What is S5.3: Normal mean hypothesis tests?

For a Normal population with known or assumed standard deviation , the sample mean satisfies XN(,^2/n).

How many marks are S5.3 questions worth?

S5.3 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has S5.3 appeared in?

We have not yet indexed a verified past-paper appearance for S5.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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.

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Normal mean hypothesis tests

Worked answers and methods for S5.3 on Edexcel A-level Maths 9MA0.

Explanation

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.
The standard error is $5/\sqrt{25}=1$ .

Worked example

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 $\overline{x}=47.3$ .

1.The standard error is $5/\sqrt{25}=1$ .
2.The acceptance interval is $50\pm2.576(1)$ , namely $47.424\leq\overline{X}\leq52.576$ .
3.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 $50$ .

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 $50$ .

Common mistakes

Don't use a two-tailed critical value for a directional alternative hypothesis.
Don't use the population standard deviation directly as the spread of the sample mean instead of dividing by the square root of sample size.

Exam tip

Standardise the sample mean with its standard error, then compare with the correct one- or two-tailed critical values.

Worked practice

Tier 1 · Easy

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)

(Total for Question 1 is 5 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Test statistic $z=2.00$ and p-value $0.0228$ . Reject $H_0$ ; there is sufficient evidence that the population mean exceeds $100$ .	5
Notes
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.

(5 marks)

Tier 2 · Standard

A Normal population has known standard deviation

10

. A random sample of

25

has mean

54.4

. Test

H_0:\mu=50

against

H_1:\mu\neq50

at the

2\%

significance level. Use the critical values

z=\pm2.326

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$z=2.20$ . Do not reject $H_0$ . There is insufficient evidence at the $2\%$ level that the population mean differs from $50$ .	5
Notes
The standard error is $10/\sqrt{25}=2$ . Hence $z=(54.4-50)/2=2.20$ . This lies between the two critical values $-2.326$ and $2.326$ , so it is not in the critical region. Do not reject $H_0$ ; there is insufficient evidence at the $2\%$ level that the population mean differs from $50$ .

(5 marks)

Tier 3 · Hard

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)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Smallest sample size $n=25$ . For $n=25$ , p-value $=0.0478$ .	7
Notes
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 $1-\Phi(1.6667)=0.0478$ .

(7 marks)

Tier 1 · Easy

A Normal population has known standard deviation

4

. A random sample of

25

has mean

51.2

. Find the test statistic for testing

H_0:\mu=50

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$z=1.50$	2
Notes
The standard error is $4/\sqrt{25}=0.8$ . Hence $z=(51.2-50)/0.8=1.50$ .

(2 marks)

Tier 2 · Standard

A Normal population has known standard deviation

2.4

. A random sample of

16

has mean

73.6

. Test

H_0:\mu=75

against

H_1:\mu\neq75

at the

5\%

level. Give the test statistic to

3

significant figures and the p-value to

4

decimal places.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Reject $H_0$ because the two-tailed p-value is $0.0196<0.05$ . $z=-2.33$ . There is sufficient evidence at the $5\%$ level that the population mean differs from $75$ .	5
Notes
The standard error is $2.4/\sqrt{16}=0.6$ , so $z=(73.6-75)/0.6=-2.333\ldots$ . The two-tailed p-value is $2P(Z\leq-2.333\ldots)=0.019630\ldots$ . Since this is below $0.05$ , reject $H_0$ and state the conclusion about the population mean.

(5 marks)

Tier 3 · Hard

Brightness readings from a display are modelled as independent Normal variables with known standard deviation

4

units. The display is tested using

H_0:\mu=50

against

H_1:\mu<50

at the

1\%

level with a sample of

36

readings. Using the critical value

-2.326

, find the critical value of the sample mean and the corresponding critical value of the total of the readings, giving each to

3

decimal places. The observed total is

1740

units. Conduct the test and find the p-value to

4

decimal places.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Reject $H_0$ because $1740<1744.176$ (equivalently $\bar{x}=48.333\ldots<48.449\ldots$ ). Critical sample mean $=48.449$ units and critical total $=1744.176$ units. p-value $=0.0062$ . There is sufficient evidence at the $1\%$ level that the population mean brightness reading is below $50$ units.	6
Notes
The standard error is $4/\sqrt{36}=2/3$ units. The lower critical mean is $50-2.326(2/3)=48.4493\ldots$ units, corresponding to total $36(48.4493\ldots)=1744.176$ units. The observed mean is $1740/36=48.3333\ldots$ units, which is in the critical region. Its test statistic is $(48.3333\ldots-50)/(2/3)=-2.5$ , so the lower-tail p-value is $P(Z\leq-2.5)=0.006209\ldots=0.0062$ .

