The shape of the lesson
By the end of the lesson, learners can
- Unbiased estimates: An estimate is unbiased if, although it varies from sample to sample, it is right on average.
- Confidence intervals for a mean: If \(X\) is normal with known \(\sigma\), then \(\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\), and in 95% of samples \(\bar{x}\) lies within \(1.96\dfrac{\sigma}{\sqrt{n}}\) of \(\mu\).
- Confidence intervals for a proportion: If \(x\) out of a large sample of \(n\) have a property, \(\hat{p} = \dfrac{x}{n}\) estimates the population proportion \(p\).
- Estimates: \(\bar{x} = \dfrac{\sum x}{n}\), \(s^{2} = \dfrac{1}{n-1}\left(\sum x^{2} - \dfrac{(\sum x)^{2}}{n}\right)\).
- Coded data: work with \(x - a\), then add \(a\) back to the mean; the variance is unchanged.
- Mean: \(\bar{x} \pm z\dfrac{\sigma}{\sqrt{n}}\); use \(s\) when \(n\) is large and \(\sigma\) is unknown.
- Proportion: \(\hat{p} \pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\), large \(n\) only.
- Interpret: a stated value inside the interval is plausible; outside, it is doubtful.
One picture every idea returns to
Unbiased estimates: \(\hat{\mu} = \bar{x}\) and \(s^{2} = \dfrac{1}{n-1}\left(\sum x^{2} - \dfrac{(\sum x)^{2}}{n}\right)\). A confidence interval for \(\mu\) is \(\bar{x} \pm z\dfrac{\sigma}{\sqrt{n}}\) (use \(s\) for \(\sigma\) when \(n\) is large); for a proportion it is \(\hat{p} \pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\). For 95%, \(z = 1.96\).
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, A range you can trust.. 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: Unbiased estimates | An estimate is unbiased if, although it varies from sample to sample, it is right on average. Model the worked example From raw data on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Confidence intervals for a mean | If \(X\) is normal with known \(\sigma\), then \(\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^{2}}{n}\right)\), and in 95% of samples \(\bar{x}\) lies within \(1.96\dfrac{\sigma}{\sqrt{n}}\) of \(\mu\). Model the worked example Normal population, known variance on the board, one line at a time. Your turn: \(X \sim N(\mu, 0.8^{2})\). A random sample of 15 has mean 24.32. Find a 98% confidence interval for \(\mu\). | Pairs |
| 21 to 30 min | Teach and check: Confidence intervals for a proportion | If \(x\) out of a large sample of \(n\) have a property, \(\hat{p} = \dfrac{x}{n}\) estimates the population proportion \(p\). Model the worked example A 90% interval for a proportion on the board, one line at a time. | Numbered Heads Together |
| 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, 21 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 Confidence intervals for a proportion, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| Read carefully whether the question wants the unbiased estimate (divide by \(n - 1\)) and do not take the square root unless the standard deviation is asked for. | Dividing by \(n\) is the usual lost mark. |
| Interpret correctly: a 95% interval does NOT mean "there is a 95% probability that \(\mu\) lies in this interval" (\(\mu\) is fixed). | It means that 95% of intervals constructed this way would contain \(\mu\) (Fig. 1). |
| Use \(\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\), not \(\dfrac{\hat{p}(1-\hat{p})}{n}\) and not \(\sqrt{np(1-p)}\) (that is the standard deviation of the COUNT, not of the proportion). | Return to Confidence intervals for a proportion: ask a pair to state the correct version and show it on the lab before moving on. |
Unbiased estimates and confidence intervals: worksheet
6 questions, 21 marks. Written in the style of the papers; no past paper question is reproduced.
Explain what is meant by saying that \(\bar{x}\) is an unbiased estimate of \(\mu\).
The times, in minutes, taken by 8 randomly chosen students to solve a puzzle were
4.2 3.8 5.1 4.6 4.9 3.5 4.4 5.0
Find unbiased estimates of the population mean and variance.
For a random sample of 60 values of \(x\), \(\sum (x - 50) = 84\) and \(\sum (x - 50)^{2} = 1260\). Find unbiased estimates of the population mean and variance.
The diameters of bolts are \(N(\mu, 0.8^{2})\) mm. A random sample of 15 bolts has mean diameter 24.32 mm. Find a 98% confidence interval for \(\mu\).
In a random sample of 400 adults, 132 said they cycle to work.
(a) Find a 95% confidence interval for the proportion of all adults who cycle to work.
(b) A council claims that 30% of adults cycle to work. Comment on this claim.
