Hypothesis Testing

Hypothesis testing gives you a formal way to decide whether sample data provide enough evidence to cast doubt on a claim about a population. In H2 Maths you will almost always be testing a claim about a population mean $\mu$ when the population variance $\sigma^2$ is known, using the $z$-test based on the sample mean $\bar{X}$.

What you need to know

• Null hypothesis $H_0$: the default claim, always written as $\mu = \mu_0$ for some specific value $\mu_0$.
• Alternative hypothesis $H_1$: what you are trying to find evidence for. If $H_1 : \mu > \mu_0$ or $H_1 : \mu < \mu_0$ the test is one-tailed; if $H_1 : \mu \neq \mu_0$ it is two-tailed.
• Test statistic: under $H_0$ the standardised sample mean follows $Z \sim \mathrm{N}(0,1)$ (exactly when $X$ is normal, approximately by the Central Limit Theorem for large $n$):

• p-value: the probability, assuming $H_0$ is true, of obtaining a test statistic at least as extreme as the observed value. The GC computes this directly once you enter the parameters.
• Decision rule: reject $H_0$ if the p-value $< \alpha$, where $\alpha$ is the significance level (commonly 0.05 or 0.01). Otherwise, do not reject $H_0$.
• Contextual conclusion: always write the conclusion in terms of the original scenario — never just say “reject $H_0$”. State whether there is sufficient evidence, at the chosen significance level, to support the alternative claim.

Worked example 1 — One-tailed z-test (right-tailed)

A manufacturer claims that the mean lifetime of a type of battery is 50 hours. A consumer group suspects the batteries last fewer hours than claimed and tests a random sample of 40 batteries, obtaining a sample mean of 47.6 hours. The population standard deviation is known to be 8 hours. Test the manufacturer’s claim at the 5% level of significance.

Solution.

Hypotheses:

The alternative hypothesis is $\mu < 50$ because the consumer group suspects the mean is lower than claimed, giving a left-tailed test.

Test statistic:

p-value (from GC, left-tailed): $p = P(Z < -1.897) = 0.0289$.

Decision: Since $0.0289 < 0.05 = \alpha$, we reject $H_0$.

Conclusion: There is sufficient evidence at the 5% significance level to conclude that the mean battery lifetime is less than 50 hours. The manufacturer’s claim is not supported by the data.

Worked example 2 — Two-tailed z-test

Using the same battery context, suppose instead a quality-control engineer simply wishes to check whether the mean lifetime has changed from 50 hours. She takes a fresh random sample of 40 batteries with sample mean 52.3 hours ($\sigma = 8$ still known). Test at the 5% level.

Solution.

Hypotheses:

Because the engineer has no prior reason to expect an increase or a decrease, the test is two-tailed.

Test statistic:

p-value (from GC, two-tailed): $p = 2 \times P(Z > 1.818) = 2 \times 0.0345 = 0.0690$.

Decision: Since $0.0690 > 0.05 = \alpha$, we do not reject $H_0$.

Conclusion: There is insufficient evidence at the 5% significance level to conclude that the mean battery lifetime has changed from 50 hours.

One-tailed vs two-tailed: Note that the same observed $\bar{x} = 52.3$ would give a one-tailed p-value of 0.0345 (significant at 5%) if $H_1 : \mu > 50$ were the correct formulation. The choice of tail must be justified by the context before seeing the data, not chosen to make the result significant.

Common mistakes

• Choosing $H_1$ after seeing the data. The direction of $H_1$ must come from the question’s wording or prior knowledge, not from the sign of $\bar{x} - \mu_0$.
• Forgetting to double the p-value for a two-tailed test. If you read $P(Z > |z_\text{calc}|)$ from the GC, multiply by 2 before comparing with $\alpha$.
• Writing a conclusion that does not reference the context. Phrases like “reject $H_0$” alone score no marks for the conclusion; you must translate back into the scenario (e.g. “there is sufficient evidence that the mean lifetime has decreased”).
• Confusing significance level with probability that $H_0$ is true. Rejecting $H_0$ at the 5% level does not mean there is a 95% chance $H_1$ is true; $\alpha$ is the probability of a Type I error (rejecting a true $H_0$), not a posterior probability.