Hypothesis tests for the mean of a normal distribution - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 17 exam-style questions on Hypothesis tests for the mean of a normal distribution for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

4 marks

The journey time, \(T\) minutes, of a train between two stations is normally distributed with mean 38.5 minutes and standard deviation 2.4 minutes. After a new timetable is introduced, the train operator claims that the mean journey time has decreased.

A random sample of 15 journeys under the new timetable has a mean journey time of 37.4 minutes.

Assuming that the standard deviation is unchanged, test, at the 5% level of significance, whether there is evidence to support the operator's claim. State your hypotheses clearly.

Model answer

\(H_0: \mu = 38.5\), \(H_1: \mu < 38.5\)
Under \(H_0\), \(\bar{T} \sim N\left(38.5, \dfrac{2.4^2}{15}\right)\)
\(P(\bar{T} \leqslant 37.4) = 0.0379\) (3 s.f.)
\(0.0379 < 0.05\), so the result is significant and \(H_0\) is rejected.
There is sufficient evidence at the 5% level to support the operator's claim that the mean journey time has decreased.

Explanation

Under \(H_0\) the sample mean of 15 journeys has distribution \(N(38.5, 2.4^2/15)\), and the probability of a sample mean of 37.4 or less is 0.0379, which is below 5%. The test is one-tailed because the claim is a decrease. The common errors are forgetting to divide the variance by n, using the sample mean as the mean of the model, and stopping at "reject \(H_0\)" without a conclusion in context.

Mark scheme

1 mark
B1 Both hypotheses: \(H_0: \mu = 38.5\) and \(H_1: \mu < 38.5\). A conclusion with no working (no probability, test statistic or critical region) scores 0 marks.
An answer of "there is evidence that the mean journey time has decreased" with no working scores 0 marks: the probability (or test statistic or critical region) must be seen. Hypotheses must be in terms of \(\mu\) and attached to \(H_0\) and \(H_1\). \(H_0: T = 38.5\) or hypotheses in terms of \(\bar{x}\) score B0 (a common error in the examiners' reports).
1 mark
M1 Selects the model \(\bar{T} \sim N\left(38.5, \dfrac{2.4^2}{15}\right)\) (variance 0.384, standard deviation 0.6197...), seen or used
Must use mean 38.5 and divide the variance by 15 (or the standard deviation by \(\sqrt{15}\)). May be implied by a correct probability, test statistic or critical value. Using the sample mean as the mean, \(N(37.4, \ldots)\), is M0. Using \(N(38.5, 2.4^2)\) (not dividing by n) is M0.
1 mark
A1 \(P(\bar{T} \leqslant 37.4) = 0.0379\) (awrt 0.0379, i.e. in [0.03785, 0.03795))
Alternatives: test statistic \(z = -1.78\) (awrt −1.78, i.e. in [−1.785, −1.775)) with critical value −1.6449; or critical region \(\bar{T} < 37.48\) (awrt 37.48, i.e. in [37.475, 37.485)) with 37.4 in it. Condone \(P(\bar{T} \geqslant 37.4) = 0.962\) only if it is compared with 0.95.
1 mark
A1 0.0379 < 0.05, so reject \(H_0\): there is sufficient evidence at the 5% level to support the operator's claim that the mean journey time has decreased
Depends on the M1 and the first A1 (not on the hypotheses). Must be in context: mention the operator's claim, or the mean journey time decreasing ("the mean is less than 38.5" alone is not enough). Contradictory statements, e.g. "significant, so there is insufficient evidence", score A0.

Question 2

5 marks

The total mass of strawberries, \(Y\) grams, produced by a plant of a certain variety in a season is normally distributed with mean 640 g and standard deviation 85 g. A grower changes the feed she gives her plants, and she believes that the mean mass of strawberries per plant has changed.

A random sample of 24 plants given the new feed produces a mean of 676 g per plant.

Assuming that the standard deviation is unchanged, test, at the 5% level of significance, whether there is evidence to support the grower's belief. State your hypotheses clearly.

Find the p-value of the test.

Model answer

\(H_0: \mu = 640\), \(H_1: \mu ≠ 640\)
Under \(H_0\), \(\bar{Y} \sim N\left(640, \dfrac{85^2}{24}\right)\)
\(P(\bar{Y} \geqslant 676) = 0.0190\) (3 s.f.)
The test is two-tailed, so compare with 0.025: \(0.0190 < 0.025\), so the result is significant and \(H_0\) is rejected.
There is sufficient evidence at the 5% level to support the grower's belief that the mean mass of strawberries per plant has changed.
p-value \(= 2 \times 0.0190 = 0.0380\)

Explanation

A belief that the mean has changed gives a two-tailed test, so the probability in the tail beyond 676 is compared with 2.5%. The p-value of a two-tailed test is that tail probability doubled, 0.0380, which is also below 5%. Examiners report that few students found the p-value correctly: common answers were the single tail probability or the significance level itself.

