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Sampling Distributions and the Central Limit Theorem

For independent observations from one population with finite mean and finite positive variance, sample means centre on the population mean and have standard error σ / √n. Their standardised distribution approaches a standard normal one as n grows.

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What a learner can do afterwards

  • Describe how sample size and population spread affect the spread of sample means
  • Calculate standard error using σ divided by √n and find a sample size for a stated target
  • Explain when a normal model fits sample means and distinguish precision from bias or guaranteed accuracy

1 · Read

Use independent random observations from one population with finite mean μ and finite, positive standard deviation σ. Imagine repeatedly taking samples of the same size n and finding each mean. These means form a sampling distribution. Its expected mean is μ. Its standard deviation is called the standard error. Sampling without replacement from a small population needs a correction to the formula used here.

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The standard error is σ / √n. With σ = 18 and n = 9, it is 18 / 3 = 6. For n = 36 it is 3, and for n = 144 it is 1.5. Multiplying n by four halves the standard error. The ratio of standard error to σ is 1 / √n. For a target standard error s, solve √n = σ / s, giving n = (σ / s)².

n = 96n = 363n = 1441.5
Each bar shows a standard error, not the population spread. Increasing sample size reduces this sampling variation.

Under these assumptions, the central limit theorem says the standardised sample mean approaches a standard normal distribution as n grows. For a large enough n, sample means are approximately normal. There is no universal cutoff of 30. Strong skew or heavy tails can require larger samples. If the population is normal, independent sample means are exactly normal for every positive sample size.

When a normal model fits the sample mean, use z = (sample mean - μ) / (σ / √n). The denominator is the standard error, not σ. For μ = 40, σ = 10 and n = 25, a mean of 44 gives z = 2. A left-cumulative normal table gives P(Z < 2) = 0.9772, so P(sample mean > 44) = 1 - 0.9772 = 0.0228.

A larger independent random sample from the same population gives a more precise mean because its standard error is smaller. This is a statement about repeated sampling. A particular large sample can still be farther from μ than a particular small sample. Increasing sample size does not repair biased selection, poor measurements or dependence between observations.

Correct SE: σ / √nReported SE: σ / nCorrect / reportedn / √n = √n
For σ > 0 and n > 1, dividing σ by n gives too small a standard error. Divide the correct result by the reported result to find the factor.

Independent sample means centre on μ and have standard error σ / √n. Larger samples reduce random variation. Normality needs suitable conditions, and size alone does not remove bias.

Lesson, worked questions and flashcards authored for this learning dataset.

Readings and extra questions authored for this learning dataset. Video references credit Khan Academy.

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Sampling Distributions and the Central Limit Theorem · Mathematics, ages 17 to 18 · LightMySky