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How do sample means let us make confident statements about a population mean?

Use the distribution of the sample mean and the central limit theorem to build confidence intervals for a population mean

WACE Specialist Unit 4 statistical inference: the sampling distribution of the sample mean, its mean and standard deviation (standard error), the central limit theorem, and constructing and interpreting confidence intervals for a population mean, with a full worked example.

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  1. What this dot point is asking
  2. From sample to population
  3. The sampling distribution of the sample mean
  4. The central limit theorem
  5. Confidence intervals for a population mean
  6. Effect of sample size and confidence level

What this dot point is asking

SCSA Unit 4 statistical inference is about reasoning from a sample to a population. You must know how the sample mean behaves across repeated samples, state and use the central limit theorem, and construct and interpret confidence intervals for a population mean.

From sample to population

Inference is the logic of reasoning backwards: a sample is observed and a statement about the whole population is made. We cannot measure every member of a population, so we draw a random sample and use its mean as an estimate of the population mean. The difficulty is that a different sample would give a different estimate, so a single number is not enough; we need to quantify the uncertainty around it. This dot point develops the machinery, the sampling distribution, the central limit theorem and the confidence interval, that turns a single sample mean into a defensible statement about the population, complete with a stated level of confidence.

The sampling distribution of the sample mean

Take repeated random samples of size nn from a population with mean μ\mu and standard deviation σ\sigma. Each sample gives a sample mean xˉ\bar{x}, and these vary from sample to sample. The random variable Xˉ\bar{X} has:

The quantity σn\dfrac{\sigma}{\sqrt{n}} is the standard error. It shrinks as nn grows, so larger samples give sample means clustered more tightly around the true mean μ\mu.

The central limit theorem

This is what makes inference possible: even if the underlying population is skewed, the sample mean behaves normally for large nn, so we can use normal-distribution probabilities and zz-values.

Confidence intervals for a population mean

An approximate confidence interval estimates μ\mu from one sample. With known (or large-sample estimated) population standard deviation σ\sigma, the interval is

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

where zz is the critical value for the chosen confidence level: z1.645z \approx 1.645 for 90%90\%, z1.96z \approx 1.96 for 95%95\%, and z2.576z \approx 2.576 for 99%99\%. The term zσnz\dfrac{\sigma}{\sqrt{n}} is the margin of error.

The correct interpretation of a 95%95\% confidence interval: if we repeated the sampling many times and built an interval each time, about 95%95\% of those intervals would contain the true mean μ\mu. It is not a 95%95\% probability that μ\mu lies in this one interval.

Effect of sample size and confidence level

A wider confidence level (say 99%99\% instead of 95%95\%) gives a larger zz and a wider interval: more confidence costs precision. A larger sample reduces the standard error and narrows the interval. Because the standard error has n\sqrt{n} in the denominator, increasing the sample size by a factor of 44 halves the width.

Exam-style practice questions

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

WACE 20248 marksCalculator-assumed. A machine fills bottles. A random sample of n=49n = 49 bottles has mean volume xˉ=498\bar{x} = 498 mL, and the population standard deviation is σ=7\sigma = 7 mL. (a) Construct a 90%90\% confidence interval for the mean fill volume. (b) The target mean is 500500 mL. Comment on whether the interval is consistent with the machine meeting target. (c) What sample size gives a 90%90\% margin of error of 11 mL?
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A full inference question with interpretation.

(a) Standard error =749=77=1= \dfrac{7}{\sqrt{49}} = \dfrac{7}{7} = 1. For 90%90\%, z=1.645z = 1.645, margin =1.645= 1.645. Interval: 498±1.645=(496.36, 499.64)498 \pm 1.645 = (496.36,\ 499.64) mL.

(b) The target 500500 mL lies above the upper limit 499.64499.64, so 500500 is not in the interval. The data suggest the machine is filling slightly below target.

(c) Require 1.645×7n11.645 \times \dfrac{7}{\sqrt{n}} \le 1, so n1.645×7=11.515\sqrt{n} \ge 1.645 \times 7 = 11.515, hence n132.6n \ge 132.6. Round up to n=133n = 133.

Markers reward the standard error, the 90%90\% critical value, the interval, the comparison with target, and rounding nn up.

WACE 20224 marksCalculator-free. Explain why the central limit theorem is needed to justify a confidence interval for a mean when the population distribution is unknown, and state the role of the standard error.
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A conceptual question on the theory.

A confidence interval xˉ±zσn\bar{x} \pm z\dfrac{\sigma}{\sqrt{n}} uses normal-distribution critical values such as z=1.96z = 1.96. These are only valid if Xˉ\bar{X} is normally distributed. When the population shape is unknown, the central limit theorem provides the justification: for a large sample, Xˉ\bar{X} is approximately normal regardless of the population's shape, so the zz-values apply.

The standard error σn\dfrac{\sigma}{\sqrt{n}} is the standard deviation of Xˉ\bar{X}; it sets the width of the interval and shrinks as nn grows, making the estimate more precise. Markers reward linking the normal critical values to the CLT and identifying the standard error as the spread of the sample mean.

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