S5 Statistical hypothesis testing — printable coverage pack

S5.1 · Apply the language of hypothesis testing via a binomial model: null/alternative hypothesis, significance level, test statistic, 1- and 2-tail tests, critical value/region, acceptance region, p-value; extend to correlation coefficients.

The null hypothesis $H_0$ gives the reference parameter value; the alternative $H_1$ states the direction or difference supported by the claim being tested.
Use a one-tailed test for a pre-specified directional alternative and a two-tailed test for any change; choose this before observing the data.
The critical region contains outcomes sufficiently unlikely under $H_0$ ; the acceptance region is its complement, and the p-value is the probability under $H_0$ of an outcome at least as extreme as observed.
For a correlation test, use $H_0:\rho=0$ and compare the sample product-moment correlation coefficient with the supplied critical value; significance does not establish causation.

Tier 1 · Easy

1. A company claims that the probability $p$ of a customer choosing its premium plan has increased from $0.40$ . State suitable hypotheses and identify the number of tails.[2 marks]
Answer
- $H_0:p=0.40$ .
- $H_1:p>0.40$ .
- This is a one-tailed test.
Method: The null uses the established value $0.40$ . The word 'increased' gives the directional alternative $p>0.40$ , so only the upper tail is relevant.

Tier 2 · Standard

1. Under $H_0$ , $X\sim\operatorname{B}(20,0.2)$ . For an upper-tailed test at the $5\%$ level, $P(X\geq7)=0.0867$ and $P(X\geq8)=0.0321$ . State the critical region, the critical value and the acceptance region.[4 marks]
Answer
- Critical region: $X\geq8$ .
- Critical value: $8$ .
- Acceptance region: $X\leq7$ .
Method: Choose the smallest upper-tail boundary whose probability under $H_0$ does not exceed $0.05$ . The boundary $7$ is too liberal because $0.0867>0.05$ , while $P(X\geq8)=0.0321<0.05$ . Hence the critical region starts at $8$ and its complement is $X\leq7$ .

Tier 3 · Hard

1. For a sample of $18$ paired observations, a two-tailed $5\%$ correlation test has critical values $-0.468$ and $0.468$ . The sample product-moment correlation coefficient is $r=-0.520$ , with p-value $0.026$ . State the hypotheses, carry out the test and interpret the p-value without claiming causation.[6 marks]
Answer
- $H_0:\rho=0$ and $H_1:\rho\neq0$ .
- Reject $H_0$ because $-0.520<-0.468$ (equivalently $0.026<0.05$ ).
- There is sufficient evidence of negative correlation in the population.
- If $\rho=0$ , the probability of a sample correlation at least this extreme in either direction is $0.026$ ; this does not prove causation.
Method: A two-tailed association test uses $H_0:\rho=0$ against $H_1:\rho\neq0$ . The observed coefficient lies in the lower critical region, and its p-value is below $0.05$ , so reject $H_0$ . State the conclusion as evidence of population correlation, not as proof that either variable causes the other.

S5.2 · Conduct a hypothesis test for the proportion in the binomial distribution and interpret results in context; a sample makes an inference about the population; the significance level is the probability of incorrectly rejecting H0.

Model the number of sample successes by $X\sim\operatorname{B}(n,p_0)$ under $H_0:p=p_0$ , provided the binomial assumptions are defensible.
Calculate the probability, under $H_0$ , of the observed result or one more extreme in the direction specified by $H_1$ ; double an appropriate tail for a symmetric two-tailed binomial test.
Reject $H_0$ when the p-value is at most the significance level; otherwise say there is insufficient evidence to reject $H_0$ , not that $H_0$ has been proved.
The significance level is the probability of incorrectly rejecting $H_0$ when it is true; discreteness often makes the actual probability of the critical region smaller than the nominal level.

Tier 1 · Easy

1. A coin is tested with $H_0:p=0.5$ against $H_1:p>0.5$ . It lands heads $10$ times in $12$ tosses. Given $P(X\geq10)=0.0193$ for $X\sim\operatorname{B}(12,0.5)$ , conduct the test at the $5\%$ level.[4 marks]
Answer
- Reject $H_0$ because $0.0193<0.05$ .
- There is sufficient evidence that the probability of heads is greater than $0.5$ .
Method: Under $H_0$ , the number of heads is $X\sim\operatorname{B}(12,0.5)$ . The upper-tail p-value for $10$ observed heads is $0.0193$ . Since this is below $0.05$ , reject $H_0$ and give the directional conclusion in context.

