Paper 6. A hypothesis test can raise a false alarm or miss a real change. Both mistakes have probabilities you can calculate: one from the critical region under the null hypothesis, the other from the acceptance region under a specific alternative value.
Type I error: rejecting \(H_0\) when it is true; \(\text{P(Type I)} = \text{P}(\text{result in critical region} \mid H_0 \text{ true})\), the actual significance level. Type II error: not rejecting \(H_0\) when it is false; \(\text{P(Type II)} = \text{P}(\text{result in acceptance region} \mid \text{a given true value})\).
Fig. 1: the sample mean X̄ ~ N(μ, 1) for a test of H₀: μ = 50 against H₁: μ > 50 at 5%, drawn to scale. Teal: Type I error, area 0.05 to the right of 51.645 under H₀. Blue: Type II error when μ is really 53, area 0.0877 to the left of 51.645
Section 01
False alarms and missed changes.
Every decision could be wrong. If you reject \(H_0\), the only possible mistake is a Type I error; if you do not reject \(H_0\), the only possible mistake is a Type II error.
\(H_0\) true
\(H_0\) false
Reject \(H_0\)
Type I error
correct
Do not reject \(H_0\)
correct
Type II error
Worked example Errors in context
A test of \(H_0: p = 0.3\) against \(H_1: p < 0.3\) is used to decide whether a seed treatment reduces the failure rate \(p\). Describe a Type I error and a Type II error in context.
Type I: rejecting \(H_0\) when it is true.
In context: concluding that the treatment reduces the failure rate when in fact it does not.
Type II: not rejecting \(H_0\) when it is false.
In context: concluding there is no evidence of a reduction when the treatment really does reduce the failure rate.
Answer: Type I: a false claim that the treatment works. Type II: failing to detect that it does work.
Examiner trap
Do not describe the errors with "accept" and "reject" mixed up. Write the definition, then translate it into the context: Cambridge often gives one mark for each.
Section 02
Probabilities with discrete distributions.
For binomial and Poisson tests, find the critical region first. \(\text{P(Type I)}\) is the probability of the critical region under \(H_0\), which is usually LESS than the nominal significance level. \(\text{P(Type II)}\) uses the acceptance region and the true value you are given.
Worked example Binomial
\(H_0: p = 0.3\), \(H_1: p < 0.3\), with \(X \sim B(20, p)\) and critical region \(X \le 2\) (5% level). Find \(\text{P(Type I)}\), and \(\text{P(Type II)}\) if in fact \(p = 0.1\).
Type I: \(\text{P}(X \le 2)\) when \(p = 0.3\) \(= 0.0355\).
Type II: the acceptance region is \(X \ge 3\). With \(p = 0.1\), \(X \sim B(20, 0.1)\).
Accidents occur at a mean rate of 6 per month. A test of \(H_0: \lambda = 6\) against \(H_1: \lambda < 6\) at the 5% level uses the number of accidents in one month. Find the critical region and \(\text{P(Type I)}\). Find \(\text{P(Type II)}\) if the rate has actually fallen to 1.5 per month.
\(H_0: p = 0.5\), \(H_1: p > 0.5\), \(X \sim B(12, p)\), 5% level. Find the critical region, \(\text{P(Type I)}\), and \(\text{P(Type II)}\) when \(p = 0.8\).
\(\text{P}(X \ge 10) = 0.0193\) and \(\text{P}(X \ge 9) = 0.0730\), so the critical region is \(X \ge 10\) and \(\text{P(Type I)} = 0.0193\). Type II: \(\text{P}(X \le 9)\) with \(p = 0.8\) \(= 0.442\).
Examiner trap
For a discrete test, \(\text{P(Type I)}\) is NOT the significance level (5%). It is the probability of the actual critical region, here \(0.0355\) or \(0.0174\).
Stage 1 · Learn
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What is a Type I error?
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What is a Type II error?
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A test for a population mean uses the normal distribution at the 2% level. What is \(\text{P(Type I)}\)?
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A test leads to rejecting \(H_0\). Which error might have been made?
Section 03
Probabilities with the normal distribution.
For a test of a mean using a continuous distribution, \(\text{P(Type I)}\) equals the significance level exactly. For \(\text{P(Type II)}\), write the critical region in terms of \(\bar{x}\), then find the probability of the acceptance region using the TRUE mean (Fig. 1).
Worked example One-tailed
\(X \sim N(\mu, 6^{2})\). A sample of 36 tests \(H_0: \mu = 50\) against \(H_1: \mu > 50\) at the 5% level. Find \(\text{P(Type II)}\) when \(\mu = 53\).
Standard error \(\dfrac{6}{6} = 1\); critical region \(\bar{x} > 50 + 1.645 = 51.645\).
Type II: \(\bar{x} \le 51.645\) when really \(\bar{X} \sim N(53, 1)\).
\(X \sim N(\mu, 10^{2})\). A sample of 25 tests \(H_0: \mu = 100\) against \(H_1: \mu \ne 100\) at the 5% level. Find \(\text{P(Type II)}\) when \(\mu = 105\).
Standard error 2. Acceptance region: \(100 \pm 1.96 \times 2\), i.e. \(96.08 < \bar{x} < 103.92\).
Find the critical value using \(H_0\)'s mean, but the Type II probability using the TRUE mean. Standardising the critical value with 50 again just gives back 0.95.
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\(X \sim B(15, p)\). A test of \(H_0: p = 0.4\) against \(H_1: p > 0.4\) at the 5% level has critical region \(X \ge 10\). Find \(\text{P(Type I)}\).
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For the same test (critical region \(X \ge 10\), \(n = 15\)), find \(\text{P(Type II)}\) when in fact \(p = 0.7\).
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\(X \sim N(\mu, 15^{2})\). A sample of 25 tests \(H_0: \mu = 200\) against \(H_1: \mu < 200\), with critical region \(\bar{x} < 194.12\). Find \(\text{P(Type II)}\) when \(\mu = 192\).
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A factory tests \(H_0\): the mean fill is 500 ml, against \(H_1\): the mean fill is less than 500 ml. Which describes a Type II error?
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