O1, O3StatisticsStretch

Normal and Correlation Tests

Q: Why divide by n rather than n?

Because variances divide by n, and the standard deviation is the square root of the variance. So ^2n becomes n on taking the root.

Two more tests complete the topic. One asks whether a $\textbf{sample mean}$ is consistent with a claimed population mean; the other asks whether an observed correlation is strong enough to be evidence of a real relationship. The logic is identical to the binomial test — only the distribution changes.

40 min Hypothesis Testing

The big picture

The genuinely new idea here is that a sample mean has its own distribution, and a narrower one than the population it came from. Averaging cancels variation: individual values scatter widely, but their mean does not, and the standard deviation of $\bar{X}$ is $\tfrac{\sigma}{\sqrt{n}}$ . That $\sqrt{n}$ is why bigger samples give more reliable estimates, and it is the quantitative version of an intuition you already have. Everything else in this lesson is the same test structure applied to a new statistic.

What you'll be able to do

State the distribution of the sample mean and explain the $\sqrt{n}$
Carry out a hypothesis test for a population mean with known variance
Carry out a test for zero correlation using critical values of $r$
Interpret conclusions in context, respecting the limits of each test

The distribution of the sample mean

If individual values follow $N\left(\mu, \sigma^{2}\right)$ , then the mean of a sample of $n$ of them follows $N\left(\mu, \tfrac{\sigma^{2}}{n}\right)$ . The centre is unchanged, but the variance is divided by $n$ .

The reason is that high and low values partially cancel when averaged. A single observation can be extreme; for the mean of 25 observations to be equally extreme, most of them would have to be extreme in the same direction, which is far less likely.

Taking the square root, the standard deviation of $\bar{X}$ is $\tfrac{\sigma}{\sqrt{n}}$ — the $\textbf{standard error}$ . Quadrupling the sample size halves it, which is the practical statement of diminishing returns in sampling.

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

Same centre, variance divided by

n

\text{s.e.} = \dfrac{\sigma}{\sqrt{n}}

The standard error of the mean.

Why this works

The $\sqrt{n}$ rather than $n$ is why precision is expensive. Going from a sample of 100 to one of 400 only halves the standard error — four times the data for twice the precision.

Testing a population mean

The structure is unchanged from the binomial test. State hypotheses about $\mu$ , assume $H_{0}$ , work out how extreme the observed sample mean is, and compare with the significance level.

The test statistic standardises the sample mean using the standard error rather than $\sigma$ itself — that is the only substantive difference from a routine normal calculation.

This test assumes a normal population with known, given or assumed population variance and a random independent sample. The AQA specification describes a test for the mean of a normal population; a large sample is not a substitute for stating the model being used.

z = \dfrac{\bar{x} - \mu}{\sigma/\sqrt{n}}

Test statistic for a sample mean.

Example 1A machine should fill bottles to a mean of 500 ml with

\sigma = 8

ml. A sample of 16 bottles has mean 495 ml. Test at

5\%

whether the machine is underfilling.

1Let

\mu

be the population mean fill.

H_{0}: \mu = 500

;

H_{1}: \mu < 500

2Under

H_{0}

\bar{X} \sim N\left(500, \tfrac{64}{16}\right) = N(500, 4)

, so the standard error is 2.

z = \dfrac{495 - 500}{2} = -2.5

4One-tailed critical value at

5\%

-1.645

, and

-2.5 < -1.645

AnswerReject

H_{0}

. There is sufficient evidence at the

5\%

level to suggest the machine is underfilling.

Tip — Divide by $\tfrac{\sigma}{\sqrt{n}}$ , not by $\sigma$ . Forgetting the $\sqrt{n}$ shrinks the test statistic and is the single most common error here.

Testing for correlation

The correlation test asks whether an observed sample correlation $r$ is large enough to indicate a genuine relationship in the population, or whether it could plausibly have arisen from chance in a sample of that size.

The hypotheses concern the $\textbf{population}$ correlation coefficient $\rho$ : $H_{0}: \rho = 0$ means no linear relationship. The alternative is $\rho \neq 0$ , $\rho > 0$ or $\rho < 0$ depending on the wording.

Read the table using the sample size and the stated one- or two-tailed level. For a positive alternative reject only for sufficiently positive $r$ ; for a negative alternative reject only for sufficiently negative $r$ . A magnitude comparison $|r|>c$ applies to a two-tailed test. Read the table heading before splitting a significance level between tails.

Sample size matters enormously. With $n = 5$ an $r$ of $0.8$ is not significant, because small samples produce large correlations by chance routinely; with $n = 30$ a much smaller $r$ is convincing.

H_{0}: \rho = 0

No linear correlation in the population.

\left|r\right| > r_{\text{crit}} \Rightarrow \text{reject } H_{0}

Two-tailed decision rule; a one-tailed rule must also match the predicted sign.

Example 1A sample of 20 pairs gives

r = 0.52

. Test at

5\%

for positive correlation, given the one-tailed critical value

0.3783

H_{0}: \rho = 0

;

H_{1}: \rho > 0

2Compare:

0.52 > 0.3783

AnswerReject

H_{0}

. There is sufficient evidence at the

5\%

level to suggest positive correlation in the population.

