Is hypothesis testing for a normal mean on AS or A Level Maths?

It is A Level only, usually taught in Year 13, so it is not on the AS exam.

Are these past paper questions?

No. All 15 questions are original, written in the style of Edexcel 9MA0 with M, A and B marks. The same content is on the AQA and OCR A Level Maths specifications, so they work for every board.

Why do I divide σ by √n?

The mean of n independent observations varies less than one observation. Its standard deviation, the standard error, is σ/√n, so use N(μ, σ²/n) for the sample mean.

Hypothesis Testing for the Mean of a Normal Distribution

Question 1easy4 marksYear 13

The time, in hours, that a brand of phone charger takes to fail under a stress test is normally distributed with mean 9.2 and standard deviation 1.6. A cheaper component is introduced and a random sample of 22 chargers has mean failure time 8.55 hours.

(a) Stating your hypotheses clearly, test at the 5% level whether the mean failure time has decreased. Assume the standard deviation is unchanged. (4)

Mark scheme and answer

(a) \(H_0:\mu=9.2,\ H_1:\mu\lt 9.2\) B1
\(P(\bar X\lt 8.55)\) where \(\bar X\sim\mathrm{N}(9.2,\ 1.6^2/22)\) M1
\(= 0.0284\) A1
\(0.0284 \lt 0.05\) so reject H₀: evidence mean failure time has decreased A1

Answer: p = 0.0284; reject H0

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 2medium6 marksYear 13

The mass of rice in a bag is normally distributed with standard deviation 2.4 g. The label states that the mean mass is 500 g. A trading standards officer suspects that the mean mass is less than 500 g. She weighs a random sample of 20 bags and finds their mean mass is 498.9 g.

(a) Test, at the 5% level of significance, whether there is evidence to support her suspicion. State your hypotheses clearly. (6)

Mark scheme and answer

(a) \(H_0: \mu = 500\), \(H_1: \mu \lt 500\) B1
\(\bar X \sim \mathrm{N}\left(500,\ \dfrac{2.4^2}{20}\right)\) M1
\(z = \dfrac{498.9-500}{2.4/\sqrt{20}} = -2.0497\) or \(P(\bar X\lt 498.9) = 0.02020\) M1A1
Compare with \(-1.6449\) or 0.05: significant, reject H₀ dM1
There is evidence that the mean mass of rice in a bag is less than 500 g A1

Answer: z = -2.050, p = 0.0202 < 0.05: evidence mean < 500 g

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 3medium6 marksYear 13

The lengths of bolts made by a machine are normally distributed with mean 25 mm and standard deviation 0.8 mm. After a service, an engineer takes a random sample of 36 bolts and finds their mean length is 25.3 mm. Assume that the standard deviation is unchanged.

(a) Test, at the 5% significance level, whether the mean length has changed. (6)

Mark scheme and answer

(a) \(H_0: \mu = 25\), \(H_1: \mu \ne 25\) B1
\(\bar X \sim \mathrm{N}\left(25,\ \dfrac{0.8^2}{36}\right)\) M1
\(z = \dfrac{25.3-25}{0.8/6} = 2.25\) A1
Critical values \(\pm1.96\) (or \(P(\bar X\gt 25.3) = 0.0122 \lt 0.025\)) M1
Reject \(H_0\) dM1
There is evidence that the mean length of bolts has changed after the service A1

Answer: z = 2.25 > 1.96: reject H0, mean has changed

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 4medium5 marksYear 13

The time, in milliseconds (ms), taken by a website to load a page is normally distributed with mean 120 ms and standard deviation 15 ms. After a software update, the developer believes the mean time has increased. A random sample of 25 page loads is taken. The standard deviation may be assumed unchanged.

(a) Find the critical region for \(\bar X\), the sample mean load time, for a test at the 1% level of significance. (4)

(b) The sample mean is 127 ms. State, with a reason, the conclusion of the test. (1)

Mark scheme and answer

(a) \(H_0: \mu=120,\ H_1: \mu\gt 120\) B1
\(\dfrac{\bar x-120}{15/5} \gt 2.3263\) M1B1
\(\bar X \gt 126.98\) A1

(b) 127 lies in the critical region (127 > 126.98) so reject \(H_0\): evidence the mean load time has increased B1ft

Answer: (a) X̄ > 126.98; (b) reject H0; evidence mean has increased

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 5medium5 marksYear 13

A random sample of 16 observations is taken from \(\mathrm{N}(\mu,\ 4^2)\). The hypotheses \(H_0: \mu = 50\) and \(H_1: \mu \ne 50\) are tested at the 5% level.

