SI1–SI6 - Sample means and the Central Limit Theorem | OCR A-Level Further Mathematics B (MEI)

SI1–SI2 turn sample data into estimates of a population's mean and variance. SI3–SI6 then treat the sample mean itself as a random variable: you will calculate its standard error, distinguish exact Normal results from the Central Limit Theorem, and approximate probabilities for means and totals.

Estimate the population, not just the spread of the list

Can averaging make a collection of very irregular measurements behave predictably? Yes, but first separate three objects: the population, the sample you actually saw, and all the other samples you might have drawn.

Suppose five measurements are $8,10,12,14,16$ . Their mean is $12$ , and their squared deviations from $12$ total $40$ . When these are a random sample from a larger population, the usual estimate of the population variance is $40/4=10$ , rather than $40/5=8$ .

Point estimate

A point estimate is a single value calculated from sample data to estimate a population parameter. Use the sample mean $\bar x$ to estimate the population mean $\mu$ :

\hat\mu=\bar x=\frac{\sum x_i}{n}

The sample mean varies between samples; the population mean being estimated is fixed.

This point-estimation rule is not separately printed in the formula booklet. The sample-variance formula below is supplied. Its divisor is $n-1$ , because the deviations are taken from the estimated mean. You need to use this result, not prove why it is unbiased.

Sample variance as a population-variance estimate

For a sample of size $n>1$ ,

s^2=\frac{\sum(x_i-\bar x)^2}{n-1} =\frac{\sum x_i^2-(\sum x_i)^2/n}{n-1}

Use $s^2$ to estimate $\sigma^2$ , and $s=\sqrt{s^2}$ to estimate the population standard deviation.

In the formula booklet.

“Unbiased” describes the variance estimator: across random samples its mean is the population variance. It does not promise that one sample gives the true value, or that a biased sample becomes representative after this calculation.

Worked example 1: estimate from a short sample

Question. A random sample is $8,10,12,14,16$ . Estimate the population mean and variance.

Plan. Find the sample mean, then divide the sum of squared deviations by $n-1$ .

\begin{aligned} \bar x&=\frac{60}{5}=12 && \text{estimate the population mean}\\ \sum(x_i-\bar x)^2&=16+4+0+4+16=40 && \text{squared deviations}\\ s^2&=\frac{40}{5-1}=10 && \text{estimate the population variance} \end{aligned}

Why this step? Dividing by $5$ describes the variance of the five-value list with equal weights; it is not the required sample-variance estimate.

Answer. Estimated mean $12$ , estimated variance $10$ .

Check. The summary formula gives $(760-60^2/5)/4=10$ too.

Change only the measurement scale: multiply every observation by $2$ . Means double, while squared deviations multiply by $4$ .

Worked example 2: changing the measurement scale

Question. Estimate the population mean and variance from the random sample $16,20,24,28,32$ .

Plan. Calculate the new summaries, retaining divisor $n-1$ .

\begin{aligned} \bar x&=\frac{120}{5}=24 && \text{new sample mean}\\ \sum x_i^2&=3040 && \text{sum of squared observations}\\ s^2&=\frac{3040-120^2/5}{4}=40 && \text{sample-variance formula} \end{aligned}

Why this step? A variance has squared measurement units, so multiplying all observations by $2$ multiplies the variance by $2^2$ .

Answer. Estimated mean $24$ , estimated variance $40$ .

Check. These are $2(12)$ and $4(10)$ , matching the scale change.

On many calculators, the sample standard deviation is labelled $s_x$ , while a different output divides by $n$ . Check the definition rather than relying on the position of the menu entry.

Watch out

Wrong: use divisor $n$ when asked to estimate a population variance from a sample.

Right: use $n-1$ . Also distinguish $s^2$ , a variance, from $s$ , a standard deviation.

Multiply the original measurements by a different scale factor.

A sample mean has its own distribution

Imagine repeatedly drawing $36$ independent measurements from a population of mean $50$ and standard deviation $12$ . Each sample has a different mean. Those means form a new distribution, centred at $50$ , with standard deviation $12/\sqrt{36}=2$ .

Sampling distribution and standard error

Before the observations are taken,

\overline X=\frac{X_1+\cdots+X_n}{n}

is a random variable. Its probability distribution is the sampling distribution of the mean. A realised value is written $\bar x$ .

The standard error of the mean is the standard deviation of this sampling distribution.

Independent observations make the mean less variable than a single observation. Expectation adds $n$ copies of $\mu$ ; variance adds $n$ copies of $\sigma^2$ , then dividing the total by $n$ divides its variance by $n^2$ .

