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QLDSpecialist MathematicsSyllabus dot point

Topic 3: Statistical inference

Construct and interpret confidence intervals for a population mean using the sample mean and standard error, choosing the appropriate confidence level, and understand the meaning of the confidence level in repeated sampling

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on confidence intervals. Covers the structure of an interval estimate, the critical value and margin of error, how confidence level and sample size affect width, the correct repeated-sampling interpretation, and a fully verified worked example with the common interpretation mistake QCAA penalises.

Generated by Claude Opus 4.76 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to build an interval estimate for an unknown population mean from a sample, choose a confidence level, compute the margin of error, and interpret the interval correctly in repeated-sampling language. Confidence intervals are a major theme of IA3 and the external assessment, and the interpretation is examined as carefully as the calculation.

The answer

Point estimate versus interval estimate

A single sample mean xˉ\bar{x} is a point estimate of the population mean μ\mu, but it almost certainly is not exactly μ\mu. A confidence interval gives a range of plausible values for μ\mu, built around xˉ\bar{x}, together with a stated level of confidence.

Structure of the interval

A confidence interval for μ\mu (with known population standard deviation σ\sigma) is

xˉ±zσn,\bar{x} \pm z^{*}\,\frac{\sigma}{\sqrt{n}},

where σn\dfrac{\sigma}{\sqrt{n}} is the standard error and zz^{*} is the critical value from the standard normal distribution for the chosen confidence level. The quantity zσnz^{*}\dfrac{\sigma}{\sqrt{n}} is the margin of error, the half-width of the interval.

Critical values

The critical value zz^{*} cuts off the central proportion of the standard normal equal to the confidence level. The standard values are:

90%:z=1.645,95%:z=1.960,99%:z=2.576.90\%: z^{*} = 1.645, \qquad 95\%: z^{*} = 1.960, \qquad 99\%: z^{*} = 2.576.

A higher confidence level needs a larger zz^{*}, so the interval is wider. There is a trade-off: more confidence costs precision.

Width, sample size and confidence level

The interval width is 2zσn2z^{*}\dfrac{\sigma}{\sqrt{n}}. It widens with a higher confidence level (larger zz^{*}) and narrows with a larger sample (the n\sqrt{n} in the denominator). To halve the margin of error you must quadruple the sample size, because of the square root.

Interpreting the confidence level

The correct interpretation refers to the method over repeated sampling. A 95%95\% confidence level means that if many samples were taken and an interval constructed from each, about 95%95\% of those intervals would contain the true mean μ\mu. It does not mean there is a 95%95\% probability that μ\mu lies in this particular interval: μ\mu is a fixed number, and any single interval either contains it or does not. The confidence is in the long-run reliability of the procedure.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2024 QCAA4 marksA company claims that the mean battery life of their latest model of smartphone is 9.5 hours. To test this claim, the battery lives of a random sample of 40 of the smartphones were measured. A sample mean of 9.31 hours and a standard deviation of 0.52 hours were calculated from this data. a) Determine an approximate 95% confidence interval for the population mean. b) Determine an approximate 99% confidence interval for the population mean. A manager comments that either confidence interval could be used to support the company's claim. c) Use your results to evaluate the reasonableness of the manager's comment.
Show worked answer →

Margin of error = z * s / sqrt(n) with n = 40, x-bar = 9.31, s = 0.52.

a) [1 mark] For 95%, z = 1.96. Margin = 1.96 * 0.52 / sqrt(40) = 0.161. Interval = 9.31 +/- 0.16 = (9.15, 9.47) hours.

b) [1 mark] For 99%, z = 2.576. Margin = 2.576 * 0.52 / sqrt(40) = 0.212. Interval = 9.31 +/- 0.21 = (9.10, 9.52) hours.

c) [2 marks] The claimed mean of 9.5 hours lies OUTSIDE the 95% interval (since 9.5 > 9.47) but INSIDE the 99% interval (9.10 to 9.52). The two intervals lead to opposite conclusions, so only the 99% interval supports the claim. The manager's comment is not reasonable: the choice of confidence level changes the conclusion, so it is wrong to say either interval could be used to support the claim.

2023 QCAA4 marksThe wait time for customers put on hold when calling complaint departments is assumed to be normally distributed. A company claims that the mean wait time for their customers is 7.6 minutes. The following data represents the wait time (minutes) from a random sample of 12 customers: 8.3, 12.7, 9.1, 7.3, 10.3, 5.4, 8.5, 10.7, 6.9, 12.5, 7.2, 11.9. a) Determine the mean of this data. The standard deviation of this data is calculated to be 2.384 minutes. b) Use an approximate 95% confidence interval for the mean to evaluate the reasonableness of the company's claim. Justify your decision using mathematical reasoning.
Show worked answer →

a) [1 mark] Sum of the 12 values = 110.8, so x-bar = 110.8 / 12 = 9.23 minutes (to 2 d.p.).

b) [3 marks] With s = 2.384, n = 12 and z = 1.96 for 95%:
margin = 1.96 * 2.384 / sqrt(12) = 1.96 * 0.6882 = 1.349.
Interval = 9.23 +/- 1.35 = (7.89, 10.58) minutes.
Decision: the claimed mean of 7.6 minutes lies BELOW the lower limit 7.89, so 7.6 is not contained in the 95% confidence interval. There is therefore evidence against the claim, so the company's claim of a 7.6 minute mean wait time is not reasonable at the 95% level.

2022 QCAA6 marksThe mass of a population of elephants is known to be normally distributed. A biologist randomly selects a number of elephants from this population and measures their masses. The mean mass of the sample is 5206 kg with a standard deviation of 356 kg. The biologist uses the data to calculate a 90% confidence interval for the population mean mass of (5159.1, 5252.9) kg. Determine a 99% confidence interval for the population mean mass based on the same data.
Show worked answer →

Step 1 - recover the sample size n from the 90% interval [2 marks]. For 90%, z = 1.645. The half-width is (5252.9 - 5159.1) / 2 = 46.9.
So 1.645 * 356 / sqrt(n) = 46.9, giving sqrt(n) = 1.645 * 356 / 46.9 = 12.49, so n = 155.9, i.e. n = 156.

Step 2 - build the 99% interval with the same x-bar = 5206 and s = 356 [3 marks]. For 99%, z = 2.576.
margin = 2.576 * 356 / sqrt(156) = 2.576 * 28.50 = 73.4.
Interval = 5206 +/- 73.4 = (5132.6, 5279.4) kg.

[1 mark] for logical structure: a higher confidence level (99% vs 90%) gives a wider interval for the same data, consistent with the result.