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Chapter 7 of 11 · ECON 1012

Hypothesis Testing

Hypothesis Testing (Module 7, Week 7) is the second core inference procedure in ECON 1012, after estimation: instead of building a range for a parameter, you decide whether the sample gives enough statistical evidence for a belief about it. The module teaches the course's six-component recipe — hypotheses, test statistic, significance level, decision rule, computed value, conclusion — used for every test from here to regression. You learn to write H₀ (which always carries the equality μ = μ₀) against a one- or two-tailed H₁, standardise with Z when σ is known or t with n − 1 degrees of freedom when it is not, and reach the same decision by rejection region or by p-value (reject H₀ when p-value < α). Type I and Type II errors round out the week, and the practice exam's case-study marking asks for every one of the six steps written out.

In this chapter

What this chapter covers

  • 01The six components of every test: hypotheses → test statistic → significance level α → decision rule → computed value → conclusion
  • 02H₀ always carries the equality (H₀: μ = μ₀); H₁ takes ≠, < or > and bears the burden of proof
  • 03Tail choice from the claim: 'differs' → μ ≠ μ₀ (two-tail) · 'more than' → μ > μ₀ · 'less than' → μ < μ₀
  • 04Test statistic: Z = (X̄ − μ₀)/(σ/√n) when σ is known; t = (X̄ − μ₀)/(s/√n) with df = n − 1 when only s is available
  • 05Rejection regions: two-tail |z₀| > z_α/2 (±1.96 at α = 0.05); one-tail uses z_α (1.645 at α = 0.05)
  • 06p-value routes: left-tail p = P(Z < z₀) · right-tail p = P(Z > z₀) · two-tail p = 2·P(Z > |z₀|); reject H₀ when p-value < α
  • 07Evidence bands: p < 0.01 overwhelming · 0.01–0.05 strong · 0.05–0.10 weak · > 0.10 no evidence
  • 08Type I error = reject a true H₀ (P = α); Type II error = don't reject a false H₀ (P = β); the two are inversely related
Worked example · free

A six-step two-tail Z-test, its p-value, and the error check

Q [12 marks]. A gym chain advertises that its members' visits last 55 minutes on average. Session lengths are normally distributed with known population standard deviation σ = 12 minutes. A random sample of n = 36 recent visits has mean x̄ = 50.5 minutes. (a) Using the six steps of hypothesis testing, test at the 5% significance level whether the mean session length differs from the advertised 55 minutes. (b) Compute the p-value and state the decision at α = 0.01. (c) Which type of error could this test have made, and what is its probability?
  • 1 mark(a) Step 1 — hypotheses: 'differs' has no direction, so the test is two-tailed. H₀: μ = 55 vs H₁: μ ≠ 55.
  • 1 mark(a) Step 2 — test statistic: σ is known, so standardise with Z = (X̄ − μ₀)/(σ/√n).
  • 2 marks(a) Steps 3 and 4 — significance level α = 0.05; two-tail decision rule: reject H₀ if z₀ < −1.96 or z₀ > 1.96 (z₀.₀₂₅ = 1.96).
  • 2 marks(a) Step 5 — value of the test statistic: standard error σ/√n = 12/√36 = 12/6 = 2, so z₀ = (50.5 − 55)/2 = −4.5/2 = −2.25.
  • 2 marks(a) Step 6 — conclusion: since z₀ = −2.25 < −1.96, reject H₀ in favour of H₁. There is sufficient evidence to infer that the mean session length differs from 55 minutes, at a significance level of 5%.
  • 2 marks(b) Two-tail p-value = 2·P(Z > |−2.25|) = 2 × (1 − 0.9878) = 2 × 0.0122 = 0.0244. Since 0.0244 < 0.05 the decision matches part (a); at α = 0.01, 0.0244 > 0.01, so H₀ is not rejected — the p-value is the smallest α at which H₀ can be rejected.
  • 2 marks(c) The test rejected H₀, so the only mistake it could have made is a Type I error — rejecting a true H₀ — with probability α = 0.05. A Type II error is impossible here: it can occur only when H₀ is not rejected.
(a) z₀ = −2.25 falls in the rejection region, so reject H₀ at α = 0.05: sufficient evidence the mean session length differs from 55 minutes. (b) p-value = 0.0244 — reject at α = 0.05 but do not reject at α = 0.01. (c) Only a Type I error is possible, with probability α = 0.05.
Sia tip — Write all six steps every time — the practice exam's case studies explicitly ask you to show and explain each one, and each step carries marks. Keep the wording clean: 'do not reject H₀', never 'accept H₀'. The p-value route pays off whenever a follow-up asks what happens at a different α.
Glossary

