What is S5.2: Binomial hypothesis tests?

Model the number of sample successes by XB(n,p_0) under H_0:p=p_0, provided the binomial assumptions are defensible.

How many marks are S5.2 questions worth?

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

Which past exam papers has S5.2 appeared in?

We have not yet indexed a verified past-paper appearance for S5.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

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.

Draft — not yet indexed

Binomial hypothesis tests

Worked answers and methods for S5.2 on Edexcel A-level Maths 9MA0.

Explanation

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.

Worked example

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.

1.The null distribution is symmetric because $p_0=0.5$ .
2.Outcomes at least as extreme as $5$ lie in the two tails, so the p-value is $2P(X\leq5)=2(0.02069)=0.04138$ .
3.This is less than $0.05$ , so reject $H_0$ and infer a difference in the population proportion.

Answer: Two-tailed p-value $=2(0.02069)=0.04138$ .; Reject $H_0$ .; There is sufficient evidence that the population success probability differs from $0.5$ .

Common mistakes

Don't conclude that $H_0$ is true when the result is not significant rather than stating insufficient evidence against it.
Don't double a one-tail probability without checking that the alternative is two-tailed and the opposite tail is symmetric.

Exam tip

State both hypotheses, compare the full p-value with the significance level and phrase the conclusion about evidence, not proof.

Worked practice

Tier 1 · Easy

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)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Reject $H_0$ because $0.0193<0.05$ . There is sufficient evidence that the probability of heads is greater than $0.5$ .	4
Notes
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.

(4 marks)

Tier 2 · Standard

A seed supplier claims that the germination probability is

0.8

. In an independent sample of

30

seeds,

19

germinate. Test

H_0:p=0.8

against

H_1:p<0.8

at the

5\%

level, given that

P(X\leq19)=0.0256

for

X\sim\operatorname{B}(30,0.8)

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	p-value $=0.0256$ Reject $H_0$ . There is sufficient evidence that the germination probability is less than $0.8$ .	5
Notes
Under $H_0$ , $X\sim\operatorname{B}(30,0.8)$ . The alternative is lower-tailed, so the p-value for observing $19$ germinations is $P(X\leq19)=0.0256$ . Since $0.0256<0.05$ , reject $H_0$ . The sample provides sufficient evidence, at the $5\%$ level, that the supplier's germination probability is below $0.8$ .

(5 marks)

Tier 3 · Hard

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)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	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.	6
Notes
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 $0.05$ .

(6 marks)

Tier 1 · Easy

A call centre claims that the proportion

p

of callers who abandon a queue has fallen from

0.12

. State suitable null and alternative hypotheses and identify the relevant tail.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$H_0:p=0.12$ $H_1:p<0.12$ Lower-tailed test	3
Notes
The null hypothesis uses the established proportion. The word 'fallen' specifies values below $0.12$ , so the alternative is directional and the lower tail is used.

(3 marks)

Tier 2 · Standard

A service has historically completed a request successfully with probability

0.25

. After a change, only

1

18

independent requests succeeds. Test

H_0:p=0.25

against

H_1:p<0.25

at the

5\%

level. Give 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 p-value is $P(X\leq1)=0.0395<0.05$ . There is sufficient evidence at the $5\%$ level that the success probability has decreased from $0.25$ .	5
Notes
Under $H_0$ , $X\sim\operatorname{B}(18,0.25)$ . The lower-tailed p-value is $P(X\leq1)=0.75^{18}+18(0.25)(0.75)^{17}=0.0394639\ldots$ . Since this is less than $0.05$ , reject $H_0$ and make the population-proportion inference in context.

(5 marks)

Tier 3 · Hard

A voice assistant is claimed to recognise a spoken command with probability

0.65

. A researcher tests

H_0:p=0.65

against

H_1:p<0.65

using

25

attempts and observes

11

recognitions. Calculate the p-value and conduct the test at the

5\%

level. The attempts consisted of five commands from each of five speakers; explain why this sampling detail may weaken the conclusion. Give 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 the p-value is $P(X\leq11)=0.0255<0.05$ ; the binomial calculation gives sufficient evidence that the recognition probability is below $0.65$ . Attempts by the same speaker may share voice characteristics or recording conditions, so their outcomes may not be independent. The effective amount of independent information may be less than $25$ attempts, so the binomial p-value and conclusion may be unreliable.	6
Notes
Assuming the binomial model, $X\sim\operatorname{B}(25,0.65)$ under $H_0$ . The lower-tail probability is $P(X\leq11)=0.0254606\ldots$ , so it is below $0.05$ and the model-based decision is to reject. However, repeated attempts from one speaker can be correlated, violating the independence assumption used to calculate that p-value.