Mark scheme

1 mark
B1 Both hypotheses: \(H_0: \mu = 640\) and \(H_1: \mu ≠ 640\). A conclusion with no working scores 0 marks.
An answer of "there is evidence that the mean has changed, p = 0.038" with no working scores 1 mark (the p-value B1 only); the model and probability are needed for the other marks. The grower believes the mean has changed, so the test is two-tailed: \(H_1: \mu > 640\) scores B0.
1 mark
M1 Selects the model \(\bar{Y} \sim N\left(640, \dfrac{85^2}{24}\right)\), seen or used
Must divide the variance by 24. May be implied by a correct probability. \(N(676, \ldots)\) is M0.
1 mark
A1 \(P(\bar{Y} \geqslant 676) = 0.0190\) (awrt 0.0190, i.e. in [0.01895, 0.01905))
Alternatives: \(z = 2.07\) (awrt 2.07, i.e. in [2.065, 2.075)) with critical value 1.96; or critical region \(\bar{Y} < 606.0\) or \(\bar{Y} > 674.0\) (awrt 674.0, i.e. in [673.95, 674.05)). Condone \(P(\bar{Y} \leqslant 676) = 0.981\) only if compared with 0.975.
1 mark
A1 0.0190 < 0.025 (two-tailed), so reject \(H_0\): there is sufficient evidence at the 5% level to support the grower's belief that the mean mass of strawberries per plant has changed
Depends on the M1 and the first A1 (not on the hypotheses). The comparison must be with 0.025 (or 0.038 with 0.05); comparing 0.0190 with 0.05 scores A0 unless 2 × 0.0190 is used. Condone "increased" for "changed". Must be in context (the grower's belief, or the mean mass of strawberries per plant). Contradictory statements score A0.
1 mark
B1 p-value \(= 2 \times 0.0190 = 0.0380\) (awrt 0.0380, i.e. in [0.03795, 0.03805))
Follow through 2 × their probability, provided their probability is less than 0.5. Accept 3.8%. An answer of 0.0190 (not doubled) scores B0: for a two-tailed test the tail probability is doubled.

Question 3

5 marks

The volume of water, \(V\) litres, used by a model of washing machine in one eco wash cycle is normally distributed with mean 42.5 litres and standard deviation 1.6 litres. After a software update, a consumer group believes that the mean volume of water used in an eco cycle has increased.

The group records the volume of water used in each of a random sample of 16 eco cycles after the update, in millilitres. The total of the 16 volumes is 692 480 ml.

Assuming that the standard deviation is unchanged, test, at the 5% level of significance, whether there is evidence to support the consumer group's belief. State your hypotheses clearly.

Model answer

\(H_0: \mu = 42.5\), \(H_1: \mu > 42.5\) (litres)
Sample mean: \(\dfrac{692\,480}{16} = 43\,280\) ml \(= 43.28\) litres
Under \(H_0\), \(\bar{V} \sim N\left(42.5, \dfrac{1.6^2}{16}\right)\), so the standard deviation of \(\bar{V}\) is 0.4
\(P(\bar{V} \geqslant 43.28) = 0.0256\) (3 s.f.)
\(0.0256 < 0.05\), so the result is significant and \(H_0\) is rejected.
There is sufficient evidence at the 5% level to support the consumer group's belief that the mean volume of water used in an eco cycle has increased after the update.

Explanation

The data are in millilitres but the model is in litres, so the sample mean of 43 280 ml must be converted to 43.28 litres (or the model converted to millilitres) before standardising. With \(\bar{V} \sim N(42.5, 0.4^2)\) the probability of a sample mean of 43.28 or more is 0.0256, below 5%, so the group's belief is supported. Mixing units makes the model meaningless, and forgetting to divide the variance by 16 gives a probability of about 0.31 and the wrong conclusion.

Mark scheme

1 mark
B1 Both hypotheses: \(H_0: \mu = 42.5\) and \(H_1: \mu > 42.5\) (in litres), or \(H_0: \mu = 42\,500\) and \(H_1: \mu > 42\,500\) (in millilitres). A conclusion with no working scores 0 marks.
An answer of "there is evidence that the mean volume of water has increased" with no working scores 0 marks: the sample mean and the probability (or test statistic or critical region) must be seen. Hypotheses must be in terms of \(\mu\) and attached to \(H_0\) and \(H_1\).
1 mark
B1 Sample mean in consistent units: \(\dfrac{692\,480}{16} = 43\,280\) ml \(= 43.28\) litres
Accept 43.28 litres, or 43 280 ml if the mean and standard deviation of the model are also in millilitres (42 500 ml and 1600 ml).
1 mark
M1 Selects the model with consistent units: \(\bar{V} \sim N\left(42.5, \dfrac{1.6^2}{16}\right)\) in litres (standard deviation 0.4), or \(N\left(42\,500, \dfrac{1600^2}{16}\right)\) in millilitres
Mixing units (e.g. 43 280 with N(42.5, ...), or millilitres with standard deviation 1.6) is M0: not converting units is a common error in the examiners' reports. Must divide the variance by 16. N(43.28, ...) is M0.
1 mark
A1 \(P(\bar{V} \geqslant 43.28) = 0.0256\) (awrt 0.0256, i.e. in [0.02555, 0.02565))
Alternatives: test statistic \(z = 1.95\) with critical value 1.6449; or critical region \(\bar{V} > 43.16\) litres (awrt 43.16, i.e. in [43.155, 43.165)), equivalently \(\bar{V} > 43\,158\) ml. Not dividing by n gives 0.313 (M0).
1 mark
A1 0.0256 < 0.05, so reject \(H_0\): there is sufficient evidence at the 5% level to support the consumer group's belief that the mean volume of water used in an eco cycle has increased after the update
Depends on the M1 and the first A1 (not on the hypotheses). Must be in context: the consumer group's belief, or the mean volume of water used. Contradictory statements, e.g. "significant, so there is insufficient evidence", score A0.