(6 marks)

Tier 2 · Standard

A Normal population has known standard deviation

8

. A random sample of

64

is used to test

H_0:\mu=100

against

H_1:\mu>100

at the

1\%

level, using critical value

z=2.326

. The recorded sample mean is

102.4

, correct to the nearest

0.1

. Find the critical value of the sample mean and determine whether the test conclusion is unaffected by the rounding.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Critical sample mean $=100+2.326(8/\sqrt{64})=102.326$ . The unrounded sample mean lies in $102.35\leq\bar{x}<102.45$ . Every value in this interval exceeds $102.326$ , so reject $H_0$ regardless of the rounding. There is sufficient evidence at the $1\%$ level that the population mean exceeds $100$ .	5
Notes
The standard error is $8/\sqrt{64}=1$ , so the upper critical boundary is $100+2.326=102.326$ . A value recorded as $102.4$ to the nearest tenth represents the half-open interval from $102.35$ to $102.45$ . Its lower endpoint already lies above the critical boundary, making the rejection decision robust to every possible unrounded value.

(5 marks)

Tier 3 · Hard

A random sample of

25

from a Normal population has mean

21.8

. For the test

H_0:\mu=20

against

H_1:\mu>20

at the

5\%

level, the population standard deviation

\sigma

is assumed known and the critical value is

1.645

. Find the greatest value of

\sigma

for which the result is significant, giving your answer to

3

significant figures. Hence state the conclusion when

\sigma=5.0

and when

\sigma=5.5

. Use unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Significance requires $\sigma\leq5.47112\ldots$ , so the greatest value is $5.47$ to $3$ significant figures. For $\sigma=5.0$ , $z=1.80>1.645$ : reject $H_0$ ; there is sufficient evidence that $\mu>20$ . For $\sigma=5.5$ , $z=1.636\ldots<1.645$ : do not reject $H_0$ ; there is insufficient evidence that $\mu>20$ .	6
Notes
The test statistic is $(21.8-20)/(\sigma/\sqrt{25})=9/\sigma$ . Rejection requires $9/\sigma\geq1.645$ , so $\sigma\leq9/1.645=5.47112\ldots$ . At $\sigma=5.0$ , the statistic is $1.8$ and lies in the critical region. At $\sigma=5.5$ , it is $1.63636\ldots$ and lies outside, so the correct non-significant conclusion uses insufficient-evidence language.

(6 marks)

Tier 3 · Hard

A random sample of

49

observations is taken from a Normal population with known standard deviation

14

. For each observation define

y=(x-100)/2

, and the sample satisfies

\sum y=112

. Test

H_0:\mu=100

against

H_1:\mu>100

. Calculate the sample mean, test statistic and p-value. State the conclusions at the

5\%

and

1\%

significance levels. Give the p-value to

4

decimal places.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\bar{x}=104.571\ldots$ and $z=2.2857\ldots$ Upper-tail p-value $=0.0111$ . At the $5\%$ level, reject $H_0$ ; there is sufficient evidence that the population mean exceeds $100$ . At the $1\%$ level, do not reject $H_0$ ; there is insufficient evidence that the population mean exceeds $100$ .	6
Notes
Since $x=2y+100$ , $\sum x=2(112)+49(100)=5124$ and $\bar{x}=5124/49=104.5714\ldots$ . The standard error is $14/\sqrt{49}=2$ , so $z=(104.5714\ldots-100)/2=2.285714\ldots$ . The upper-tail probability is $0.011135\ldots=0.0111$ , which lies below $0.05$ but above $0.01$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A random sample of

36

observations from a Normal population with known standard deviation

6

is used to test

H_0:\mu=48

against

H_1:\mu>48

at the

1\%

level. The recorded total is

1836

, but one recorded value

74

is confirmed to be an error for

47

. Calculate the test statistic before and after correction. Conduct the test using the corrected data and find its p-value to

4

decimal places. Explain how the error would have changed the conclusion.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Before correction, $\bar{x}=51$ and $z=3.00$ . The corrected total is $1809$ , so $\bar{x}=50.25$ and $z=2.25$ . The corrected upper-tail p-value is $0.0122$ . Do not reject $H_0$ at the $1\%$ level; there is insufficient evidence that the population mean exceeds $48$ . The erroneous value would have produced rejection because $z=3.00$ has p-value below $0.01$ ; the data error therefore reverses the decision.	7
Notes
The standard error is $6/\sqrt{36}=1$ . Before correction the mean is $1836/36=51$ , giving $z=(51-48)/1=3$ . Correct the total by $1836-74+47=1809$ , so the mean is $50.25$ and $z=2.25$ . The upper-tail probability is $1-\Phi(2.25)=0.012224\ldots=0.0122$ , above $0.01$ , so the corrected result is not significant at the stated level.

(7 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for S5.3. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in S5 Statistical hypothesis testing

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