Mean, variance and standard error of a sample mean

For $n$ independent observations from a population with mean $\mu$ and finite variance $\sigma^2$ ,

E(\overline X)=\mu,\qquad \operatorname{Var}(\overline X)=\frac{\sigma^2}{n},\qquad \operatorname{SE}(\overline X)=\frac{\sigma}{\sqrt n}

If $\sigma$ is unknown, estimate the standard error by $s/\sqrt n$ . These moment results do not require a Normal population.

Not in the formula booklet: learn it.

A standard deviation describes variation between individual observations. A standard error describes variation between their sample means. The similar-looking formula does not make them the same quantity.

Worked example 3: repeated samples of thirty-six

Question. A population has mean $50$ and variance $144$ . Independent random samples of size $36$ are taken. Find the mean, variance and standard error of their sample means.

Plan. Retain the population mean and divide the population variance by the sample size.

\begin{aligned} E(\overline X)&=50 && \text{sample means centre on }\mu\\ \operatorname{Var}(\overline X)&=\frac{144}{36}=4 && \text{independent mean variance}\\ \operatorname{SE}(\overline X)&=\sqrt4=2 && \text{standard deviation of the means} \end{aligned}

Why this step? The variance $144$ belongs to individual observations. It must be reduced before calculating the spread of the means.

Answer. Mean $50$ , variance $4$ , standard error $2$ .

Check. The standard error is $12/\sqrt{36}=2$ , smaller than the individual standard deviation $12$ .

Change only the sample size, multiplying it by $4$ . This halves the standard error, rather than dividing it by $4$ .

Worked example 4: four times as many observations

Question. Repeat the calculation for independent samples of size $144$ from the same population.

Plan. Replace $n$ in the variance and standard-error formulae.

\begin{aligned} E(\overline X)&=50 && \text{unchanged population mean}\\ \operatorname{Var}(\overline X)&=\frac{144}{144}=1 && \text{new sample size}\\ \operatorname{SE}(\overline X)&=1 && \text{square root of variance} \end{aligned}

Why this step? The square root in the denominator makes precision improve more slowly than sample size grows.

Answer. Mean $50$ , variance $1$ , standard error $1$ .

Check. $1$ is half the previous standard error $2$ .

If only one sample is available and the population variance is unknown, use its $s$ to estimate the standard error. This remains an estimate: a second sample can give a different $s$ .

Exam tip

When asked what standard error means, say “the standard deviation of the sampling distribution of the mean”. Giving only $s/\sqrt n$ explains how it is estimated, not what it is.

Now the spread is estimated from a sample rather than known for the population.

Watch out

Wrong: $\operatorname{SE}(\overline X)=12/36$ for samples of size $36$ . The variance divides by $n$ , but the standard deviation divides by $\sqrt n$ . Correct: $\operatorname{Var}(\overline X)=144/36=4$ and $\operatorname{SE}(\overline X)=12/6=2$ .

Identify whether a given spread is a variance or standard deviation before using it below.

Normal populations give exact Normal sample means

You already know

Independent Normal combinations are Normal. In particular, for independent $N(\mu,\sigma^2)$ observations,

\overline X\sim N\left(\mu,\frac{\sigma^2}{n}\right)

This holds for every positive sample size, as established in the preceding Normal-combinations lesson.

Here the new sampling-distribution language makes the interpretation clear. A small sample does not prevent an exact Normal mean when the population itself is Normal. You do not need the Central Limit Theorem to justify this case.

Worked example 5: an exact mean probability

Question. Individual measurements are independent with distribution $N(120,6^2)$ . Find the probability that the mean of $9$ measurements exceeds $123$ .

Plan. State the exact distribution of the mean, then calculate its upper tail.

\begin{aligned} \overline X&\sim N\left(120,\frac{36}{9}\right)=N(120,4) && \text{Normal population}\\ P(\overline X>123)&=P\left(Z>\frac{123-120}{2}\right) && \text{standard error }2\\ &=P(Z>1.5)=0.066807\ldots && \text{standard Normal tail} \end{aligned}

Why this step? Use the mean's standard error $2$ , not the individual standard deviation $6$ .

Answer. Probability $0.0668$ .

Check. The threshold is $1.5$ standard errors above the mean, so an upper tail below $0.1$ is sensible.

Change only the sample size from $9$ to $36$ , keeping the threshold fixed.

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SI1–SI6 - Sample means and the Central Limit Theorem

Estimate the population, not just the spread of the list

A sample mean has its own distribution

Normal populations give exact Normal sample means

Continue this lesson.

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