Key terms

Null hypothesis (H₀)
The statement that the parameter equals the hypothesised value (H₀: μ = μ₀). Equality is always part of H₀, and it is presumed true unless the sample provides sufficient evidence against it.
Alternative hypothesis (H₁)
The claim that carries the burden of proof, taking one of three forms: μ ≠ μ₀ (two-tail), μ < μ₀ (left-tail) or μ > μ₀ (right-tail). Its direction fixes where the rejection region sits; the course also writes it H_A.
Significance level α
The probability of rejecting the null hypothesis when it is true — i.e. the probability of a Type I error. Chosen before testing (commonly 0.05) and used to set the critical value(s) of the rejection region.
Rejection region
The set of test-statistic values that lead to rejecting H₀: z₀ > z_α for a right-tail test, z₀ < −z_α for a left-tail test, |z₀| > z_α/2 for a two-tail test; analogous rules use t critical values with df = n − 1 when σ is unknown.
p-value
The evidence against the null hypothesis — the lower the p-value, the stronger the evidence — equivalently the minimum significance level required to reject H₀. Decision rule: reject H₀ when p-value < α.
Type I and Type II errors
A Type I error rejects a true H₀ (probability α); a Type II error fails to reject a false H₀ (probability β). For a given sample size the two probabilities are inversely related — lowering α raises β.
FAQ

Hypothesis Testing FAQ

When do I use t instead of Z in an ECON 1012 hypothesis test?

Ask one question: is the population standard deviation σ known? If σ (or σ²) is given, standardise with Z = (X̄ − μ₀)/(σ/√n). If you only have the sample standard deviation s, use t = (X̄ − μ₀)/(s/√n) with df = n − 1 and read the critical value from the t table — both Z and t tables are provided in the final exam. MCQ options are regularly built on exactly this choice, so make it your first written line.

Why can't I write 'accept H₀' when the test does not reject?

Because failing to reject only means the sample lacked sufficient evidence against H₀ — it does not demonstrate that H₀ is true. H₀ is presumed true from the start and H₁ carries the burden of proof, so the only two conclusions are 'reject H₀ in favour of H₁' or 'do not reject H₀', each paired with a plain-English sentence: 'there is (not) sufficient evidence to infer that …, at a significance level of 5%'.

How do I decide between a one-tailed and a two-tailed test?

Read the direction of the claim. 'Differs', 'changed', 'is not' → two-tail (H₁: μ ≠ μ₀, with α split into two tails of α/2). 'More than', 'increased', 'exceeds' → right-tail (H₁: μ > μ₀). 'Less than', 'decreased', 'fell' → left-tail (H₁: μ < μ₀). The null hypothesis is identical in all three cases — H₀: μ = μ₀, because equality always belongs to H₀ — only the alternative and the rejection region move.

What does the p-value actually tell me?

Two equivalent readings from the module: it measures the evidence against H₀ (the lower, the stronger), and it is the minimum significance level at which H₀ would be rejected — so reject when p-value < α. The course's evidence bands: p < 0.01 overwhelming, 0.01–0.05 strong, 0.05–0.10 weak, above 0.10 no evidence. For a two-tail test remember to double the tail area: p = 2·P(Z > |z₀|).

Studying with AI? Sia — free AI economics tutor works through ECON 1012 step by step.

Study strategy

Exam move

Marks in this module are won by discipline, not difficulty. Write all six steps every time — the practice exam's case-study marking instructs you to show and explain each one; skipped steps are skipped marks. The two slips that cost most: reaching for Z when σ is unknown (only s given → t with df = n − 1), and standardising with σ instead of the standard error σ/√n. Never write 'accept H₀' — the course wording is 'do not reject', paired with a plain-English sentence stating the significance level. Remember a Type I error is only possible when you reject and a Type II only when you fail to reject, with P(Type I) = α. Finish with the Module 7 quiz on myLearning — the exam MCQs mirror the module quizzes.

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