Practise The language of hypothesis testing. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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State, with a reason, the conclusion the student should draw about the null hypothesis.
Reject \(H_0\), because \(0.03 \lt 0.05\).
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
A student carries out a hypothesis test at the 5% level of significance and obtains a \(p\)-value of 0.03.
(a) State, with a reason, the conclusion the student should draw about the null hypothesis.[1]
(b) The student writes: “This means there is a 3% chance that the null hypothesis is true.” Explain why this interpretation is not correct.[1]
(c) Explain what is meant by the critical region of a hypothesis test.[1]
(d) Explain why the significance level should be chosen before the data are collected.[1]
Show the answer and mark scheme
(a)Answer: Reject \(H_0\), because \(0.03 \lt 0.05\).
B1 for reject \(H_0\) since \(0.03 \lt 0.05\) (the result is significant)
Worked solution: The \(p\)-value 0.03 is less than the significance level 0.05, so the result is significant and \(H_0\) is rejected.
(b)Answer: The \(p\)-value is the probability, assuming \(H_0\) is true, of a result at least as extreme as the one observed; it is not the probability that \(H_0\) is true.
B1 for the \(p\)-value is calculated assuming \(H_0\) is true: it is the probability of a result as extreme as (or more extreme than) the one observed, not the probability that \(H_0\) is true
Worked solution: The \(p\)-value is worked out by assuming \(H_0\) is true. It is the probability of getting a result at least as extreme as the observed one if \(H_0\) is true, not the probability that \(H_0\) itself is true.
(c)Answer: The set of values of the test statistic for which the null hypothesis would be rejected.
B1 for the range (set) of values of the test statistic that would lead to rejecting \(H_0\)
Worked solution: The critical region is the set of values of the test statistic that are so unlikely under \(H_0\) (at the chosen significance level) that, if the observed value falls in it, \(H_0\) is rejected.
(d)Answer: So that the level cannot be chosen, after seeing the data, to give the result the tester wants.
B1 for to avoid bias: choosing the level after seeing the result could make any desired conclusion possible
Worked solution: If the significance level were chosen after seeing the data, it could be picked to give whichever conclusion the tester preferred. Fixing it in advance keeps the test fair.
Question 2Medium3 marks
The length of a bolt made by a machine is modelled by a normal distribution with mean \(\mu\) mm. It is claimed that \(\mu = 40\). After the machine is serviced, an engineer wants to check whether the mean length has changed. A random sample is to be used to test this.
(a) Write down suitable hypotheses.[2]
(b) Explain why the hypotheses are written in terms of \(\mu\) rather than the sample mean.[1]
Show the answer and mark scheme
(a)Answer: \(H_0: \mu = 40, \ H_1: \mu \ne 40\)
B1 for \(H_0: \mu = 40\)
B1 for \(H_1: \mu \ne 40\)
(b)Answer: The test uses the sample to make an inference about the population: \(\mu\) is the population mean, whereas the sample mean is known exactly once the sample is taken.
B1 for the hypotheses being about the population parameter, with the sample used to make an inference about it
Question 3Hard4 marks
A seed company states that 75% of its seeds germinate. A gardener wants to find out whether seeds from a new batch have a different germination rate. A random sample of 30 seeds is taken, and \(X\) is the number that germinate. A test of \(H_0: p = 0.75, \ H_1: p \ne 0.75\) at the 5% level has critical region \(X \le 17 \text{ or } X \ge 28\).
(a) Find the probability of incorrectly rejecting \(H_0\) using this test.[3]
(b) Explain why this probability is less than 0.05.[1]
Show the answer and mark scheme
(a)Answer: \(0.0216 + 0.0106 = 0.0322\)
M1 for \(P(X \le 17)\) or \(P(X \ge 28)\)
M1 for adding the two tail probabilities
A1 for awrt 0.0322
(b)Answer: \(X\) is discrete, so each tail can only take certain probabilities; each tail is chosen to be as large as possible without exceeding 0.025, so the total is less than 0.05.
B1 for the discreteness of the binomial distribution, with each tail not exceeding 0.025