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Sampling and the sample mean

Choose a sampling model, estimate population parameters and distinguish exact normal results from CLT approximations.

Learn one idea at a time

Choose a unit. Each has worked examples and practice.

  1. Sample fairly and estimate

    Distinguish the population, sampling frame, parameter and statistic.

  2. Calculate unbiased estimates

    Connect the formulas to raw, frequency and shifted data.

  3. Model a sample mean

    Choose exact normal results or an appropriate large-sample approximation.

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By the end, you can

  • Separate a population parameter from a sample statistic
  • Find the mean and standard error of a sample mean
  • Use unbiased variance estimation and the CLT with care

3 focused units · about 35 minutes for the full chapter. Allow extra time for practice.

The central idea

Population parameters are fixed; sample statistics vary across repeated samples. For iid observations with mean μ\mu and finite variance σ2\sigma^2, E(Xˉ)=μE(\bar X)=\mu and Var⁡(Xˉ)=σ2/n\operatorname{Var}(\bar X)=\sigma^2/n. An exact normal population gives an exact normal mean; otherwise a suitable large-sample CLT approximation may apply. Use n−1n-1 in the unbiased variance estimate, and keep sampling bias separate from sampling variability.

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Worked example

Independent component measurements are normally distributed with population mean 100 and known standard deviation 6. Let XX be one measurement and let Xˉ\bar X be the mean of nine independent measurements. Then X∼N(100,36)X\sim N(100,36), while Xˉ∼N(100,36/9)=N(100,4)\bar X\sim N(100,36/9)=N(100,4) exactly.

For a mean above 104, the standardised boundary is (104−100)/2=2(104-100)/2=2, giving P(Xˉ>104)≈0.0228P(\bar X>104)\approx0.0228.

For one measurement above 104, it is (104−100)/6=2/3(104-100)/6=2/3, giving P(X>104)≈0.252P(X>104)\approx0.252.

Sketch both curves around the same centre: the mean has the smaller spread. No CLT approximation is needed because the population is normal.

The standard deviation of a mean is $\sigma/\sqrt n$, not $\sigma/n$. More observations reduce the variability of the mean; they do not change the distribution of a single measurement.

Check your understanding

Five independent randomly selected observations from a population of unspecified shape give ∑x=50\sum x=50 and ∑x2=516\sum x^2=516. Estimate the population mean and variance. Can these summaries justify calculating an exact normal probability for the sample mean? If the same sampling process is used with 20 observations, what happens to the theoretical standard error, and would that repair a biased sampling frame?

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Practice questions

27 examples and practice questions, in 5 sections.

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