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AS & A Level · Practice questions · Hypothesis tests for a population mean

Standardise, compare, conclude.

Six original Cambridge-style Paper 6 questions on hypothesis tests for a population mean, 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

\(z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}\), then compare.

01
Recall
[2 marks]

A test about a population mean uses a sample of size 80 from a population whose distribution is unknown. Explain why a test based on the normal distribution can still be used.

  • The sample is large, so by the Central Limit Theorem the sample mean is approximately normally distributed, ✓
  • whatever the distribution of the population; \(s^{2}\) from the sample estimates \(\sigma^{2}\). ✓
02
Calculation
[4 marks]

\(X \sim N(\mu, 5^{2})\). A random sample of 16 values has mean 32.4. Test at the 5% level whether \(\mu > 30\).

  • \(H_0: \mu = 30\); \(H_1: \mu > 30\). ✓
  • \(z = \dfrac{32.4 - 30}{5/\sqrt{16}} = 1.92\). ✓
  • \(1.92 > 1.645\): reject \(H_0\). ✓
  • There is sufficient evidence at the 5% level that \(\mu > 30\). ✓
03
Calculation
[4 marks]

Bags of flour have masses \(N(\mu, 0.12^{2})\) kg. A random sample of 10 bags has mean 1.43 kg. Test at the 5% level whether the mean mass differs from 1.5 kg.

  • \(H_0: \mu = 1.5\); \(H_1: \mu \ne 1.5\). ✓
  • \(z = \dfrac{1.43 - 1.5}{0.12/\sqrt{10}} = -1.845\). ✓
  • Critical values \(\pm 1.96\); \(|-1.845| < 1.96\): do not reject \(H_0\). ✓
  • There is insufficient evidence at the 5% level that the mean mass differs from 1.5 kg. (At 10% the same data would lead to rejection.) ✓
04
Analysis
[4 marks]

\(X \sim N(\mu, 6^{2})\). A random sample of 36 values is used to test \(H_0: \mu = 50\) against \(H_1: \mu > 50\) at the 1% level.

(a) Find the critical region for \(\bar{x}\).

(b) The sample mean is 51.9. State the conclusion.

  • (a) Standard error \(\dfrac{6}{6} = 1\); critical value 2.326, so reject when \(\bar{x} > 50 + 2.326 = 52.326\). ✓
  • Critical region \(\bar{x} > 52.3\) (3 s.f.). ✓
  • (b) \(51.9\) is not in the critical region: do not reject \(H_0\). ✓
  • There is insufficient evidence at the 1% level that \(\mu > 50\). ✓
05
Analysis
[5 marks]

A random sample of 80 students' test scores gives \(\sum x = 5720\) and \(\sum x^{2} = 412\,300\). Test at the 5% level whether the mean score is greater than 70.

  • \(\bar{x} = 71.5\); \(s^{2} = \dfrac{1}{79}\left(412\,300 - \dfrac{5720^{2}}{80}\right) = \dfrac{3320}{79} = 42.03\). ✓
  • \(H_0: \mu = 70\); \(H_1: \mu > 70\). By the CLT, \(\bar{X} \approx N\!\left(70, \dfrac{42.03}{80}\right)\). ✓
  • \(z = \dfrac{71.5 - 70}{\sqrt{42.03/80}} = 2.070\). ✓
  • \(2.070 > 1.645\): reject \(H_0\). There is sufficient evidence at the 5% level that the mean score is greater than 70. ✓
06
Analysis
[7 marks]

A packet of biscuits is labelled 250 g. Masses are \(N(\mu, 8^{2})\) g. A trading inspector suspects that packets are underweight on average, and weighs a random sample of 20 packets, finding a mean of 246.1 g.

(a) Test at the 1% level whether the inspector's suspicion is justified.

(b) Find the set of significance levels at which the inspector would conclude that packets are underweight.

  • (a) \(H_0: \mu = 250\); \(H_1: \mu < 250\). ✓
  • \(z = \dfrac{246.1 - 250}{8/\sqrt{20}} = -2.180\). ✓
  • Critical value \(-2.326\); \(-2.180 > -2.326\): do not reject \(H_0\). There is insufficient evidence at the 1% level that the mean mass is below 250 g. ✓
  • (b) \(\text{P}(Z < -2.180) = 1 - 0.9854 = 0.0146\). ✓
  • \(H_0\) is rejected for any significance level greater than \(1.46\%\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.