(c) Using the same estimate of the proportion, find the smallest sample size for which the width of the 95% interval is at most 0.05.
A random sample of 100 calls to a help desk had durations \(x\) minutes with \(\sum x = 4250\) and \(\sum x^{2} = 184\,000\).
(a) Find unbiased estimates of the population mean and variance.
(b) Find a 95% confidence interval for the mean call duration, and explain why it is valid even though call durations are not normally distributed.
(c) Ten such 95% intervals are found from independent samples. Find the probability that all ten contain the true mean.
Answer key and marking notes
Individual sample means vary, but on average (over all possible samples) they equal \(\mu\): \(\text{E}(\bar{X}) = \mu\).
- \(\sum x = 35.5\), \(\sum x^{2} = 159.87\).
- \(\bar{x} = \dfrac{35.5}{8} = 4.4375\) (4.44 to 3 s.f.).
- \(s^{2} = \dfrac{1}{7}\left(159.87 - \dfrac{35.5^{2}}{8}\right) = \dfrac{1}{7}(159.87 - 157.53125) = 0.334\) (3 s.f.).
- Mean of \(x - 50\): \(\dfrac{84}{60} = 1.4\), so \(\bar{x} = 51.4\).
- \(s^{2} = \dfrac{1}{59}\left(1260 - \dfrac{84^{2}}{60}\right) = \dfrac{1}{59}(1260 - 117.6) = \dfrac{1142.4}{59}\).
- \(s^{2} = 19.4\) (3 s.f.); subtracting 50 does not change the variance.
- \(z = 2.326\) (1% in each tail).
- Margin: \(2.326 \times \dfrac{0.8}{\sqrt{15}} = 0.480\).
- Interval: \((23.8, 24.8)\) mm (3 s.f.).
- (a) \(\hat{p} = 0.33\); margin \(1.96\sqrt{\dfrac{0.33 \times 0.67}{400}} = 1.96 \times 0.023\,51 = 0.0461\). Interval \((0.284, 0.376)\).
- (b) 0.30 lies inside the interval, so the data do not contradict the council's claim.
- (c) Width \(= 2 \times 1.96\sqrt{\dfrac{0.2211}{n}} \le 0.05\), so \(n \ge \left(\dfrac{3.92}{0.05}\right)^{2} \times 0.2211 = 1359.004\).
- Smallest \(n = 1360\).
- (a) \(\bar{x} = 42.5\); \(s^{2} = \dfrac{1}{99}\left(184\,000 - \dfrac{4250^{2}}{100}\right) = \dfrac{3375}{99} = 34.1\) (3 s.f.).
- (b) Margin \(1.96\sqrt{\dfrac{34.09}{100}} = 1.144\); interval \((41.4, 43.6)\) minutes.
- \(n = 100\) is large, so by the Central Limit Theorem \(\bar{X}\) is approximately normal whatever the distribution of durations.
- (c) Each interval contains \(\mu\) with probability 0.95, independently: \(0.95^{10} = 0.599\) (3 s.f.).
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.
For a random sample of 10 values, \(\sum x = 84\) and \(\sum x^{2} = 736\). Find an unbiased estimate of the population variance.
- A\(3.38\)
- B\(3.04\)
- C\(1.84\)
- D\(11.2\)
The lengths of components are \(N(\mu, 2.4^{2})\) mm. A random sample of 16 components has mean 31.5 mm. Find a 95% confidence interval for \(\mu\).
- A\((30.3, 32.7)\)
- B\((26.8, 36.2)\)
- C\((30.5, 32.5)\)
- D\((31.2, 31.8)\)
In a random sample of 180 households, 63 own a dog. Find a 95% confidence interval for the proportion of all households that own a dog.
- A\((0.280, 0.420)\)
- B\((0.348, 0.352)\)
- C\((0.292, 0.408)\)
- D\((0.258, 0.442)\)
A 95% confidence interval for a population mean \(\mu\) is calculated from a random sample. Which statement correctly describes what "95%" means?
- AIf many random samples were taken and an interval calculated from each, about 95% of the intervals would contain \(\mu\)
- B95% of the values in the population lie inside the interval
- C95% of the values in the sample lie inside the interval
- DAbout 95% of all possible sample means lie inside this particular interval
A random sample of 64 values gives \(\sum x = 3328\) and \(\sum x^{2} = 176\,500\). Find a 98% confidence interval for the population mean.
- A\((49.9, 54.1)\)
- B\((50.2, 53.8)\)
- C\((49.6, 54.4)\)
- D\((34.8, 69.2)\)
A population has standard deviation 6. Find the smallest sample size for which a 95% confidence interval for the population mean has width less than 2.
