Paper 6. A sample gives a single best guess for the population mean or proportion, but a range is more honest. Unbiased estimates give the guess; a confidence interval wraps it in a margin that captures the true value in a known proportion of samples.
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%, \(z = 1.96\).
Fig. 1: illustration of twelve 95% confidence intervals from different random samples. Each interval has the same width; their centres vary. In the long run 95% of such intervals contain μ and 5% (about 1 in 20) miss it
Section 01
Unbiased estimates.
An estimate is unbiased if, although it varies from sample to sample, it is right on average. The sample mean is unbiased for \(\mu\). The plain sample variance (dividing by \(n\)) is slightly too small on average, so we divide by \(n - 1\) instead.
Answer: \(\bar{x} = 24.8\), \(s^{2} = 18.3\) (3 s.f.). Dividing by 50 would give the biased value \(17.96\).
Examiner trap
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.
Section 02
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 \(\mu\). Turning that round gives the interval \(\bar{x} \pm 1.96\dfrac{\sigma}{\sqrt{n}}\). If \(\sigma\) is unknown but \(n\) is large, use \(s\) and the Central Limit Theorem.
Confidence level
90%
95%
98%
99%
\(z\)
1.645
1.96
2.326
2.576
Worked example Normal population, known variance
Lengths of rods are \(N(\mu, 3.2^{2})\) mm. A random sample of 20 rods has mean 48.6 mm. Find a 95% confidence interval for \(\mu\).
Standard error: \(\dfrac{3.2}{\sqrt{20}} = 0.7155\).
Margin: \(1.96 \times 0.7155 = 1.402\).
Interval: \(48.6 \pm 1.402\).
Answer: \((47.2, 50.0)\) mm (3 s.f.)
Worked example Large sample, unknown variance
A random sample of 80 commuting times (minutes) gives \(\sum x = 2936\) and \(\sum x^{2} = 108\,530\). Find a 99% confidence interval for the mean commuting time.
\(n\) is large, so by the CLT \(\bar{X}\) is approximately normal; use \(s\) for \(\sigma\). Margin: \(2.576\sqrt{\dfrac{9.858}{80}} = 0.904\).
Interval: \(36.7 \pm 0.904\).
Answer: \((35.8, 37.6)\) minutes (3 s.f.)
Your turn
\(X \sim N(\mu, 0.8^{2})\). A random sample of 15 has mean 24.32. Find a 98% confidence interval for \(\mu\).
Margin \(2.326 \times \dfrac{0.8}{\sqrt{15}} = 0.480\), so \((23.8, 24.8)\).
Examiner trap
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).
Stage 1 · Learn
Check your understanding
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When finding an unbiased estimate of the population variance, what do you divide \(\sum x^{2} - \dfrac{(\sum x)^{2}}{n}\) by?
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Which \(z\) value is used for a 95% confidence interval?
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A 95% confidence interval for \(\mu\) is \((12.1, 13.5)\). Which statement is correct?
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With everything else unchanged, the sample size is multiplied by 4. What happens to the width of the confidence interval?
Section 03
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 \(p\). The proportion is approximately normal with variance \(\dfrac{p(1-p)}{n}\), estimated with \(\hat{p}\), giving \(\hat{p} \pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\).
Worked example A 90% interval for a proportion
In a random sample of 150 voters, 54 support a new policy. Find a 90% confidence interval for the proportion of all voters who support it.
\(\hat{p} = \dfrac{54}{150} = 0.36\).
Standard error: \(\sqrt{\dfrac{0.36 \times 0.64}{150}} = 0.039\,19\).
Margin: \(1.645 \times 0.039\,19 = 0.0645\).
Interval: \(0.36 \pm 0.0645\).
Answer: \((0.296, 0.424)\) (3 s.f.)
Worked example Finding the confidence level
A 36-item random sample from \(N(\mu, 6^{2})\) gives the interval \((21.3, 24.7)\) for \(\mu\). Find the confidence level.
Half-width: \(\dfrac{24.7 - 21.3}{2} = 1.7\).
\(z \times \dfrac{6}{\sqrt{36}} = 1.7\), so \(z = 1.7\).
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).
Stage 2 · Exam
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A random sample of 40 values has \(\sum x = 1860\) and \(\sum x^{2} = 87\,340\). Find an unbiased estimate of the population variance.
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\(X \sim N(\mu, 5^{2})\). A random sample of 25 has mean 72.4. Find a 95% confidence interval for \(\mu\).
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In a random sample of 120 students, 48 walk to school. Find a 99% confidence interval for the proportion of all students who walk.
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A random sample of 36 from \(N(\mu, 6^{2})\) gives the confidence interval \((21.3, 24.7)\). What is the confidence level?
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