reject region in real units. $\text{SE}=\frac{15}{\sqrt{25}}=3$. We reject when $z\ge 1.645$, i.e. when $\bar{X}\ge {\mu }_0+1.645\cdot \text{SE}=100+1.645(3)=104.935$. Why this step? $\beta$ is about the real-unit cutoff, so we translate the $z$-cutoff back into an $\bar{X}$ threshold. Here $1.645$ is the one-tailed 5% critical z.

Step 2 — evaluate under the truth. If really ${\mu }_1=106$, the mean $\bar{X}$ is Normal centred at $106$ with the same $\text{SE}=3$. A miss (Type II) happens when $\bar{X}<104.935$: $\beta =P\left(Z<\frac{104.935-106}{3}\right)=P(Z<-0.355)\approx \mathrm{0.361.}$ Why this step? $\beta =P(\text{fail to reject}\mid H_0\text{ false})$ — measured under the true distribution, not the null. See figure s04: the amber "true" bell overlaps the accept region.

Step 3. Power $=1-\beta \approx 1-0.361=0.639$.

Verify: $z$-cutoff back-check: $(104.935-100)/3=1.645$ ✓. Miss z: $(104.935-106)/3=-0.355$, and $P(Z<-0.355)\approx 0.361$ ✓. Power $\approx 64\%$ — modest, as expected for a 2-SE-sized true effect. Larger $n$ or effect would raise it (Statistical Power & Sample Size, Confidence Intervals).

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Ex 8 — Case H: real-world word problem (a proportion, the coin)

Forecast: 62 heads out of 100 — fluke or bias?

Step 1. $H_0:p=0.5$, $H_1:p>0.5$.

Step 2 — SE for a proportion. Under $H_0$, each flip has variance $p_0(1-p_0)$, so $\text{SE}=\sqrt{\frac{p_0(1-p_0)}{n}}=\sqrt{\frac{0.5\cdot 0.5}{100}}=\sqrt{0.0025}=0.05$. Why this step? We compute SE using the null value $p_0=0.5$, because the whole test lives inside the "if $H_0$ were true" world.

Step 3. $z=\frac{\hat{p}-p_0}{\text{SE}}=\frac{0.62-0.5}{0.05}=\frac{0.12}{0.05}=2.4$.

Step 4. $p=P(Z\ge 2.4)\approx 0.0082<0.05\Rightarrow$ reject $H_0$: evidence of bias toward heads.

Verify: SE $=0.05$, $z=2.4$, tail $P(Z\ge 2.4)\approx 0.0082$ ✓. The Central Limit Theorem justifies treating $\hat{p}$ as approximately Normal here since $n{p}_0=50$ and $n(1-p_0)=50$ are both large.

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Ex 9 — Case I: the exam twist (they hand you the p-value)

Forecast: does $p=0.03$ measure the probability the null is true? Decide before reading.

Step 1 — name the trap. The claim confuses $P(\text{data}\mid H_0)$ with $P(H_0\mid \text{data})$. This is the prosecutor's fallacy. Why this step? $p$ is computed assuming $H_0$ true — you cannot then read it as $H_0$'s own probability; that would require a prior, which lives in Bayesian Inference.

Step 2 — state it right. "$p=0.03$" means: *if $H_0$ were true, data this extreme (or more) would occur only 3% of the time." It says nothing directly about $P(H_0)$.

Step 3 — the correct decision. Since $0.03\le 0.05$, we reject $H_0$. Correct verdict, wrong reasoning by the classmate.

Verify: Logic check only — the numeric comparison $0.03\le 0.05$ is true, so "reject" is the right action; the interpretation is the error. ✓

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Recall Self-test: which cell is this?

"Testing whether a mean is different (either way) with known $\sigma$." ::: Case A — two-tailed z (Ex 1). "Data lands exactly on ${\mu }_0$; what is the two-tailed p-value?" ::: Case D — $p=1$, fail to reject (Ex 4). "Same effect size, ever-larger $n$: what happens to $p$?" ::: Case F — $p\to 0$; a fixed effect always becomes significant (Ex 6). "Given the true mean, find the chance of missing it." ::: Case G — compute $\beta$, power $=1-\beta$ (Ex 7). "Someone says p is the probability $H_0$ is true." ::: Case I — prosecutor's fallacy; $p=P(\text{data}\mid H_0)$ (Ex 9).