Hypothesis tests for a binomial proportion - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 19 exam-style questions on Hypothesis tests for a binomial proportion 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

3 marks

In the past, 30% of patients at a hospital clinic missed their appointments. The clinic starts sending text reminders, and a manager tests \(H_0: p = 0.3\) against \(H_1: p < 0.3\), where p is the proportion of patients who miss their appointments. She uses a random sample of 30 patients and the critical region \(X \leqslant 4\), where X is the number of these patients who miss their appointments.

Find the actual significance level of the test, and explain what it means in this context.

Model answer

Under \(H_0\), \(X \sim B(30, 0.3)\)
Actual significance level \(= P(X \leqslant 4) = 0.0302\)
This is the probability of rejecting \(H_0\) when it is true: the probability of concluding that the text reminders have reduced the proportion of missed appointments when in fact 30% of patients still miss them.

Explanation

The actual significance level is the probability that the test statistic falls in the critical region when the null hypothesis is true, here \(P(X \leqslant 4)\) for \(B(30, 0.3)\). It is the risk of wrongly concluding that the proportion has fallen. A common error is to describe it as the probability that the null hypothesis is true.

Mark scheme

1 mark
M1 Uses \(X \sim B(30, 0.3)\) to find \(P(X \leqslant 4)\); or a correct final answer of 0.0302 with no working
An answer of 0.0302 with no working scores 2 marks (M1 A1); the explanation is marked separately. \(P(X = 4)\) or \(P(X \leqslant 5)\) (0.0766) scores M0.
1 mark
A1 Actual significance level = 0.0302 (awrt 0.0302, i.e. [0.03015, 0.03025)) or 3.02%
Booklet value 0.0302 (exact value 0.030154...).
1 mark
B1 It is the probability of rejecting \(H_0\) when it is true: the probability of concluding that the proportion of patients who miss appointments has fallen when it is in fact still 0.3
Must be in context (missed appointments, or the reminders appearing to work) for the mark. "The probability of incorrectly rejecting the null hypothesis" without context scores B0. Do not accept "the probability that \(H_0\) is true".

Question 2

4 marks

An online shop finds that 10% of the people who receive its discount emails make a purchase. It redesigns the email, and the marketing manager claims that the new design will increase this proportion. The new email is sent to a random sample of 25 people, and 5 of them make a purchase.

Test the manager's claim at the 5% level of significance. State your hypotheses clearly.

Model answer

\(H_0: p = 0.1\), \(H_1: p > 0.1\), where p is the proportion of people receiving the new email who make a purchase
Under \(H_0\), \(X \sim B(25, 0.1)\)
\(P(X \geqslant 5) = 1 - P(X \leqslant 4) = 1 - 0.9020 = 0.0980\)
\(0.0980 > 0.05\), so the result is not significant: do not reject \(H_0\).
There is insufficient evidence at the 5% level to support the manager's claim that the new design increases the proportion of people who make a purchase.

Explanation

The claim is of an increase, so find the upper-tail probability \(P(X \geqslant 5) = 1 - P(X \leqslant 4)\) under \(H_0\). At 0.0980 it is above 5%, so there is not enough evidence for the claim; the critical region \(X \geqslant 6\) leads to the same decision. The conclusion must be in context, and "insufficient evidence" is the safest wording.

Mark scheme

1 mark
B1 \(H_0: p = 0.1\), \(H_1: p > 0.1\). A conclusion with no working scores 0 marks; correct hypotheses alone score this mark
An answer of "there is insufficient evidence to support the claim" with no working scores 0 marks. Both hypotheses must be in terms of p (or \(\pi\)) and attached to \(H_0\) and \(H_1\). Hypotheses in words, or in terms of x, X or \(\bar{x}\), score B0. Allow 10% for 0.1.
1 mark
M1 Uses \(X \sim B(25, 0.1)\) and writes or uses \(P(X \geqslant 5) = 1 - P(X \leqslant 4)\)
\(1 - P(X \leqslant 5)\) (0.0334) or \(P(X = 5)\) scores M0. Critical-region route: \(X \geqslant 6\) (\(P(X \geqslant 6) = 0.0334\), \(P(X \geqslant 5) = 0.0980\)); both approaches to the critical region give \(X \geqslant 6\).
1 mark
A1 \(P(X \geqslant 5) = 0.0980\) (awrt 0.0980, i.e. [0.09795, 0.09805)), or critical region \(X \geqslant 6\) with its probability
Exact value 0.097993...; booklet \(1 - 0.9020 = 0.0980\). Allow 0.098.
1 mark
A1 Correct conclusion in context with no contradiction, e.g. "There is insufficient evidence at the 5% level to support the manager's claim that the new design increases the proportion of people who make a purchase"
Depends on the M1 and the first A1. Also accept a definite statement in context such as "the proportion who make a purchase is still 10%" or "the new design has not increased the proportion". A contradiction (e.g. "not significant, so supports the claim") scores A0.

Question 3

5 marks

A GP surgery finds that 80% of patients attend their booked appointments. The practice manager believes that the proportion attending has fallen since online booking was introduced. She will test her belief at the 5% level of significance using a random sample of 20 appointments, where X is the number of these appointments that the patients attend.

