The shape of the lesson
By the end of the lesson, learners can
- The z test, known variance: Use this when the population is normal (or the sample is large) and \(\sigma\) is given.
- Two-tailed tests and large samples: With no suggested direction, use \(H_1: \mu \ne \mu_0\) and critical values \(\pm z\).
- Writing a full answer: Paper 6 mark schemes reward each stage separately.
- Hypotheses: \(H_0: \mu = \mu_0\); \(H_1\) with \(<\), \(>\) or \(\ne\).
- Distribution: \(\bar{X} \sim N\!\left(\mu_0, \dfrac{\sigma^{2}}{n}\right)\); say why it is normal.
- Unknown \(\sigma\), large \(n\): use the unbiased \(s^{2}\).
- Compare: \(z\) with the critical value, or the tail probability with the level.
- Critical region for \(\bar{x}\): \(\mu_0 \pm z\dfrac{\sigma}{\sqrt{n}}\).
- Conclude in context.
One picture every idea returns to
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.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Is the mean really what they claim?. Paper 6. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: The z test, known variance | Use this when the population is normal (or the sample is large) and \(\sigma\) is given. Model the worked example A one-tailed test on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Two-tailed tests and large samples | With no suggested direction, use \(H_1: \mu \ne \mu_0\) and critical values \(\pm z\). Model the worked example Large sample, unknown variance, two-tailed on the board, one line at a time. 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. | Pairs |
| 21 to 30 min | Teach and check: Writing a full answer | Paper 6 mark schemes reward each stage separately. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 26 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Writing a full answer, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Divide by \(\dfrac{\sigma}{\sqrt{n}}\), not by \(\sigma\). | With \(\sigma\) alone the bottling example gives \(z = -0.55\) and the wrong decision. |
| 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. |
| "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. |
Hypothesis tests for a population mean: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
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.
\(X \sim N(\mu, 5^{2})\). A random sample of 16 values has mean 32.4. Test at the 5% level whether \(\mu > 30\).
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.
\(X \sim N(\mu, 6^{2})\). 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 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.
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.
Answer key and marking notes
- 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 \(\sigma^{2}\).
- \(H_0: \mu = 30\); \(H_1: \mu > 30\).
- \(z = \dfrac{32.4 - 30}{5/\sqrt{16}} = 1.92\).
- \(1.92 > 1.645\): reject \(H_0\).
- There is sufficient evidence at the 5% level that \(\mu > 30\).
- \(H_0: \mu = 1.5\); \(H_1: \mu \ne 1.5\).
- \(z = \dfrac{1.43 - 1.5}{0.12/\sqrt{10}} = -1.845\).
- 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.)
- (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 \(\mu > 50\).
- \(\bar{x} = 71.5\); \(s^{2} = \dfrac{1}{79}\left(412\,300 - \dfrac{5720^{2}}{80}\right) = \dfrac{3320}{79} = 42.03\).
- \(H_0: \mu = 70\); \(H_1: \mu > 70\). 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.
- (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 \(1.46\%\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What is the critical value of \(z\) for a two-tailed test of a population mean at the 2% significance level?
- A\(\pm 2.326\)
- B\(\pm 2.054\)
- C\(\pm 2.576\)
- D\(\pm 1.96\)
A machine is set to fill bags with a mean mass of 500 g. An engineer takes a random sample of bags to check whether the setting has changed. Which hypotheses should she use, where \(\mu\) is the mean mass of all bags filled?
- A\(H_0: \mu = 500\), \(H_1: \mu \ne 500\)
- B\(H_0: \mu = 500\), \(H_1: \mu > 500\)
- C\(H_0: \bar{x} = 500\), \(H_1: \bar{x} \ne 500\)
- D\(H_0: \mu \ne 500\), \(H_1: \mu = 500\)
\(X \sim N(\mu, 8^{2})\). A random sample of 16 values has mean 53.6. Test \(H_0: \mu = 50\) against \(H_1: \mu > 50\) at the 1% significance level. Which is correct?
- A\(z = 1.8\); do not reject \(H_0\)
- B\(z = 1.8\); reject \(H_0\)
- C\(z = 0.45\); do not reject \(H_0\)
- D\(z = 7.2\); reject \(H_0\)
In a test of \(H_0: \mu = 40\) against \(H_1: \mu < 40\), the test statistic is \(z = -1.88\). Which is correct at the 5% significance level?
- A\(\Phi(-1.88) = 0.0301 < 0.05\), so reject \(H_0\)
- B\(\Phi(-1.88) = 0.0301 > 0.025\), so do not reject \(H_0\)
- C\(\Phi(1.88) = 0.9699 > 0.05\), so do not reject \(H_0\)
- D\(2 \times 0.0301 = 0.0602 > 0.05\), so do not reject \(H_0\)
A random sample of 50 values of \(X\) gives \(\sum x = 1285\) and \(\sum x^{2} = 33\,600\). Test at the 5% significance level whether the population mean is greater than 25. Which is correct?