(a) Find the critical region for \(\bar X\). (4)

(b) State the actual significance level of this test. (1)

Mark scheme and answer

(a) \(\bar X\sim\mathrm{N}(50,\ 1^2)\) M1
\(\bar x = 50 \pm 1.96\times1\) M1B1
\(\bar X \lt 48.04\) or \(\bar X \gt 51.96\) A1

(b) 5% (continuous distribution) B1

Answer: (a) X̄ < 48.04 or X̄ > 51.96; (b) 5%

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 6medium6 marksYear 13

A student tests whether the mean volume of juice in cartons differs from 1 litre. The volume is normally distributed with standard deviation 0.02 litres. He writes:

\(H_0: \bar x = 1,\ H_1: \bar x \gt 1\). My sample of 10 cartons has mean 1.013, so \(z = 0.65\). This is less than 1.6449 so I accept that \(H_0\) is true.

(a) Identify three errors in the student's work. (3)

(b) Carry out the test correctly at the 5% level. (3)

Mark scheme and answer

(a) Hypotheses should be in terms of the population mean \(\mu\), not \(\bar x\) B1
Test should be two-tailed (\(H_1: \mu \ne 1\)) since he is testing for a difference B1
Test statistic wrong: \(z = \dfrac{0.013}{0.02/\sqrt{10}}\) (he has not divided \(\sigma\) by \(\sqrt n\)) / cannot "accept \(H_0\) is true" – only insufficient evidence to reject B1

(b) \(z = \dfrac{0.013}{0.02/\sqrt{10}} = 2.055\) M1A1
\(\gt 1.96\) so reject \(H_0\): evidence mean volume differs from 1 litre A1

Answer: (a) hypotheses in μ; should be two-tailed; z incorrect / cannot accept H0; (b) z = 2.055 > 1.96: reject H0

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 7medium6 marksYear 13

The birth mass of babies in a region is normally distributed with mean 3.5 kg and standard deviation 0.6 kg. A health visitor believes that babies born in one hospital have a higher mean mass. A random sample of 45 babies born at the hospital has mean mass 3.68 kg.

(a) Test the health visitor's belief at the 2.5% level of significance. Assume the standard deviation is 0.6 kg. (5)

(b) Explain what is meant by the significance level in this context. (1)

Mark scheme and answer

(a) \(H_0:\mu=3.5,\ H_1:\mu\gt 3.5\) B1
\(z = \dfrac{3.68-3.5}{0.6/\sqrt{45}} = 2.012\) M1A1
c.v. 1.96 (or \(p = 0.0221\)) B1
Reject H₀: evidence that the mean birth mass at this hospital is higher A1

(b) The probability of concluding that the mean birth mass is higher when in fact it is 3.5 kg (probability of incorrectly rejecting \(H_0\)) B1

Answer: (a) z = 2.012, reject H0; (b) P(reject H0 | H0 true) = 0.025

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 8medium5 marksYear 13

Last year the time spent by students on a weekly online homework task was normally distributed with mean 42.0 minutes and standard deviation 5.5 minutes. This year a random sample of 50 students spent a mean time of 40.9 minutes.

(a) Investigate, at the 10% level of significance, whether the mean time has changed. Assume the standard deviation is unchanged. (5)

Mark scheme and answer

(a) \(H_0:\mu=42,\ H_1:\mu\ne42\) B1
\(z = \dfrac{40.9-42}{5.5/\sqrt{50}} = -1.414\) M1A1
Critical value \(-1.6449\) (or \(p = 0.157\) two-tailed) B1
Do not reject H₀: insufficient evidence at 10% that the mean time has changed A1

Answer: z = -1.414; do not reject H0

Board notes: "Investigate ... whether" is typical AQA phrasing. Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 9medium4 marksYear 13

A test of \(H_0: \mu = 85\) against \(H_1: \mu \lt 85\) is carried out at the 2.5% level using a random sample of 64 observations from a normal distribution with standard deviation 10.

(a) Find the critical region for \(\bar X\). (3)

(b) The sample mean is 82.9. State the conclusion of the test. (1)

Mark scheme and answer

(a) \(\bar X\sim\mathrm{N}(85,\ 1.25^2)\) M1
\(85 - 1.96\times1.25\) M1
\(\bar X \lt 82.550\) A1

(b) 82.9 is not in the critical region: insufficient evidence to reject \(H_0\) B1ft

Answer: (a) X̄ < 82.550; (b) do not reject H0

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 10hard6 marksYear 13

A nutritionist believes that a new diet lowers the resting heart rate of adults. Before the diet, resting heart rates were normally distributed with mean 72 beats per minute and standard deviation 6 beats per minute. After following the diet, a random sample of 30 adults had a mean resting heart rate of 70.1 beats per minute. Assume the standard deviation is unchanged.

