Sampling variation and distributions of sample means | Year 12

01 / LEARN

Understand the idea.

Imagine repeating the sampling process

A sampling distribution describes how a statistic varies across repeated samples obtained in the same way. It differs from the distribution of individuals in one sample. The population can remain unchanged while different samples give different sample means.

For independent observations from the same population with finite variance, the mean of $\bar{X}$ is $μ$ and its standard deviation is $σ / \sqrt{n}$ . This standard deviation is called the standard error; distinguish it from the population standard deviation $σ$ .

\begin{aligned} E (\bar{X}) & = μ \\ SD (\bar{X}) & = \frac{σ}{\sqrt{n}} \end{aligned}

Know why normality is being used

A normal population gives an exactly normal sample mean at every sample size. For independent random samples from a non-normal population with finite variance, the central limit theorem supports a normal approximation as sample size increases. Strong skew can require a larger sample.

A larger sample reduces random sampling variation, but cannot remove systematic selection bias. Before finding a sampling probability, justify the sampling assumptions, use the standard error rather than the individual spread, and state whether the normal calculation is exact or approximate.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A probability for a sample mean

A population has mean 100 and SD 20. For 25 independent observations from a normal population, find P(mean>108).

Worked solution

1
Calculate the standard error.
$SE (\bar{X}) = \frac{20}{\sqrt{25}} = 4$
Independent averaging reduces the spread of the mean relative to individual observations.
2
State the exact distribution and standardise the boundary.
$\bar{X} \sim N (100, 16), z = \frac{108 - 100}{4} = 2$
The population is normal, so the sample mean is exactly normal even for this finite sample.
3
Evaluate the upper tail.
$P (\bar{X} > 108) = 1 - Φ (2) \approx 0.0228$
The event lies above, rather than below, the standardised boundary.

Answer

0.0228.

EXAMPLE 02

What a bigger sample changes

A population has SD 18. Compare the standard errors for independent random samples of 36 and 144. Explain what changes.

Worked solution

1
Calculate the standard error for thirty-six observations.
${SE}_{36} = \frac{18}{\sqrt{36}} = \frac{18}{6} = 3$
The standard error uses the square root of the number of independent observations.
2
Calculate it for one hundred and forty-four observations.
${SE}_{144} = \frac{18}{\sqrt{144}} = \frac{18}{12} = 1.5$
The larger sample averages away more random variation.
3
Compare the changes in size and spread.
$\frac{144}{36} = 4, \frac{1.5}{3} = \frac{1}{2}$
Quadrupling sample size halves the standard error; the population standard deviation remains eighteen.

Answer

3 and 1.5; the sample means become less variable.

03 / YOUR TURN

Try it yourself.

Check your understanding

What sample size is needed to halve the standard error, all else unchanged?

Show answer & reasoning

Four times the sample size.

Since standard error is $σ / \sqrt{n}$ , replacing $n$ by $4 n$ gives $σ / \sqrt{4 n} = \frac{1}{2} σ / \sqrt{n}$ . The sample size must be quadrupled.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

A population has SD 15. Find the standard error for $n = 25$ independent observations.
2 marks
A population has mean 42. What is the mean of the sampling distribution of a random sample mean?
2 marks
Is the mean of 4 normal observations exactly normal or only approximately normal?
2 marks
Why may a sample of 30 be insufficient for a very skewed population?
2 marks
A sample mean has SE 2 at $n = 100$ . What is its SE at $n = 25$ with the same population SD?
2 marks
A website surveys only volunteers. Would increasing its sample from 500 to 5000 remove self-selection bias?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

A population has SD 15. Find the standard error for $n = 25$ independent observations.
2 marks
$\frac{15}{\sqrt{25}} = 3$ .
3.
A population has mean 42. What is the mean of the sampling distribution of a random sample mean?
2 marks
Under independent identically distributed sampling, the sample mean is unbiased for the population mean.
42.
Is the mean of 4 normal observations exactly normal or only approximately normal?
2 marks
Independent normal linear combinations are normal.
Exactly normal, assuming independence.
Why may a sample of 30 be insufficient for a very skewed population?
2 marks
Convergence depends on shape and tails, not only a fixed $n$ threshold.
The approximation can still be poor.
A sample mean has SE 2 at $n = 100$ . What is its SE at $n = 25$ with the same population SD?
2 marks
Reducing $n$ by a factor of four doubles SE.
4.
A website surveys only volunteers. Would increasing its sample from 500 to 5000 remove self-selection bias?
2 marks
The same selection mechanism can remain biased even with a much larger sample.
No.

Checking saved progress…

Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: continuous random variables, normal distributions and sampling distributions of sample means. Original lessons and practice.

Official curriculum source ↗

In this lesson

Understand the idea.

Imagine repeating the sampling process

Know why normality is being used

See the method in action.

A probability for a sample mean

Calculate the standard error.

State the exact distribution and standardise the boundary.

Evaluate the upper tail.

What a bigger sample changes

Calculate the standard error for thirty-six observations.

Calculate it for one hundred and forty-four observations.

Compare the changes in size and spread.

Try it yourself.

Check your understanding

Keep practising