Hypothesis Testing for GATE DA: z-test, t-test & Chi-Squared Solved Step by Step

By Piyush Wairale — Instructor, BS Data Science program at IIT Madras · IIT Madras alumnus · 10,000+ GATE students mentored · Last updated: August 2026

Hypothesis testing is where the Probability & Statistics syllabus ends — and where many aspirants leave marks behind, not because the mathematics is hard but because the procedure feels unfamiliar under time pressure. The remedy is repetition of one fixed script. This deep dive complements the P&S pillar guide and the distributions cheat sheet (the normal, t and chi-squared tables live there) with the worked-problem depth that pillar posts can’t fit. For lectures and tests, see the Probability & Statistics for GATE DA Course.

The 5-Step Framework (Use It Every Single Time)

1. State the hypotheses. H₀ is the status quo with the equality (μ = μ₀); H₁ is what you suspect (μ ≠ μ₀ for two-tailed, μ > μ₀ or μ < μ₀ for one-tailed). The equality always lives in H₀.
2. Fix the significance level α — typically 0.05 or 0.01; it is the Type-I-error probability you tolerate.
3. Compute the test statistic — z, t or chi-squared, per the selection rules below.
4. Find the rejection region from the critical value(s), or compute the p-value.
5. Decide and phrase it correctly: “reject H₀” or “fail to reject H₀” — never “accept H₀”. Absence of evidence against the null is not proof of it.

Type I, Type II and Power

Two ways to be wrong, and GATE tests the vocabulary directly. A Type I error rejects a true H₀ — a false alarm — and P(Type I) = α, the significance level you chose. A Type II error fails to reject a false H₀ — a miss — with probability β. The power of a test is 1 − β, the probability of correctly detecting a real effect. The trade-offs that appear as MCQ statements: lowering α (being stricter) raises β for a fixed sample; increasing sample size n lowers β (raises power) at fixed α; and power grows with the true effect size. A courtroom mnemonic keeps them straight: H₀ = “innocent”; Type I convicts the innocent; Type II acquits the guilty.

One-Tailed vs Two-Tailed — and the Critical Values

The direction of H₁ sets the tails. “Has the mean changed?” → two-tailed, split α across both tails: at α = 0.05 reject when |z| > 1.96; at α = 0.01 when |z| > 2.576. “Has the mean increased?” → one-tailed (right), all of α in one tail: at α = 0.05 reject when z > 1.645; at α = 0.01 when z > 2.326. Memorise the trio 1.645 / 1.96 / 2.576 — GATE questions are built around them, and picking the wrong one flips the final answer even when every computation is right. For t-tests the same logic applies with critical values from the t-table at n−1 degrees of freedom (the question supplies them); chi-squared tests are typically right-tailed by construction.

Choosing the Test in Ten Seconds

The keywords do the work: “standard deviation of the population is known” → z; “sample standard deviation” with n ≤ 30 → t; “die/coin/frequency table/proportions across categories” → chi-squared. The reference distributions themselves are covered in the distributions cheat sheet.

Worked Problem 1: z-Test on a Mean

Question. A machine fills bottles with mean 500 ml and known σ = 8 ml. A sample of 64 bottles has mean 497.5 ml. At α = 0.05, has the mean changed?

Solution. H₀: μ = 500; H₁: μ ≠ 500 (two-tailed — “changed”). Statistic: z = (497.5 − 500)/(8/√64) = (−2.5)/(1) = −2.5. Critical values at α = 0.05 two-tailed: ±1.96. Since |−2.5| = 2.5 > 1.96, reject H₀ — the filling mean has shifted. (p-value ≈ 2 × P(Z < −2.5) = 2 × 0.0062 = 0.0124 < 0.05, same verdict.) Note the two places marks are lost: forgetting √64 in the standard error, and using 1.645 because “the sample mean went down” — the hypothesis wording, not the data, sets the tails.

Worked Problem 2: t-Test With a Small Sample

Question. A trainer claims students average more than 70 in a test. Nine students score a sample mean of 74 with sample standard deviation s = 6. At α = 0.05 (t-critical for 8 d.o.f., one-tailed: 1.860), is the claim supported?

Solution. σ is unknown, n = 9 is small → t-test. H₀: μ = 70; H₁: μ > 70 (one-tailed — “more than”). Statistic: t = (74 − 70)/(6/√9) = 4/2 = 2.0. Degrees of freedom: n − 1 = 8; critical value 1.860. Since 2.0 > 1.860, reject H₀ — the data support the claim at the 5% level. Had the question asked at α = 0.01 (t₀.₀₁,₈ = 2.896), we would fail to reject: the same data can support a claim at one level and not another, which is precisely what the significance level means.

Worked Problem 3: Chi-Squared Goodness-of-Fit

Question. A die is rolled 60 times, giving face counts 6, 8, 11, 14, 9, 12. At α = 0.05 (χ²-critical for 5 d.o.f.: 11.07), is the die fair?

Solution. H₀: the die is fair, so each expected count is E = 60/6 = 10. Statistic: Σ(O−E)²/E = (16 + 4 + 1 + 16 + 1 + 4)/10 = 42/10 = 4.2. Degrees of freedom: k − 1 = 5; critical value 11.07. Since 4.2 < 11.07, fail to reject H₀ — the observed wobble is well within chance for 60 rolls. Habits that save marks: expected counts come from H₀ (not the data), the statistic sums over all categories, and the test is right-tailed — a big statistic means a bad fit.

Reading p-Values Correctly

The p-value is the probability, assuming H₀ is true, of observing a test statistic at least as extreme as the one you got. Decision rule: p < α → reject. Two interpretation traps GATE plants in MSQ options: the p-value is not the probability that H₀ is true, and it is not the probability the result happened “by chance” in any loose sense — it is a tail probability under a specific assumption. Also remember the duality: rejecting at level α with a two-tailed z-test is equivalent to μ₀ falling outside the (1−α) confidence interval — a connection question that has appeared across GATE papers.

The Mistakes That Cost Marks

• Wrong tails: using 1.96 for a one-tailed test (or 1.645 for two-tailed). Read H₁, not the data direction.
• Forgetting √n: the standard error is σ/√n (or s/√n) — dividing by σ alone is the most common numeric slip.
• Wrong degrees of freedom: n−1 for the t and variance tests, k−1 for goodness-of-fit, (r−1)(c−1) for independence tables.
• “Accepting” H₀: the correct conclusion is “fail to reject” — GATE has used this phrasing distinction in options.
• Swapping α and β: α is set by you (Type I); β depends on the true effect and sample size (Type II).
• Misreading the p-value as P(H₀ is true) — it is a tail probability computed under H₀.

How GATE Tests Hypothesis Testing

• Compute-and-decide NATs (2 marks): exactly the three worked problems above — statistic, compare, decide.
• Test-selection MCQs (1 mark): which test fits the scenario; the ten-second table settles it.
• Vocabulary MSQs (1–2 marks): true/false statements about Type I/II, power, p-values and significance levels.
• Critical-value reasoning (1 mark): what happens to the rejection region when α changes or tails change.
• CI-test duality (1–2 marks): connecting a two-tailed test with the corresponding confidence interval from the P&S pillar.

FAQs on Hypothesis Testing for GATE DA

Three tests, one script, five critical numbers. Run the 5-step framework on twenty problems — mixing tails, levels and test types — and hypothesis testing turns from the syllabus’s most intimidating topic into its most mechanical marks. Continue the probability sub-cluster with the Bayes deep dive and distributions cheat sheet, or return to the full syllabus roadmap.