(6 marks)

Tier 2 · Standard

A delivery service claims that the proportion

p

of orders arriving on time is

0.40

. In a random sample of

40

orders, let

X

be the number arriving on time. A lower-tailed

5\%

test uses the critical region

X\leq10

, whose probability under

H_0

0.0352

. State the hypotheses and conduct the test when

10

orders arrive on time. State the actual significance level.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$H_0:p=0.40$ and $H_1:p<0.40$ Reject $H_0$ because $10$ lies in the critical region $X\leq10$ . There is sufficient evidence at the $5\%$ level that the population proportion arriving on time is below $0.40$ . Actual significance level $=0.0352$ , or $3.52\%$ .	5
Notes
The claim is tested against a decrease, so the alternative is lower-tailed. Under $H_0$ , $X\sim\operatorname{B}(40,0.40)$ . The observed count equals the critical value and is therefore inside the rejection region. Reject $H_0$ and make the inference about the population proportion. The null probability of that region is the actual significance level $0.0352$ .

(5 marks)

Tier 3 · Hard

A provider claims that the population proportion

p

of successful installations is

0.60

. One random sample contains

8

successes from

20

installations and a second independent random sample contains

15

successes from

30

installations. Assuming both samples concern the same population probability, test

H_0:p=0.60

against

H_1:p<0.60

at the

5\%

level. You may use

P(X\leq23)=0.03141

for

X\sim\operatorname{B}(50,0.60)

. Explain why pooling the samples would be invalid if they came from customer groups with different success probabilities.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	The pooled count is $23$ successes from $50$ installations, so the p-value is $0.03141$ . Reject $H_0$ because $0.03141<0.05$ . There is sufficient evidence that the common population success probability is below $0.60$ . Pooling requires one constant success probability across all $50$ independent trials. If the customer groups have different probabilities, the pooled count is not binomial with one parameter, and the p-value and conclusion need not be valid; analyse the groups separately.	6
Notes
Under the stated common-probability assumption, independent binomial counts combine: the number of successes is $X\sim\operatorname{B}(20+30,0.60)$ and the observed total is $8+15=23$ . The lower-tail p-value $0.03141$ is below $0.05$ , so reject. If group membership changes $p$ , the constant-probability condition fails even though each result is still success or failure, so a single pooled binomial test conceals the group structure.

(6 marks)

Tier 3 · Hard

A platform claims that the population probability

p

of a user completing a task is

0.40

. In a random sample of

25

independent users,

16

complete the task. Test

H_0:p=0.40

against

H_1:p>0.40

at the

5\%

level. You may use

P(X=16)=0.008843

and

P(X\geq16)=0.013169

for

X\sim\operatorname{B}(25,0.40)

. State which probability is the p-value and explain why.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The p-value is $P(X\geq16)=0.0132$ to $4$ decimal places, not $P(X=16)$ . The upper-tailed p-value includes the observed result and every result more extreme in the direction of $H_1$ . Reject $H_0$ because $0.013169<0.05$ . There is sufficient evidence at the $5\%$ level that the population completion probability exceeds $0.40$ .	5
Notes
Under $H_0$ , $X\sim\operatorname{B}(25,0.40)$ . The directional alternative makes large counts evidence against the null, so the tail from $16$ through $25$ is required. Its probability is $0.013169$ , whereas the single-point probability omits more extreme outcomes. Comparison with $0.05$ gives rejection and the conclusion is stated about the population parameter.

(5 marks)

Q10

Tier 3 · Hard

10.

A provider claims that the population success probability is

0.75

. A fixed random sample of

60

independent cases is used to test

H_0:p=0.75

against

H_1:p<0.75

at the

5\%

level. The preliminary record shows

39

successes, but a source check confirms that one of these cases was misclassified and the verified number of successes is

38

. Calculate the exact lower-tail p-value using each count, giving both values to

4

decimal places. Conduct the test using the verified data and explain how the confirmed correction changes the conclusion. The sample size remains

60

throughout.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	For the preliminary count, $P(X\leq39)=0.0541$ to $4$ decimal places, so it would not lead to rejection at the $5\%$ level. For the verified count, $P(X\leq38)=0.0298$ to $4$ decimal places. Reject $H_0$ because $0.0298<0.05$ ; there is sufficient evidence that the population success probability is below $0.75$ . The confirmed correction changes $X$ from $39$ to $38$ without changing $n=60$ , moving the exact lower-tail p-value from above to below the significance level and reversing the decision.	7
Notes
Under $H_0$ , the fixed sample count is $X\sim\operatorname{B}(60,0.75)$ . Since the alternative is lower-tailed, include the observed count and every smaller count. Thus $P(X\leq39)=\sum_{r=0}^{39}\binom{60}{r}(0.75)^r(0.25)^{60-r}=0.0541439\ldots$ , whereas $P(X\leq38)=0.0298009\ldots$ . Only the verified count gives a p-value below $0.05$ , so the evidence statement is about the population parameter $p$ using the corrected data.

(7 marks)

Verified exam appearances

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Other points in S5 Statistical hypothesis testing

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