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AS & A Level · Practice questions · Type I and Type II errors

Size up the risks of a test.

Six original Cambridge-style Paper 6 questions on Type I and Type II errors, with full worked solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Critical region first, then the right distribution.

01
Recall
[2 marks]

A drug trial tests \(H_0\): the drug has no effect, against \(H_1\): the drug lowers blood pressure. Describe a Type I error and a Type II error in context.

  • Type I: concluding the drug lowers blood pressure when it has no effect. ✓
  • Type II: concluding there is no evidence of an effect when the drug really does lower blood pressure. ✓
02
Calculation
[5 marks]

Calls arrive at a mean rate of 5 per hour. A test of \(H_0: \lambda = 5\) against \(H_1: \lambda < 5\) at the 10% level uses the number of calls in one hour.

(a) Find the critical region and \(\text{P(Type I)}\).

(b) Find \(\text{P(Type II)}\) if the mean rate is actually 2 per hour.

  • (a) \(\text{P}(X \le 1) = 6e^{-5} = 0.0404 \le 0.10\); \(\text{P}(X \le 2) = 18.5e^{-5} = 0.1247 > 0.10\). ✓
  • Critical region \(X \le 1\); \(\text{P(Type I)} = 0.0404\). ✓
  • (b) Accept when \(X \ge 2\), with \(X \sim \text{Po}(2)\): \(1 - 3e^{-2} = 1 - 0.4060 = 0.594\) (3 s.f.). ✓
03
Calculation
[4 marks]

\(X \sim B(12, p)\). A test of \(H_0: p = 0.5\) against \(H_1: p > 0.5\) is carried out at the 5% level. Find the critical region, \(\text{P(Type I)}\), and \(\text{P(Type II)}\) when \(p = 0.8\).

  • \(\text{P}(X \ge 10) = 0.0193 \le 0.05\); \(\text{P}(X \ge 9) = 0.0730 > 0.05\). Critical region \(X \ge 10\). ✓
  • \(\text{P(Type I)} = 0.0193\). ✓
  • Type II: \(\text{P}(X \le 9)\) for \(B(12, 0.8)\) \(= 0.442\) (3 s.f.). ✓
04
Calculation
[5 marks]

\(X \sim N(\mu, 10^{2})\). A random sample of 25 is used to test \(H_0: \mu = 100\) against \(H_1: \mu \ne 100\) at the 5% level. Find \(\text{P(Type II)}\) when \(\mu = 105\).

  • Standard error \(\dfrac{10}{5} = 2\). Acceptance region \(100 \pm 1.96 \times 2\): \(96.08 < \bar{x} < 103.92\). ✓
  • Under \(\mu = 105\): \(z_1 = -4.46\), \(z_2 = -0.54\). ✓
  • \(\text{P(Type II)} = \Phi(-0.54) - \Phi(-4.46) = 0.2946 - 0.0000 = 0.295\) (3 s.f.). ✓
05
Analysis
[3 marks]

A test at the 5% level of whether a coin is biased gives a result in the acceptance region.

(a) Which type of error could have been made? Explain.

(b) Explain why choosing a 1% significance level instead would make a Type I error less likely but a Type II error more likely.

  • (a) Type II only: \(H_0\) was not rejected, so the possible mistake is failing to reject a false \(H_0\) (the coin is biased but the test missed it). ✓
  • (b) A 1% level shrinks the critical region, so a true \(H_0\) is rejected less often (Type I less likely). ✓
  • The acceptance region grows, so a false \(H_0\) is more likely to be kept (Type II more likely). ✓
06
Analysis
[7 marks]

The lifetimes of a type of battery are \(N(\mu, 3^{2})\) hours. The manufacturer claims \(\mu = 40\). A consumer group will test a random sample of 16 batteries at the 5% level to see whether the mean is less than 40 hours.

(a) Find the critical region for the sample mean.

(b) State \(\text{P(Type I)}\).

(c) Find \(\text{P(Type II)}\) if in fact \(\mu = 38\).

(d) State one way to reduce \(\text{P(Type II)}\) without changing \(\text{P(Type I)}\).

  • (a) Standard error \(\dfrac{3}{4} = 0.75\). Reject when \(\bar{x} < 40 - 1.645 \times 0.75 = 38.766\). ✓
  • Critical region \(\bar{x} < 38.8\) (3 s.f.). ✓
  • (b) \(\text{P(Type I)} = 0.05\). ✓
  • (c) Type II: \(\bar{x} \ge 38.766\) with \(\bar{X} \sim N(38, 0.75^{2})\): \(z = \dfrac{38.766 - 38}{0.75} = 1.022\), so \(1 - 0.8466 = 0.153\) (3 s.f.). ✓
  • (d) Increase the sample size: the standard error falls, the two distributions overlap less, and the 5% level is unchanged. ✓

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