Paper 6. A die lands on six too often; a junction has fewer accidents after new lights. A hypothesis test asks how surprising the observation would be if nothing had changed, and only a result surprising enough to fall below the significance level counts as evidence.
Assume \(H_0\) is true. Find the probability of a result at least as extreme as the one observed (in the direction of \(H_1\)), or find the critical region. If that probability is below the significance level, the result lies in the critical region: reject \(H_0\). Always conclude in context and without certainty: "there is (in)sufficient evidence that ...".
Fig. 1: B(20, 0.3), drawn to scale. For a lower-tail test at the 5% level the critical region is X ≤ 2, because P(X ≤ 2) = 0.0355 ≤ 0.05 but P(X ≤ 3) = 0.1071 > 0.05; the actual significance level is 3.55%
Section 01
The language of a test.
Every test follows the same script. Learn the words exactly, because Paper 6 asks for them by name.
Null hypothesis \(H_0\): the parameter has its usual or claimed value, e.g. \(H_0: p = 0.3\) or \(H_0: \lambda = 6\). Always an equals sign.
Alternative hypothesis \(H_1\): what you suspect: \(p < 0.3\) or \(p > 0.3\) (one-tailed), or \(p \ne 0.3\) (two-tailed, when no direction is suggested).
Test statistic: the quantity you observe, e.g. the number of sixes \(X\).
Significance level: the threshold probability, e.g. 5%. In a two-tailed test it is split equally, 2.5% in each tail.
Critical (rejection) region: values of the test statistic that lead to rejecting \(H_0\); the rest is the acceptance region.
Worked example A one-tailed binomial test
A die is thrown 30 times and gives 9 sixes. Test, at the 5% significance level, whether the die is biased towards six.
\(H_0: p = \tfrac{1}{6}\); \(H_1: p > \tfrac{1}{6}\) (one-tailed). Under \(H_0\), \(X \sim B(30, \tfrac{1}{6})\).
Probability of 9 or more sixes: \(\text{P}(X \ge 9) = 1 - \text{P}(X \le 8) = 1 - 0.9803 = 0.0197\).
\(0.0197 < 0.05\), so the result is in the critical region: reject \(H_0\).
Answer: There is sufficient evidence at the 5% level that the die is biased towards six.
Worked example Finding a critical region
Usually 30% of seeds fail to germinate. A new treatment is tried on 20 seeds. Find the critical region for a test at the 5% level of whether the treatment reduces the failure rate, and state the actual significance level.
\(H_0: p = 0.3\); \(H_1: p < 0.3\). Under \(H_0\), \(X \sim B(20, 0.3)\), where \(X\) is the number that fail.
So the critical region is \(X \le 2\), and the probability of landing in it when \(H_0\) is true is \(0.0355\).
Answer: Critical region \(X \le 2\); actual significance level \(3.55\%\)
Examiner trap
Compare the TAIL probability \(\text{P}(X \ge 9)\) with 5%, never the single probability \(\text{P}(X = 9)\). And never write "\(H_0\) is true" or "this proves the die is biased": a test only gives evidence.
Section 02
Two-tailed tests and Poisson tests.
In a two-tailed test, compare the tail probability with HALF the significance level. For a Poisson mean, the method is the same with \(X \sim \text{Po}(\lambda)\), scaling \(\lambda\) to the period observed.
Worked example Two-tailed
A coin is tossed 25 times and shows 18 heads. Test at the 5% level whether the coin is biased.
\(H_0: p = 0.5\); \(H_1: p \ne 0.5\) (two-tailed, so 2.5% in each tail). \(X \sim B(25, 0.5)\).
18 is above the expected 12.5, so use the upper tail: \(\text{P}(X \ge 18) = 0.0216\).
\(0.0216 < 0.025\): reject \(H_0\).
Answer: There is sufficient evidence at the 5% level that the coin is biased.
Worked example A Poisson test
Accidents at a junction occurred at a mean rate of 6 per month. After new lights were installed, there were 2 accidents in a month. Test at the 5% level whether the rate has decreased.
Answer: There is insufficient evidence at the 5% level that the accident rate has decreased.
Your turn
A machine breaks down at a mean rate of 3.2 per month. After a service it breaks down once in the next 2 months. Test at the 2% level whether the rate has fallen.
\(H_0: \lambda = 6.4\) (for 2 months), \(H_1: \lambda < 6.4\). \(\text{P}(X \le 1) = 7.4e^{-6.4} = 0.0123 < 0.02\): reject \(H_0\); there is sufficient evidence that the breakdown rate has fallen.
Examiner trap
Write hypotheses about the PARAMETER (\(p\) or \(\lambda\)), not about the observed count, and scale \(\lambda\) to the period of the observation: 2 months at 3.2 per month means \(H_0: \lambda = 6.4\).
Stage 1 · Learn
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A company claims that 30% of its customers are new. Which is the null hypothesis for testing this claim?
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When is a two-tailed test appropriate?
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In a two-tailed test at the 10% level, what do you compare each tail probability with?
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Which is the best conclusion when a test result lies in the critical region?
Section 03
Using a normal approximation.
When \(n\) is large (binomial with \(np, nq > 5\)) or \(\lambda > 15\) (Poisson), use a normal approximation with a continuity correction for the tail probability, exactly as in the approximation lessons.
Worked example Poisson with a large mean
Accidents used to occur at 6 per month. Over 3 months after new lights there were 8 accidents. Test at the 5% level whether the rate has decreased.
\(H_0: \lambda = 18\) (for 3 months); \(H_1: \lambda < 18\). \(\lambda > 15\), so use \(Y \sim N(18, 18)\).
Answer: There is sufficient evidence at the 5% level that the accident rate has decreased.
Worked example Binomial with large \(n\)
A firm claims 40% of customers buy a warranty. In a random sample of 200 customers, 95 bought one. Test at the 1% level whether the proportion is higher than claimed.
\(H_0: p = 0.4\); \(H_1: p > 0.4\). \(X \sim B(200, 0.4) \approx N(80, 48)\).
\(1 - \Phi(2.093) = 1 - 0.9818 = 0.0182 > 0.01\): do not reject \(H_0\).
Answer: There is insufficient evidence at the 1% level that more than 40% of customers buy a warranty.
Examiner trap
The continuity correction points AWAY from the tail: \(X \ge 95\) becomes \(Y > 94.5\), and \(X \le 8\) becomes \(Y < 8.5\).
Stage 2 · Exam
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It is claimed that 25% of bulbs are faulty. In a sample of 40, 4 are faulty. For a test of \(H_1: p < 0.25\), which probability should be compared with the significance level?
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\(X \sim B(15, 0.4)\) under \(H_0: p = 0.4\). Find the critical region for a test of \(H_1: p > 0.4\) at the 5% level.
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Calls arrive at a mean rate of 4.5 per hour. In one hour after an advert there are 9 calls. Which value should be compared with the significance level for \(H_1: \lambda > 4.5\)?
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Under \(H_0\), the number of faults in 4 weeks is \(\text{Po}(18)\). 28 faults are observed. Using a normal approximation, find the probability used to test \(H_1: \lambda > 18\).
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