Tier 2 · Standard

1. A process is tested using $H_0:p=0.5$ against $H_1:p\neq0.5$ . In $20$ independent trials there are $5$ successes. Given $P(X\leq5)=0.02069$ under $X\sim\operatorname{B}(20,0.5)$ , conduct a two-tailed test at the $5\%$ level.[5 marks]
Answer
- Two-tailed p-value $=2(0.02069)=0.04138$ .
- Reject $H_0$ .
- There is sufficient evidence that the population success probability differs from $0.5$ .
Method: The null distribution is symmetric because $p_0=0.5$ . Outcomes at least as extreme as $5$ lie in the two tails, so the p-value is $2P(X\leq5)=2(0.02069)=0.04138$ . This is less than $0.05$ , so reject $H_0$ and infer a difference in the population proportion.

Tier 3 · Hard

1. A one-tailed test uses $H_0:p=0.2$ against $H_1:p>0.2$ with a sample of $15$ . Under $H_0$ , $P(X\geq6)=0.0611$ and $P(X\geq7)=0.0181$ . Find the $5\%$ critical region. If $6$ successes are observed, state the conclusion and explain the actual probability of a Type I error.[6 marks]
Answer
- Critical region: $X\geq7$ .
- With $X=6$ , do not reject $H_0$ ; there is insufficient evidence that $p>0.2$ .
- The actual probability of rejecting a true $H_0$ is $P(X\geq7)=0.0181$ , which is below the nominal $5\%$ level.
Method: A critical region must have probability at most $0.05$ under $H_0$ . Since $P(X\geq6)=0.0611$ is too large but $P(X\geq7)=0.0181$ is acceptable, use $X\geq7$ . The observed value $6$ is outside this region, so there is insufficient evidence to reject $H_0$ . Because the binomial distribution is discrete, the attainable Type I error probability is $0.0181$ , not exactly $0.05$ .

S5.3 · Conduct a statistical hypothesis test for the mean of a Normal distribution with known, given or assumed variance and interpret the results in context.

For a Normal population with known or assumed standard deviation $\sigma$ , the sample mean satisfies $\overline{X}\sim\operatorname{N}(\mu,\sigma^2/n)$ .
Under $H_0:\mu=\mu_0$ , standardise the observed mean using $Z=\frac{\overline{x}-\mu_0}{\sigma/\sqrt n}$ .
Use the tail or tails specified by $H_1$ , compare the p-value with the significance level, and give the conclusion in the language of the population mean.
The method relies on a random, independent sample from a Normal population and uses a known, given or assumed variance rather than estimating it within this specified test.

Tier 1 · Easy

1. A Normal population has known standard deviation $12$ . Test $H_0:\mu=100$ against $H_1:\mu>100$ using a random sample of $36$ with mean $104$ , at the $5\%$ level.[5 marks]
Answer
- Test statistic $z=2.00$ and p-value $0.0228$ .
- Reject $H_0$ ; there is sufficient evidence that the population mean exceeds $100$ .
Method: Under $H_0$ , $\overline{X}\sim\operatorname{N}(100,12^2/36)$ , so the standard error is $12/6=2$ . Thus $z=(104-100)/2=2.00$ . The upper-tail p-value is $P(Z\geq2)=0.0228<0.05$ , so reject $H_0$ and state the conclusion about the population mean.

Tier 2 · Standard

1. A Normal population has known standard deviation $5$ . For a sample of $25$ , test $H_0:\mu=50$ against $H_1:\mu\neq50$ at the $1\%$ level. The critical standard Normal values are $\pm2.576$ . Find the critical values of the sample mean and decide what to conclude if $\overline{x}=47.3$ .[6 marks]
Answer
- Critical sample-mean values are approximately $47.424$ and $52.576$ .
- Reject $H_0$ because $47.3<47.424$ .
- There is sufficient evidence at the $1\%$ level that the population mean differs from $50$ .
Method: The standard error is $5/\sqrt{25}=1$ . The acceptance interval is $50\pm2.576(1)$ , namely $47.424\leq\overline{X}\leq52.576$ . Since $47.3$ lies below the lower boundary, it is in the critical region, so reject $H_0$ and conclude that the mean differs from $50$ .

Tier 3 · Hard

1. A Normal population has known standard deviation $6$ . To test $H_0:\mu=80$ against $H_1:\mu>80$ at the $5\%$ level, find the smallest sample size $n$ for which an observed mean of $82$ would lead to rejection. Use the critical value $1.645$ , then find the p-value for this minimum $n$ .[7 marks]
Answer
- Smallest sample size $n=25$ .
- For $n=25$ , p-value $=0.0478$ .
Method: Rejection requires $\frac{82-80}{6/\sqrt n}\geq1.645$ . Hence $\sqrt n\geq\frac{1.645(6)}{2}=4.935$ , so $n\geq24.354\ldots$ and the smallest integer is $25$ . For $n=25$ , $z=2/(6/5)=1.6667$ , giving the upper-tail p-value $1-\Phi(1.6667)=0.0478$ .