Hypothesis tests
Null and alternative hypotheses, significance levels, one- and two-tailed tests, critical regions, tests for a binomial proportion, a Poisson mean and a normal mean, and Type I and Type II errors with their probabilities.
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In 30 seconds
- H₀ states the accepted value (p = 0.3, λ = 6, μ = 50); H₁ states what you suspect (<, > or ≠).
- Reject H₀ if the result is in the critical region, i.e. the probability of a result this extreme (assuming H₀) is less than the significance level. For two-tailed tests, compare with half the level in each tail.
- Discrete tests (binomial, Poisson) use cumulative probabilities; normal tests use a z-value.
- Type I error: rejecting H₀ when it is true; its probability is the actual significance level. Type II error: accepting H₀ when it is false.
By the end you can
- State hypotheses in terms of a population parameter.
- Carry out tests for a binomial proportion, a Poisson mean and a normal mean, one- or two-tailed.
- Find critical regions and actual significance levels for discrete tests.
- Understand Type I and Type II errors and calculate their probabilities.
- Write conclusions in context, without overstating them.
1The language of tests
Hypotheses and conclusions
- Hypotheses use parameters: "H₀: p = 0.3, H₁: p < 0.3", not "H₀: X = 2".
- Conclusions are cautious: "there is evidence at the 5% level that the proportion has decreased", or "insufficient evidence that …". Never "this proves".
- Significance level: the probability of rejecting H₀ when it's true. Smaller levels need stronger evidence.
2Discrete tests
Binomial
A shop claims 30% of customers pay by card. In a sample of 20, only 2 do. Test at the 5% level whether the proportion is less than 30%.
- H₀: p = 0.3; H₁: p < 0.3 B1
- under H₀, X ~ B(20, 0.3): P(X ≤ 2) = 0.0355 M1 A1
- 0.0355 < 0.05, so reject H₀ M1
- there is evidence at the 5% level that the proportion paying by card is less than 30% A1
For this test, find the critical region and the probability of a Type I error. If in fact p = 0.1, find the probability of a Type II error.
- P(X ≤ 2) = 0.0355 < 0.05 but P(X ≤ 3) = 0.1071 > 0.05 M1
- critical region X ≤ 2; P(Type I) = 0.0355 A1 A1
- Type II: accept H₀ although p = 0.1: P(X ≥ 3 ǀ p = 0.1) = 1 − 0.6769 = 0.323 M1 A1
Poisson
A road had an average of 6 accidents a week. After a speed limit was introduced, there were 2 accidents in a week. Test at the 5% level whether the rate has decreased.
- H₀: λ = 6; H₁: λ < 6 B1
- P(X ≤ 2) = e−6(1 + 6 + 18) = 0.0620 M1 A1
- 0.0620 > 0.05, so do not reject H₀ M1
- insufficient evidence at the 5% level that the accident rate has decreased A1
3Normal mean tests
z-tests
Lengths are known to be normal with σ = 6. A sample of 36 has mean 48.1. Test at the 5% level whether the mean is less than 50.
- H₀: μ = 50; H₁: μ < 50 B1
- z = (48.1 − 50)/(6/6) = −1.9 M1 A1
- −1.9 < −1.645, so reject H₀ M1
- evidence at the 5% level that the mean length is less than 50 A1
- Two-tailed at 5%: critical values ±1.96. With x̄ = 51.5 instead, z = 1.5, between −1.96 and 1.96, so there's insufficient evidence that μ ≠ 50.
Test yourself — cover the answers first
Quick check
- State the meaning of a Type I error.
- What critical z-values are used in a two-tailed 5% test?
- Write hypotheses for "has the mean increased from 20?".
- A test statistic falls in the critical region. What do you do?
- Why is the actual significance level of a binomial test often below 5%?
Exam-style
- X ~ B(15, 0.5) under H₀: p = 0.5 vs H₁: p > 0.5. Find the critical region at the 5% level.
- Under H₀: λ = 6, with critical region X ≤ 1, find P(Type I error).
- σ = 4, n = 25, test H₀: μ = 30 vs H₁: μ > 30 at 1%. Find the smallest x̄ that leads to rejection.
Show answers
- rejecting H₀ when it is true
- ±1.96
- H₀: μ = 20, H₁: μ > 20
- reject H₀
- X is discrete, so the tail probability jumps
- P(X ≥ 12) = 0.0176, P(X ≥ 11) = 0.0592 ⇒ X ≥ 12
- e−6(1 + 6) = 0.0174
- 30 + 2.326 × 4/5 = 31.86
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