A hypothesis test asks one question: is this sample surprising enough to doubt a claim about the population? In Year 1 you tested a proportion $p$ using the binomial distribution. Year 2 (DfE content area O · Statistical hypothesis testing) applies exactly the same logic to two new situations.
Normal mean: $\mathrm{H}_0: \mu = \mu_0$ — test statistic the sample mean $\bar{x}$, with $\bar{X} \sim \mathrm{N}\!\left(\mu_0,\ \dfrac{\sigma^2}{n}\right)$
This is a recap of the Year 1 vocabulary, which is examined again in Year 2. For the full treatment see The Language of Hypothesis Testing.
| Term | Meaning |
|---|---|
| Null hypothesis $\mathrm{H}_0$ | The default claim, assumed true for the test; always states a single value of the parameter. |
| Alternative hypothesis $\mathrm{H}_1$ | What we look for evidence of: $>$, $<$ (one-tailed) or $\neq$ (two-tailed). |
| Test statistic | The value calculated from the sample: $r$ or $\bar{x}$ in Year 2. |
| Significance level | The probability threshold for rejecting $\mathrm{H}_0$, e.g. $5\%$. |
| Critical value | The boundary of the critical region. |
| Critical region | The set of values of the test statistic that lead to rejecting $\mathrm{H}_0$, written as inequalities, e.g. $r \geqslant 0.5494$ or $\bar{X} \leqslant 115.35$. |
| $p$-value | The probability, assuming $\mathrm{H}_0$, of a result at least as extreme as the one observed. |
| Population parameter | $\rho$ (population correlation) or $\mu$ (population mean). Hypotheses are about these, never about $r$ or $\bar{x}$. |
| Correlation test | Normal mean test | |
|---|---|---|
| Parameter | $\rho$ | $\mu$ |
| $\mathrm{H}_0$ | $\rho = 0$ | $\mu = \mu_0$ |
| Test statistic | $r$ | $\bar{x}$, or $z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$ |
| Critical value from | Table of critical values of $r$ for sample size $n$ | Inverse normal |
| Assumption | Random sample from a bivariate normal population | Normal population with known, unchanged $\sigma$; random sample |
For $14$ randomly chosen days, the correlation coefficient between daily rainfall and the number of umbrellas sold is $r = 0.58$. Test at the $2.5\%$ level whether there is positive correlation.
$X \sim \mathrm{N}(\mu,\ 2^2)$. A sample of $16$ has mean $51.1$. Test $\mathrm{H}_0: \mu = 50$ against $\mathrm{H}_1: \mu > 50$ at the $1\%$ level.
The two Year 2 tests, and the page that covers each. Both belong to DfE content area O.
- OTesting for Correlation$\mathrm{H}_0: \rho = 0$ with one- and two-tailed alternatives, critical values of $r$, and $p$-values.
- OTesting a Normal Mean$\bar{X} \sim \mathrm{N}(\mu,\ \sigma^2/n)$; one- and two-tailed tests, critical regions and $p$-values.
For the underlying distributions, see Regression and Exponential Models and The Normal Distribution. The Year 1 binomial test is in Binomial Hypothesis Tests.
Correlation $\mathrm{H}_0$
$\rho = 0$; $\mathrm{H}_1$: $\rho > 0$, $\rho < 0$ or $\rho \neq 0$.
Correlation decision
Reject if $|r| \geqslant$ critical value for $n$ (in the right direction).
Sample mean
$\bar{X} \sim \mathrm{N}(\mu,\ \sigma^2/n)$
Mean test statistic
$z = \dfrac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$
Critical $z$ (5%)
One tail $1.6449$; two tails $\pm 1.96$.
Critical $z$ (1%)
One tail $2.3263$; two tails $\pm 2.5758$.
$p$-value
Reject $\mathrm{H}_0$ if $p <$ significance level (per tail).
Critical region
Write as an inequality, e.g. $\bar{X} \geqslant 25.33$.
Conclusion
In context, with the level, non-assertive.
Ten questions drawing on the whole Year 2 Hypothesis Testing section. Critical values of $r$ are on the Testing for Correlation page.
Explain what is meant by (a) the significance level of a test, (b) the critical region.
▶ Show solution
(a) The probability threshold below which a result is considered too unlikely under $\mathrm{H}_0$ — so it is the probability of rejecting $\mathrm{H}_0$ when it is in fact true.
(b) The set of values of the test statistic for which $\mathrm{H}_0$ is rejected.
Packets of rice should have mean mass $500$ g. A consumer group believes the mean has changed. Write suitable hypotheses.
▶ Show solution
$\mathrm{H}_0: \mu = 500$, $\mathrm{H}_1: \mu \neq 500$, where $\mu$ is the mean mass in grams of all packets. Two-tailed, since "changed" has no direction.
For $12$ countries, $r = -0.52$ between average income and infant mortality. Test at the $5\%$ level for negative correlation.
