Paper 6. A bottling plant claims 500 ml per bottle; a sample averages 497.8 ml. The sample mean has a known normal distribution if the claim is true, so we can measure exactly how unusual 497.8 is, using either a test statistic or a critical value for the sample mean.
Under \(H_0: \mu = \mu_0\), \(\bar{X} \sim N\!\left(\mu_0, \dfrac{\sigma^{2}}{n}\right)\), so the test statistic is \(z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}\). Compare \(z\) with the critical value (e.g. \(1.645\) one-tailed at 5%, \(\pm 1.96\) two-tailed at 5%). For a large sample with unknown \(\sigma\), use the unbiased estimate \(s\) and the Central Limit Theorem.
Fig. 1: the standard normal distribution under H₀, drawn to scale. For a lower-tail test at 5%, the rejection region is z < −1.645 (shaded). The test statistic −1.905 from the bottling example lies inside it, so H₀ is rejected
Section 01
The z test, known variance.
Use this when the population is normal (or the sample is large) and \(\sigma\) is given. Standardise the observed sample mean using the standard error \(\dfrac{\sigma}{\sqrt{n}}\), then compare with the critical value, or find the tail probability and compare with the significance level. Both methods give the same decision.
Significance level
One-tailed
Two-tailed
10%
1.282
\(\pm 1.645\)
5%
1.645
\(\pm 1.96\)
2%
2.054
\(\pm 2.326\)
1%
2.326
\(\pm 2.576\)
Worked example A one-tailed test
Volumes of drink in bottles are \(N(\mu, 4^{2})\) ml. The company claims \(\mu = 500\). A random sample of 12 bottles has mean 497.8 ml. Test at the 5% level whether the mean is less than claimed.
\(H_0: \mu = 500\); \(H_1: \mu < 500\).
Under \(H_0\), \(\bar{X} \sim N\!\left(500, \dfrac{16}{12}\right)\).
Critical value \(-1.645\); \(-1.905 < -1.645\), so reject \(H_0\) (Fig. 1). (Equivalently \(\Phi(-1.905) = 0.0284 < 0.05\).)
Answer: There is sufficient evidence at the 5% level that the mean volume is less than 500 ml.
Worked example A critical region for the sample mean
\(X \sim N(\mu, 6^{2})\). A sample of 36 is used to test \(H_0: \mu = 50\) against \(H_1: \mu > 50\) at the 5% level. Find the critical region for \(\bar{x}\).
Standard error: \(\dfrac{6}{\sqrt{36}} = 1\).
Reject when \(\dfrac{\bar{x} - 50}{1} > 1.645\).
So \(\bar{x} > 50 + 1.645 = 51.645\).
Answer: Critical region \(\bar{x} > 51.645\) (51.6 to 3 s.f.)
Examiner trap
Divide by \(\dfrac{\sigma}{\sqrt{n}}\), not by \(\sigma\). With \(\sigma\) alone the bottling example gives \(z = -0.55\) and the wrong decision.
Section 02
Two-tailed tests and large samples.
With no suggested direction, use \(H_1: \mu \ne \mu_0\) and critical values \(\pm z\). If the population is not known to be normal, or \(\sigma\) is unknown, a large sample lets you use the unbiased estimate \(s^{2}\) in place of \(\sigma^{2}\), with the Central Limit Theorem justifying the normal distribution of \(\bar{X}\).
Worked example Large sample, unknown variance, two-tailed
A random sample of 60 journey times (minutes) gives \(\sum x = 3150\) and \(\sum x^{2} = 168\,450\). Test at the 5% level whether the mean journey time differs from 54 minutes.
Critical values \(\pm 1.645\); \(-1.845 < -1.645\), so reject \(H_0\).
Answer: There is sufficient evidence at the 10% level that the mean mass is not 1.5 kg.
Your turn
\(X \sim N(\mu, 5^{2})\). A random sample of 16 has mean 32.4. Test \(H_0: \mu = 30\) against \(H_1: \mu > 30\) at the 2.5% level.
\(z = \dfrac{32.4 - 30}{5/4} = 1.92\). Critical value 1.96; \(1.92 < 1.96\), so do not reject \(H_0\): insufficient evidence at the 2.5% level that the mean exceeds 30.
Examiner trap
In a two-tailed test at 5%, the critical value is 1.96, not 1.645. Using the one-tailed value makes you reject too easily.
Stage 1 · Learn
Check your understanding
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Which is the test statistic for \(H_0: \mu = \mu_0\) with known \(\sigma\) and sample size \(n\)?
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What is the critical value of \(z\) for a one-tailed test of \(H_1: \mu > \mu_0\) at the 5% level?
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What are the critical values for a two-tailed test at the 1% level?
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A population is not normal and \(\sigma\) is unknown. A sample of 100 is taken. What allows a \(z\) test for \(\mu\)?
Section 03
Writing a full answer.
Paper 6 mark schemes reward each stage separately. A complete answer has all five parts below, in this order.
Hypotheses about \(\mu\), with the population mean named in context if asked.
Distribution of \(\bar{X}\) under \(H_0\), with the reason it is normal (population normal, or large \(n\) and the CLT).
Test statistic \(z\) (or the tail probability) to at least 3 s.f.
Comparison with the critical value (or the significance level), written explicitly: "\(-1.905 < -1.645\)".
Conclusion in context, as evidence, not proof.
Examiner trap
"Accept \(H_0\)" overstates the case. Prefer "do not reject \(H_0\); there is insufficient evidence that ...". Failing to find evidence of a change is not evidence that there was no change.
Stage 2 · Exam
Exam-style questions
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\(X \sim N(\mu, 12^{2})\). A random sample of 36 has mean 83.5. Test \(H_0: \mu = 80\) against \(H_1: \mu > 80\) at the 5% level.
Exam style+20 XP
\(X \sim N(\mu, 15^{2})\). A sample of 25 is used to test \(H_0: \mu = 200\) against \(H_1: \mu < 200\) at the 2.5% level. Find the critical region for \(\bar{x}\).
Exam style+20 XP
A random sample of 50 has \(\bar{x} = 12.6\) and unbiased variance estimate \(s^{2} = 6.48\). Test \(H_0: \mu = 12\) against \(H_1: \mu \ne 12\) at the 5% level.
Exam style+20 XP
In a two-tailed test of a population mean, the test statistic is \(z = 2.12\). Find the smallest significance level at which \(H_0\) would be rejected.
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