- A\(139\)
- B\(138\)
- C\(35\)
- D\(98\)
In a random sample of 120 people, a proportion \(\hat{p}\) agree with a statement. An approximate confidence interval for the population proportion, calculated from this sample, is \((0.315, 0.485)\). Find the confidence level.
- A\(94.3\%\)
- B\(97.1\%\)
- C\(5.74\%\)
- D\(47.1\%\)
Five researchers each take an independent random sample and each calculates a 95% confidence interval for the same population mean \(\mu\). Find the probability that at least 4 of the 5 intervals contain \(\mu\).
- A\(0.977\)
- B\(0.774\)
- C\(0.815\)
- D\(0.204\)
Quiz key
\(\sum x^{2} - \dfrac{84^{2}}{10} = 736 - 705.6 = 30.4\), and \(s^{2} = \dfrac{30.4}{9} = 3.38\). The tempting \(3.04\) divides by \(n = 10\), which gives the biased sample variance. \(1.84\) is \(\sqrt{3.38}\), an estimate of the standard deviation, and \(11.2\) is \(\dfrac{\sum x^{2}}{9} - \bar{x}^{2}\), which mixes the two formulas.
\(\dfrac{2.4}{\sqrt{16}} = 0.6\), so the interval is \(31.5 \pm 1.96 \times 0.6 = 31.5 \pm 1.176\), giving \((30.3, 32.7)\). The tempting \((30.5, 32.5)\) uses 1.645, which is the 90% value. \((26.8, 36.2)\) uses \(\sigma\) instead of \(\dfrac{\sigma}{\sqrt{n}}\), and \((31.2, 31.8)\) divides \(\sigma\) by \(n\) instead of \(\sqrt{n}\).
\(\hat{p} = 0.35\). \(\sqrt{\dfrac{0.35 \times 0.65}{180}} = 0.035\,55\), so the interval is \(0.35 \pm 1.96 \times 0.035\,55 = 0.35 \pm 0.0697\), giving \((0.280, 0.420)\). The tempting \((0.348, 0.352)\) forgets the square root. \((0.292, 0.408)\) uses 1.645 (90%) and \((0.258, 0.442)\) uses 2.576 (99%).
The confidence level describes the METHOD: 95% of intervals constructed this way contain \(\mu\). The interval is about the mean, not individual values, so it does not contain 95% of the population or of the sample; it is much narrower than that. The tempting last option is wrong because sample means are centred on \(\mu\), not on this sample's \(\bar{x}\), so this particular interval need not contain 95% of them.
\(\bar{x} = 52\). \(s^{2} = \dfrac{1}{63}\left(176\,500 - \dfrac{3328^{2}}{64}\right) = \dfrac{3444}{63} = 54.67\), so \(s = 7.394\). Interval: \(52 \pm 2.326 \times \dfrac{7.394}{8} = 52 \pm 2.150\), giving \((49.9, 54.1)\). The tempting \((50.2, 53.8)\) uses the 95% value 1.96; \((49.6, 54.4)\) uses the 99% value 2.576; \((34.8, 69.2)\) forgets to divide \(s\) by \(\sqrt{64}\).
Width \(= 2 \times 1.96 \times \dfrac{6}{\sqrt{n}} < 2\) gives \(\sqrt{n} > 11.76\), so \(n > 138.3\) and the smallest sample size is 139. The tempting 35 sets the half-width (not the width) less than 2. 138 rounds down, giving a width just above 2, and 98 uses the 90% value 1.645.
\(\hat{p} = 0.4\) and the half-width is 0.085. \(\sqrt{\dfrac{0.4 \times 0.6}{120}} = 0.044\,72\), so \(z = \dfrac{0.085}{0.044\,72} = 1.901\). \(\Phi(1.901) = 0.9713\), so the level is \(2 \times 0.9713 - 1 = 0.943\), i.e. \(94.3\%\). The tempting \(97.1\%\) stops at \(\Phi(z)\) and does not account for the two tails. \(5.74\%\) is the total area OUTSIDE the interval, and \(47.1\%\) is only the area between 0 and \(z\).
The number of intervals containing \(\mu\) is \(B(5, 0.95)\). \(\text{P}(\ge 4) = 0.95^{5} + 5 \times 0.95^{4} \times 0.05 = 0.7738 + 0.2036 = 0.977\). The tempting \(0.774\) is all 5 only, and \(0.204\) is exactly 4 only. \(0.815\) is \(0.95^{4}\), which leaves out both the fifth interval and the 5 arrangements.
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.