6.4 - The distribution of a sample mean

6.4 - The distribution of a sample mean

Individual measurements vary, and so do the means of repeated samples. In this part of 6.4, you will find the centre and spread of a sample mean, use its exact normal distribution for a normal population, and apply the Central Limit Theorem when a large sample makes normal approximation appropriate.

A mean as a random variable

If you multiply a sample size by four, does the sample mean become four times as precise? Its standard deviation halves. The square root in this relationship explains why obtaining a much more stable average can require many more observations.

Before a sample is taken, write its independent observations as X1,…,XnX_1,\ldots,X_n, each from the same population with mean μ\mu and variance σ2\sigma^2. Their average

X‾=X1+⋯+Xnn\overline X=\frac{X_1+\cdots+X_n}{n}

is itself a random variable. Different samples give different observed values, written xˉ\bar x.

The earlier linear-combination rules give

E(X‾)=1n(nμ)=μVar⁡(X‾)=1n2(nσ2)=σ2n.\begin{aligned} \mathrm E(\overline X)&=\frac1n(n\mu)=\mu\\ \operatorname{Var}(\overline X)&=\frac1{n^2}(n\sigma^2)=\frac{\sigma^2}{n}. \end{aligned}

The nn independent variance contributions add before the total is divided by n2n^2. No normal-population assumption was used for these two moment results.

Mean and variance of a sample mean

For nn independent observations from the same population, with finite mean μ\mu and variance σ2\sigma^2,

E(X‾)=μ,Var⁡(X‾)=σ2n,SD⁡(X‾)=σn.\mathrm E(\overline X)=\mu,\qquad \operatorname{Var}(\overline X)=\frac{\sigma^2}{n},\qquad \operatorname{SD}(\overline X)=\frac{\sigma}{\sqrt n}.

The standard deviation of the sample mean is called its standard error.

The mean and variance appear in the formula booklet's sample-mean normal approximation. They also hold without normality; standard error is the square root of that variance.

The population standard deviation describes individual measurements. The standard error describes the means of samples of the specified size. Their units match, but they measure different variation.

Worked example 1: the spread of an average

Question. A population has mean 3030 and variance 144144. For a random sample of 99 independent observations, find the expectation, variance and standard error of its mean.

Plan. Keep the expectation at the population mean, divide variance by nn, then take a square root.

E(X‾)=30,Var⁡(X‾)=144/9=16,SE⁡=4.\mathrm E(\overline X)=30,\qquad \operatorname{Var}(\overline X)=144/9=16,\qquad \operatorname{SE}=4.

Answer. Expectation 3030, variance 1616, standard error 44.

Check. The individual standard deviation is 1212. Dividing by 9=3\sqrt9=3 gives 44, agreeing with the square root of the calculated variance.

Increase only the sample size, from 99 to 3636. Individual measurements retain the same spread.

Worked example 2: quadrupling the sample size

Question. For the same population, find the expectation, variance and standard error of a mean from 3636 independent observations.

Plan. Replace nn by 3636 in the sample-mean formulae.

E(X‾)=30,Var⁡(X‾)=144/36=4,SE⁡=2.\mathrm E(\overline X)=30,\qquad \operatorname{Var}(\overline X)=144/36=4,\qquad \operatorname{SE}=2.

Answer. Expectation 3030, variance 44, standard error 22.

Check. The sample size is four times larger, its variance is one quarter as large, and its standard error is half as large. The population standard deviation is still 1212.

For a sample without replacement from a large finite population, independence may be an approximation justified when the sample is a small fraction of that population. Use the supplied model. Recording one observation repeatedly does not produce independent information.

Watch out

The variance of a mean is σ2/n\sigma^2/n, not σ2/n2\sigma^2/n^2. A sum of nn independent observations has variance nσ2n\sigma^2; averaging then divides that by n2n^2.

Keep the population SD and standard error distinct in this check.

Normal populations

If the population is normal, its independent sample mean is exactly normal for every positive sample size. This follows from the earlier result that a linear combination of independent normal variables is normal. You do not need a large sample to establish normality in this case.

Sample mean from a normal population

If X1,…,XnX_1,\ldots,X_n are independent, each with distribution N(μ,σ2)\mathrm N(\mu,\sigma^2), then

X‾∼N(μ,σ2n).\overline X\sim\mathrm N\left(\mu,\frac{\sigma^2}{n}\right).

Standardise its value aa using

z=a−μσ/n.z=\frac{a-\mu}{\sigma/\sqrt n}.

The sample-mean normal form is in the formula booklet under the CLT heading. Its exactness for a normal population and the standardisation step are not stated there separately: learn the distinction.

[DIAGRAM: asset_name: Individual measurements and sample means; asset_slug: sample-mean-spread; caption: Averaging leaves the centre fixed and reduces the standard deviation; description: Normal density curves for individual measurements N(100,64), means of sixteen with N(100,4), and means of sixty-four with N(100,1). All share centre one hundred. The three standard deviations are labelled eight, two and one. Each curve has total area one, and the narrowest curve is tallest.]
Averaging leaves the centre fixed and reduces the standard deviation

Use the normal tables just as before, but divide by the standard error of the mean. There is no continuity correction for the continuous measurements in the examples.

Worked example 3: an exact normal probability

Question. Individual measurements have distribution N(100,64)\mathrm N(100,64). A random sample of 1616 independent measurements is taken. Find the probability that its mean exceeds 102102.

Plan. Identify the mean's normal distribution, then standardise with 8/168/\sqrt{16}.

X‾∼N(100,64/16)=N(100,4)P(X‾>102)=1−Φ(102−1002)=1−Φ(1)=1−0.8413=0.1587.\begin{aligned} \overline X&\sim\mathrm N(100,64/16)=\mathrm N(100,4)\\ \mathrm P(\overline X>102) &=1-\Phi\left(\frac{102-100}{2}\right)\\ &=1-\Phi(1)=1-0.8413=0.1587. \end{aligned}

Answer. 0.1590.159, to 33 significant figures.

Check. A sample total has distribution N(1600,1024)\mathrm N(1600,1024). The event is total >1632>1632, giving z=(1632−1600)/32=1z=(1632-1600)/32=1, the same standardised boundary.

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