Question 4

6 marks

Rhys plays an online game. The time, \(R\) seconds, that he takes to react to a signal in the game is normally distributed with mean 0.31 s and standard deviation 0.06 s.

Find the probability that one randomly chosen reaction time is less than 0.24 s.

Rhys uses a training program, and he then believes that his mean reaction time has decreased. A random sample of 18 of his reaction times after the training has a mean of 0.284 s. Assuming that the standard deviation is unchanged, test, at the 5% level of significance, whether there is evidence to support Rhys's belief. State your hypotheses clearly.

Rhys says that the result of his test proves that the training program has reduced his mean reaction time. Comment on his statement.

Model answer

\(P(R < 0.24) = 0.122\) (3 s.f.)
\(H_0: \mu = 0.31\), \(H_1: \mu < 0.31\)
Under \(H_0\), \(\bar{R} \sim N\left(0.31, \dfrac{0.06^2}{18}\right)\)
\(P(\bar{R} \leqslant 0.284) = 0.0330\) (3 s.f.)
\(0.0330 < 0.05\), so the result is significant and \(H_0\) is rejected.
There is sufficient evidence at the 5% level to support Rhys's belief that his mean reaction time has decreased.
Rhys's statement is too strong: the test only gives evidence at the 5% level. Even if his mean reaction time had not changed, there was a probability of 0.05 of a significant result, and the test cannot rule out other causes, such as extra practice at the game.

Explanation

A single reaction time uses \(R \sim N(0.31, 0.06^2)\), but the test uses the mean of 18 times, whose standard deviation is \(\dfrac{0.06}{\sqrt{18}}\). The probability 0.0330 is below 5%, so there is evidence of a decrease, but a significant result is evidence rather than proof, because the significance level is the probability of rejecting \(H_0\) when it is true. Mixing up the distributions of a single value and of the sample mean is the commonest error in this kind of question.

Mark scheme

1 mark
B1 \(P(R < 0.24) = 0.122\) (awrt 0.122, i.e. in [0.1215, 0.1225)); a correct probability with no working scores this mark
An answer of 0.122 with no working scores 1 mark (this B1); the test needs the model and the probability for its marks. Using the mean of 18 for this probability (giving about 0.0000004) scores B0.
1 mark
B1 Both hypotheses: \(H_0: \mu = 0.31\) and \(H_1: \mu < 0.31\)
Hypotheses in terms of \(\mu\), attached to \(H_0\) and \(H_1\). Accept 310 ms for 0.31 s (exact values from the question).
1 mark
M1 Selects the model \(\bar{R} \sim N\left(0.31, \dfrac{0.06^2}{18}\right)\), seen or used
Must divide the variance by 18. \(N(0.284, \ldots)\) is M0. Accept exact equivalents of the variance (0.0002).
1 mark
A1 \(P(\bar{R} \leqslant 0.284) = 0.0330\) (awrt 0.0330, i.e. in [0.03295, 0.03305))
Alternatives: \(z = -1.84\) (awrt −1.84, i.e. in [−1.845, −1.835)) with critical value −1.6449, or the critical region \(\bar{R} < 0.287\) with 0.284 in it.
1 mark
A1 0.0330 < 0.05, so reject \(H_0\): there is sufficient evidence at the 5% level to support Rhys's belief that his mean reaction time has decreased
Depends on the M1 and the first A1. Must be in context (Rhys's belief, or his mean reaction time). Contradictory statements score A0. Accept the equivalent comparison of their z or critical value.
1 mark
B1 A valid comment, e.g. the test does not prove anything: it only gives evidence at the 5% level, and if his mean reaction time had not changed there would still be a probability of 0.05 (the significance level) of a significant result; or the test cannot show the cause: something other than the program (e.g. extra practice at the game) may have reduced his reaction time
One valid comment is enough. Accept "significant does not mean certain" with a reason (sampling variation, or the 5% chance of rejecting \(H_0\) when it is true), or a correlation-not-cause style point about other factors. B0 for simply agreeing with Rhys, or for "the sample is too small" with no further reason.

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