- A\(z = 1.44\); do not reject \(H_0\)
- B\(z = 1.44\); reject \(H_0\)
- C\(z = 0.204\); do not reject \(H_0\)
- D\(z = 10.2\); reject \(H_0\)
\(X \sim N(\mu, 12^{2})\). A random sample of 36 values is used to test \(H_0: \mu = 80\) against \(H_1: \mu \ne 80\) at the 5% significance level. Find the critical region for \(\bar{x}\).
- A\(\bar{x} < 76.08\) or \(\bar{x} > 83.92\)
- B\(\bar{x} < 76.71\) or \(\bar{x} > 83.29\)
- C\(\bar{x} < 56.48\) or \(\bar{x} > 103.52\)
- D\(\bar{x} > 83.92\)
\(X \sim N(\mu, 4^{2})\). A random sample of 40 values has mean 41.3. A test of \(H_0: \mu = 40\) against \(H_1: \mu \ne 40\) is carried out. Find the smallest significance level at which \(H_0\) would be rejected.
- A\(3.98\%\)
- B\(1.99\%\)
- C\(96.0\%\)
- D\(74.5\%\)
\(X \sim N(\mu, 5^{2})\). A random sample of \(n\) values has mean 31.2. A test of \(H_0: \mu = 30\) against \(H_1: \mu > 30\) is carried out at the 5% significance level. Find the smallest value of \(n\) for which \(H_0\) is rejected.
- A\(47\)
- B\(46\)
- C\(67\)
- D\(1175\)
Quiz key
1% in each tail, so \(\Phi(z) = 0.99\) and \(z = 2.326\). The tempting \(2.054\) puts all 2% in one tail (\(\Phi(z) = 0.98\)), which is the one-tailed 2% value. \(2.576\) is the two-tailed 1% value and \(1.96\) the two-tailed 5% value.
No direction is suggested, so the test is two-tailed: \(H_0: \mu = 500\), \(H_1: \mu \ne 500\). The tempting \(\mu > 500\) is one-tailed and would miss a fall in the mean. Hypotheses are about the population mean \(\mu\), not the sample mean \(\bar{x}\), and \(H_0\) must contain the equals sign.
\(z = \dfrac{53.6 - 50}{8/\sqrt{16}} = \dfrac{3.6}{2} = 1.8\). Since \(1.8 < 2.326\), do not reject \(H_0\): there is insufficient evidence at the 1% level that \(\mu > 50\). The tempting "reject" compares 1.8 with the 5% value 1.645 instead of the 1% value. \(z = 0.45\) divides by \(\sigma\) instead of \(\dfrac{\sigma}{\sqrt{n}}\), and \(z = 7.2\) divides by \(\dfrac{\sigma}{n} = 0.5\).
\(\text{P}(Z < -1.88) = 1 - \Phi(1.88) = 1 - 0.9699 = 0.0301\), which is less than 0.05, so reject \(H_0\): there is evidence at the 5% level that \(\mu < 40\). (Equivalently, \(-1.88 < -1.645\).) The tempting comparison with 0.025 and the doubled probability \(0.0602\) both treat the test as two-tailed. \(0.9699\) is the wrong tail.
\(\bar{x} = 25.7\), \(s^{2} = \dfrac{1}{49}\left(33\,600 - \dfrac{1285^{2}}{50}\right) = \dfrac{575.5}{49} = 11.74\). \(z = \dfrac{25.7 - 25}{\sqrt{11.74/50}} = 1.444\). The critical value is 1.645 and \(1.444 < 1.645\), so do not reject \(H_0\): insufficient evidence at the 5% level that the mean is greater than 25. The tempting "reject" has the right \(z\) but the wrong decision. \(z = 0.204\) divides by \(s\) instead of \(\dfrac{s}{\sqrt{50}}\), and \(z = 10.2\) divides by \(\dfrac{s}{50}\).
\(\dfrac{12}{\sqrt{36}} = 2\). Reject \(H_0\) when \(\bar{x} < 80 - 1.96 \times 2 = 76.08\) or \(\bar{x} > 80 + 1.96 \times 2 = 83.92\). The tempting \(76.71\) and \(83.29\) use 1.645, the one-tailed value. \(56.48\) and \(103.52\) use \(\sigma = 12\) instead of \(\dfrac{\sigma}{\sqrt{n}}\), and \(\bar{x} > 83.92\) alone ignores the lower tail of a two-tailed test.
\(z = \dfrac{41.3 - 40}{4/\sqrt{40}} = 2.055\). \(\Phi(2.055) = 0.9801\), so each tail is 0.0199 and the two-tailed probability is \(2 \times 0.0199 = 0.0398\). \(H_0\) is rejected at any level above \(3.98\%\). The tempting \(1.99\%\) is only one tail, which would be right for a one-tailed test. \(96.0\%\) is the area between \(-z\) and \(z\), and \(74.5\%\) divides by \(\sigma\) instead of \(\dfrac{\sigma}{\sqrt{n}}\).
\(\dfrac{1.2\sqrt{n}}{5} > 1.645\) gives \(\sqrt{n} > 6.854\), so \(n > 46.98\) and the smallest value is 47. The tempting 46 rounds down, giving \(z = 1.628 < 1.645\). 67 uses the two-tailed value 1.96, and 1175 uses the variance 25 in place of the standard deviation 5.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.