(a) Find the \(p\)-value for a test of \(H_0: \mu = 72\) against \(H_1: \mu \lt 72\). (3)

(b) State, with a reason, whether the result is significant at the 5% level and at the 1% level. (2)

(c) Write down the smallest significance level at which \(H_0\) would be rejected. (1)

Mark scheme and answer

(a) \(\bar X\sim\mathrm{N}(72,\ 6^2/30)\) M1
\(P(\bar X\lt 70.1) = P(Z\lt -1.7345)\) M1
\(= 0.0414\) A1

(b) \(0.0414 \lt 0.05\): significant at 5% – evidence the diet lowers heart rate B1
\(0.0414 \gt 0.01\): not significant at 1% B1

(c) 4.14% B1ft

Answer: (a) 0.0414; (b) significant at 5%, not at 1%; (c) 4.14%

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 11hard4 marksYear 13

The weight of a species of songbird is known to be normally distributed with standard deviation 3.5 g. A biologist wants to test \(H_0: \mu = 18\) against \(H_1: \mu \gt 18\) at the 5% level. She expects the sample mean to be 18.9 g.

(a) Find the smallest sample size for which a sample mean of 18.9 g would lead to rejection of \(H_0\). (4)

Mark scheme and answer

(a) \(\dfrac{18.9-18}{3.5/\sqrt n} \gt 1.6449\) M1B1
\(\sqrt n \gt 6.3968\) dM1
\(n \gt 40.92\) so \(n = 41\) A1

Answer: n = 41

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 12hard6 marksYear 13

A club runner's times for a 2 km time trial were normally distributed with mean 604 seconds and standard deviation 9 seconds. After a new training programme, her times for 12 randomly chosen trials were

612, 598, 605, 590, 601, 595, 588, 603, 597, 592, 600, 594

Assume the times are still normally distributed with standard deviation 9 seconds.

(a) Test, at the 5% level, whether there is evidence that her mean time has decreased. (6)

Mark scheme and answer

(a) \(H_0: \mu = 604,\ H_1: \mu \lt 604\) B1
\(\bar x = \dfrac{7175}{12} = 597.92\) B1
\(z = \dfrac{597.92-604}{9/\sqrt{12}} = -2.341\) M1A1
\(p = 0.00960\); compare with 0.05 or \(z\) with \(-1.6449\) dM1
Reject H₀: evidence her mean time has decreased A1

Answer: x̄ = 597.92, z = -2.341, reject H0

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 13hard5 marksYear 13

The breaking strength, in newtons, of a type of cord is normally distributed with standard deviation 12 N. A manufacturer claims a new process has increased the mean breaking strength from 200 N. A random sample of 40 lengths of cord is tested.

(a) Find the set of values of the sample mean \(\bar x\) for which the claim would be supported at the 5% level but not at the 1% level. (5)

Mark scheme and answer

(a) \(\sigma/\sqrt{40} = 1.897\) B1
5%: \(\bar x \gt 200 + 1.6449\times1.897 = 203.12\) M1A1
1%: \(\bar x \gt 200 + 2.3263\times1.897 = 204.41\) A1
\(203.12 \lt \bar x \le 204.41\) A1

Answer: 203.12 < x̄ ≤ 204.41

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 14hard7 marksYear 13

The noise level, in decibels, near a busy road is normally distributed with mean 88 dB and standard deviation 4.2 dB. Residents believe that noise levels have increased since a new junction opened. Fifteen randomly timed readings give \(\sum x = 1348\).

(a) Test, at the 5% level of significance, whether there is evidence to support the residents' belief. Assume the standard deviation is unchanged. (6)

(b) State one assumption about the readings that is needed for the test to be valid. (1)

Mark scheme and answer

(a) \(H_0:\mu=88,\ H_1:\mu\gt 88\) B1
\(\bar x = 89.87\) B1
\(z = \dfrac{89.87-88}{4.2/\sqrt{15}} = 1.721\) M1A1
Compare with 1.6449 (\(p = 0.0426\)) dM1
Reject H₀: evidence that the mean noise level has increased A1

(b) The readings are a random sample / independent of each other B1

Answer: (a) x̄ = 89.87, z = 1.721; reject H0; (b) independent random readings

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

Question 15challenging4 marksYear 13

A random sample of 25 observations is taken from a normal distribution with unknown mean \(\mu\) and known standard deviation \(\sigma\). A two-tailed test of \(H_0: \mu = 250\) at the 5% level leads to rejection of \(H_0\) if and only if \(\bar x \lt 247\) or \(\bar x \gt 253\).

(a) Find the value of \(\sigma\). (3)

(b) Using the same critical region, a one-tailed test of \(H_0: \mu = 250\) against \(H_1: \mu \gt 250\) rejects \(H_0\) when \(\bar x \gt 253\). State the significance level of this test. (1)

Mark scheme and answer

(a) \(\dfrac{253-250}{\sigma/5} = 1.96\) M1B1
\(\sigma = 7.653\) A1

(b) \(P(\bar X \gt 253) = 0.025\), i.e. 2.5% B1

Answer: (a) σ = 7.653; (b) 2.5%

Board notes: Edexcel expects hypotheses in terms of \(\mu\), a test statistic or critical value, a comparison and a conclusion in context with a non-assertive phrase ("there is evidence ..."). AQA mark schemes usually require the conclusion to refer to the population parameter and the significance level; OCR requires both a statistical conclusion and a contextual one.

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