Hypothesis tests for a meanEdexcel International A Level Further Maths: Revision notes

A hypothesis test decides whether sample data give enough evidence against a claim about a population parameter. The claim is the null hypothesis $H_0$ (for a mean, $H_0:\mu=\mu_0$). The alternative hypothesis $H_1$ says what you are looking for: $\mu>\mu_0$ or $\mu<\mu_0$ (one-tailed), or $\mu\neq\mu_0$ (two-tailed). Hypotheses are always about the population parameter $\mu$, never the sample mean. The steps are:

1. State $H_0$ and $H_1$ and the significance level.
2. State the distribution of the statistic assuming $H_0$ is true.
3. Calculate the test statistic and compare it with the critical value (or find the $p$-value).
4. State whether $H_0$ is rejected, and give the conclusion in the context of the question. The significance level is the probability of rejecting $H_0$ when it is true.

If $X\sim N(\mu,\sigma^2)$ with $\sigma$ known, then under $H_0$, $\bar X\sim N\left(\mu_0,\frac{\sigma^2}{n}\right)$ and the test statistic is $Z=\frac{\bar X-\mu_0}{\sigma/\sqrt n}\sim N(0,1).$ Critical values of $z$:

• one-tailed, $5\%$: $1.645$ (upper) or $-1.645$ (lower); one-tailed, $1\%$: $2.326$;
• two-tailed, $5\%$: $\pm1.96$; two-tailed, $1\%$: $\pm2.576$. Worked example: lifetimes $N(\mu,80^2)$, $H_0:\mu=1200$, $H_1:\mu<1200$, $n=16$, $\bar x=1165$. Then $z=\frac{1165-1200}{80/4}=-1.75$. This is below $-1.645$, so reject $H_0$ at the $5\%$ level: there is evidence that the mean lifetime is below $1200$ hours. It is not below $-2.326$, so $H_0$ is not rejected at the $1\%$ level.

The critical region (rejection region) is the set of values of the test statistic for which $H_0$ is rejected. It can also be written for $\bar X$ itself: for a two-tailed $5\%$ test with $\mu_0=48$, $\sigma=6$, $n=100$, reject $H_0$ if $\bar X<48-1.96\times0.6=46.8$ or $\bar X>48+1.96\times0.6=49.2$. Alternatively find the $p$-value, the probability, assuming $H_0$ is true, of a result at least as extreme as the one observed, and reject $H_0$ if $p<$ the significance level. For a two-tailed test, double the tail probability: $z=2.25$ gives $p=2\times0.0122=0.0244<0.05$. Always finish in context: 'there is evidence that the mean volume has changed from $500$ ml' or 'there is insufficient evidence that the mean waiting time is longer'. Never say that $H_0$ has been proved true.

The Central Limit Theorem states that if $X$ has mean $\mu$ and variance $\sigma^2$, then for a large sample size $n$ (roughly $n\geq30$) the sample mean $\bar X$ is approximately $N\left(\mu,\frac{\sigma^2}{n}\right)$, whatever the shape of the distribution of $X$. So the same test statistic can be used for populations that are not Normal, provided $n$ is large, and the same applies to confidence intervals. The result is approximate. Example: waiting times with a skewed distribution and $\sigma=3.5$; $H_0:\mu=5.8$, $H_1:\mu>5.8$; $n=49$, $\bar x=6.4$. Under $H_0$, $\bar X\approx N(5.8,0.25)$ and $z=\frac{0.6}{0.5}=1.2<1.645$, so there is insufficient evidence at the $5\%$ level that the mean wait is longer.

When $\sigma^2$ is unknown, use the unbiased estimate $s^2=\frac{1}{n-1}\left(\sum x^2-n\bar x^2\right)$. If $n$ is large, $\frac{\bar X-\mu}{S/\sqrt n}$ can be treated as $N(0,1)$, so the test is the same as before with $s$ in place of $\sigma$: $z=\frac{\bar x-\mu_0}{s/\sqrt n}.$ Example: $n=100$, $\sum x=4930$, $\sum x^2=246\,613$, $H_0:\mu=48$, $H_1:\mu\neq48$. Then $\bar x=49.3$, $s^2=\frac{246\,613-100(49.3)^2}{99}=36$, $z=\frac{1.3}{6/10}=2.17>1.96$, so reject $H_0$ at the $5\%$ level: there is evidence that the mean differs from $48$. For a small sample with unknown variance this approximation is not reliable (a different distribution would be needed, and that is not required here), so check that $n$ is large and say so.