▶ Show solution
$\mathrm{H}_0: \rho = 0$, $\mathrm{H}_1: \rho < 0$. Critical region ($n = 12$, $5\%$ one tail): $r \leqslant -0.4973$.
$-0.52 \leqslant -0.4973$: reject $\mathrm{H}_0$. Sufficient evidence of negative correlation between income and infant mortality.
For $16$ pairs, $r = 0.43$. Test at the $5\%$ level whether there is any correlation.
▶ Show solution
$\mathrm{H}_0: \rho = 0$, $\mathrm{H}_1: \rho \neq 0$. Two tails at $5\%$, so use the $2.5\%$ column: $n = 16$ gives $0.4973$.
$0.43 < 0.4973$: do not reject $\mathrm{H}_0$. Insufficient evidence of correlation.
$X \sim \mathrm{N}(\mu,\ 3^2)$. A sample of $25$ has mean $101.1$. Test $\mathrm{H}_0: \mu = 100$ against $\mathrm{H}_1: \mu > 100$ at the $5\%$ level.
▶ Show solution
$\bar{X} \sim \mathrm{N}(100,\ 0.6^2)$ under $\mathrm{H}_0$.
$\mathrm{P}(\bar{X} \geqslant 101.1) = 1 - \Phi(1.833) = 0.0334 < 0.05$
Reject $\mathrm{H}_0$: sufficient evidence at the $5\%$ level that $\mu > 100$.
$X \sim \mathrm{N}(\mu,\ 4^2)$. Find the critical region for a test of $\mathrm{H}_0: \mu = 64$ against $\mathrm{H}_1: \mu \neq 64$ at the $1\%$ level using a sample of $16$.
▶ Show solution
$\sigma/\sqrt{n} = 1$; $0.5\%$ in each tail, $z = \pm 2.5758$.
Critical region: $\bar{X} \leqslant 61.42$ or $\bar{X} \geqslant 66.58$.
A one-tailed test gives a $p$-value of $0.07$. State the conclusion at the $5\%$ and $10\%$ levels.
▶ Show solution
$5\%$: $0.07 > 0.05$, do not reject $\mathrm{H}_0$.
$10\%$: $0.07 < 0.10$, reject $\mathrm{H}_0$.
The evidence is moderate: convincing at a lenient level, not at a stricter one.
A student writes "$\mathrm{H}_0: r = 0$". Explain the error.
▶ Show solution
$r$ is the sample correlation coefficient — it is known exactly once the data are collected, so there is nothing to test about it. Hypotheses are about the unknown population correlation coefficient: $\mathrm{H}_0: \rho = 0$.
Which test (correlation, normal mean, or the Year 1 binomial test) suits each claim? (a) "More than $20\%$ of customers return their purchase." (b) "Taller people tend to have longer arm spans." (c) "The new fertiliser increases the mean yield per plant."
▶ Show solution
(a) Binomial test of a proportion: $\mathrm{H}_0: p = 0.2$, $\mathrm{H}_1: p > 0.2$.
(b) Correlation test: $\mathrm{H}_0: \rho = 0$, $\mathrm{H}_1: \rho > 0$.
(c) Normal mean test (yields continuous, assumed normal with known $\sigma$): $\mathrm{H}_1: \mu > \mu_0$.
A farmer records the amount of fertiliser used and the yield for $10$ fields, obtaining $r = 0.68$.
(a) Test at the $1\%$ level whether there is positive correlation. (b) Would the conclusion change at the $5\%$ level? (c) Historically, yields per field are $\mathrm{N}(6.2,\ 0.9^2)$ tonnes. After a change of seed, $15$ fields have mean yield $6.65$ tonnes. Assuming the standard deviation is unchanged, test at the $5\%$ level whether the mean yield has increased. (d) Find the critical region for part (c).
▶ Show solution
(a) $\mathrm{H}_0: \rho = 0$, $\mathrm{H}_1: \rho > 0$. Critical value ($n = 10$, $1\%$): $0.7155$. $0.68 < 0.7155$: do not reject $\mathrm{H}_0$; insufficient evidence at $1\%$.
(b) At $5\%$ the critical value is $0.5494$; $0.68 \geqslant 0.5494$, so $\mathrm{H}_0$ would be rejected — yes, the conclusion changes.
(c) $\mathrm{H}_0: \mu = 6.2$, $\mathrm{H}_1: \mu > 6.2$. $\sigma/\sqrt{n} = 0.9/\sqrt{15} = 0.2324$.
$z = \dfrac{0.45}{0.2324} = 1.936$; $p = 1 - \Phi(1.936) = 0.0264 < 0.05$. Reject $\mathrm{H}_0$: sufficient evidence at the $5\%$ level that the mean yield has increased.
(d) $6.2 + 1.6449 \times 0.2324 = 6.58$, so the critical region is $\bar{X} \geqslant 6.58$ (2 d.p.). Indeed $6.65 \geqslant 6.58$ ✓