Why this works

Rejecting $H_{0}$ gives evidence of population linear correlation under the model, rather than certainty that a relationship exists. It does not establish causation. The usual Pearson test assumes a bivariate normal population and random independent pairs.

What each test does and does not show

Both tests are about the $\textbf{population}$ , using the sample as evidence. A significant result says the observed data would be unlikely if the null were true — nothing stronger.

The mean test assumes the population variance is known, which is a substantial assumption; in practice it is usually estimated, which leads to methods beyond this specification.

The correlation test detects only $\textbf{linear}$ relationships, so a non-significant $r$ does not rule out a strong curved relationship. And with a large sample, a correlation too small to be practically interesting can still be statistically significant — significance and importance are different questions.

Example 1A study of 1000 pairs finds

r = 0.08

, significant at

5\%

. Comment on its practical importance.

1With

n = 1000

the critical value is very small, so significance is easy to attain.

2An

r

0.08

describes an extremely weak relationship.

3Statistical significance concerns whether an effect exists, not whether it matters.

AnswerThe result gives statistical evidence of a weak population linear association under the model. The small coefficient suggests limited predictive value, and the test does not establish causation or guarantee practical importance.

Tip — When a question gives a large sample and a small $r$ , it is usually inviting exactly this comment. Say that the result is significant but the correlation is too weak to be useful.

Think like an examiner

Divide by the standard error $\tfrac{\sigma}{\sqrt{n}}$ , and show that step separately.
For correlation, state hypotheses about $\rho$ , the population parameter, not about $r$ .
Quote the critical value you used and show the comparison.
Interpret in context and never claim causation from a significant correlation.

Common misconceptions

Standardising a sample mean with

\sigma

instead of

\tfrac{\sigma}{\sqrt{n}}

The sample mean has a smaller spread. Dividing by

\sigma

understates the test statistic.

Writing

H_{0}: r = 0

Hypotheses concern the population coefficient

\rho

;

r

is the sample evidence.

Using a two-tailed critical value for a one-tailed correlation test.

The tables list both. Choose the column matching the tail from the wording.

Concluding a significant correlation is practically important.

With a large

n

, tiny correlations reach significance. Comment on the size as well.

Mean and correlation tests

\bar{X} \sim N(\mu, \tfrac{\sigma^{2}}{n})

Distribution of the sample mean.

\text{s.e.} = \tfrac{\sigma}{\sqrt{n}}

Standard error.

z = \tfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}

Test statistic.

H_{0}: \rho = 0

No population correlation.

|r| > r_{\text{crit}}

Reject condition.

Stretch yourself

A supplier claims its cables have a mean breaking strength of 1000 N with standard deviation 40 N. A sample of 25 cables has mean 985 N. Test at $1\%$ whether the mean is below the claim. Then state how large a sample would be needed for a 15 N shortfall to be significant at the same level.

Hint — For the second part, set the test statistic equal to the critical value and solve for $n$ .

Reveal solution

SolutionLet

\mu

be the population mean breaking strength.

H_{0}: \mu = 1000

against

H_{1}: \mu < 1000

. Under

H_{0}

, the standard error is

\tfrac{40}{\sqrt{25}} = 8

, so

z = \tfrac{985-1000}{8} = -1.875

. The one-tailed

1\%

critical value is

-2.326

. Since

-1.875 > -2.326

, do not reject

H_{0}

— there is insufficient evidence at the

1\%

level that the mean is below 1000 N. For the second part, we need

\tfrac{-15}{40/\sqrt{n}} \leq -2.326

, i.e.

\tfrac{15\sqrt{n}}{40} \geq 2.326

, giving

\sqrt{n} \geq 6.203

and

n \geq 38.5

. So a sample of 39 would suffice. The same 15 N shortfall is unconvincing at

n=25

and convincing at

n=39

— a direct illustration that significance depends on sample size as much as on effect size.

\blacksquare

Questions students ask

Why divide by

\sqrt{n}

rather than

n

Because variances divide by

n

, and the standard deviation is the square root of the variance. So

\tfrac{\sigma^{2}}{n}

becomes

\tfrac{\sigma}{\sqrt{n}}

on taking the root.

What if the population is not normal?

For a large sample the mean is approximately normal anyway, because averaging many independent values tends towards normality. For small samples from a non-normal population the test is unreliable.

Does a significant correlation prove causation?

No. It provides evidence that a linear relationship exists in the population. The cause could run either way, or a third variable could drive both.

Why does the critical value of

r

depend on sample size?

Small samples produce large correlations by chance quite easily, so a high bar is needed. As

n

grows, chance correlations shrink and a smaller

r

becomes convincing.

Key takeaways

The sample mean follows $N\left(\mu, \tfrac{\sigma^{2}}{n}\right)$ , with standard error $\tfrac{\sigma}{\sqrt{n}}$ .
Test a mean with $z = \tfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ and compare against the critical value.
Correlation tests use $H_{0}: \rho = 0$ and tabulated critical values of $r$ .
Critical values of $r$ fall as sample size rises.
Significance is not importance, and a significant correlation still does not show causation.

How this fits the course

Build on

Correlation and Regression

Leads to

Mechanicssoon

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

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