State suitable hypotheses and, by considering the number of appointments missed, find the critical region for X, with its probability as close as possible to, but not more than, 5%. State the actual significance level of the test.

[For \(Y \sim B(20, 0.2)\): \(P(Y \leqslant 6) = 0.9133\), \(P(Y \leqslant 7) = 0.9679\), \(P(Y \leqslant 8) = 0.9900\)]

Model answer

\(H_0: p = 0.8\), \(H_1: p < 0.8\), where p is the proportion of appointments attended
Let Y = number of appointments missed \(= 20 - X\). Under \(H_0\), \(Y \sim B(20, 0.2)\), and a fall in p means large values of Y.
\(P(Y \geqslant 8) = 1 - P(Y \leqslant 7) = 1 - 0.9679 = 0.0321 < 0.05\)
\(P(Y \geqslant 7) = 1 - P(Y \leqslant 6) = 1 - 0.9133 = 0.0867 > 0.05\)
So reject \(H_0\) when \(Y \geqslant 8\), i.e. \(X \leqslant 12\).
Critical region: \(X \leqslant 12\); actual significance level = 0.0321

Explanation

The booklet has no p above 0.5, so work with the number missed, \(Y = 20 - X \sim B(20, 0.2)\); a fall in attendance means an increase in Y. The critical region \(Y \geqslant 8\) has probability 0.0321, and it converts to \(X \leqslant 12\). A common error is to convert \(Y \geqslant 8\) into \(X \leqslant 13\).

Mark scheme

1 mark
B1 \(H_0: p = 0.8\), \(H_1: p < 0.8\), where p is the proportion of appointments attended. Answers of \(X \leqslant 12\) and 0.0321 with no working score 2 marks for the critical region and level
Answers of \(X \leqslant 12\) and 0.0321 with no working score 2 marks (special case: the A1 for \(X \leqslant 12\) and the B1 for 0.0321); \(X \leqslant 12\) alone scores 1 mark. Correct hypotheses score this B1 in addition. Allow hypotheses in terms of the proportion missed: \(H_0: q = 0.2\), \(H_1: q > 0.2\), if q is defined. Both hypotheses must be in terms of p (or \(\pi\)) and attached to \(H_0\) and \(H_1\). Hypotheses in words, or in terms of x, X or \(\bar{x}\), score B0.
1 mark
M1 Uses the number missed, \(Y = 20 - X \sim B(20, 0.2)\), with upper-tail probabilities, e.g. \(P(Y \geqslant 8) = 1 - 0.9679 = 0.0321\) (or \(X \sim B(20, 0.8)\) on a calculator, \(P(X \leqslant 12)\))
May be implied by 0.0321 or 0.0867 seen.
1 mark
A1 \(Y \geqslant 8\) (since \(P(Y \geqslant 8) = 0.0321\) and \(P(Y \geqslant 7) = 0.0867\))
Accept the equivalent statement in terms of X directly if the probabilities are correct.
1 mark
A1 Critical region for X: \(X \leqslant 12\)
Depends on the M1. Accept \(X < 13\) or {0, 1, ..., 12}. \(X \leqslant 13\) scores A0.
1 mark
B1 Actual significance level 0.0321 (or 3.21%) ft
B1ft: the probability of their critical region (provided it is less than 0.5). Booklet value \(1 - 0.9679 = 0.0321\); exact value 0.032142... Do not accept 0.05.

Question 4

6 marks

A factory makes microchips. For a good batch, the proportion of faulty microchips is 0.02. To check a batch, a random sample of n microchips is tested, and the batch is rejected if 2 or more of the sample are faulty.

This rule is a test of \(H_0: p = 0.02\) against \(H_1: p > 0.02\), where p is the proportion of faulty microchips in the batch, with critical region \(X \geqslant 2\).

Explain what the significance level of this test means in context. Find the largest value of n for which the significance level of the test is at most 5%.

Model answer

The significance level is the probability of rejecting \(H_0\) when it is true: the probability of rejecting a good batch (one with proportion faulty 0.02) because 2 or more of the sample are faulty.
With \(X \sim B(n, 0.02)\): significance level \(= P(X \geqslant 2) = 1 - 0.98^n - 0.02n \times 0.98^{n - 1}\)
n = 18: \(1 - 0.98^{18} - 18 \times 0.02 \times 0.98^{17} = 0.0495 \leqslant 0.05\)
n = 19: \(1 - 0.98^{19} - 19 \times 0.02 \times 0.98^{18} = 0.0546 > 0.05\)
The significance level increases with n, so the largest value is n = 18.

Explanation

For a fixed critical region the significance level is \(P(X \geqslant 2)\) under \(H_0\), and it grows as the sample size grows, because larger samples are more likely to contain 2 faulty chips by chance. Evaluating it for n = 18 and n = 19 shows that 18 is the largest sample keeping it at or below 5%. The explanation must be in context: it is the risk of rejecting a good batch.

Mark scheme

1 mark
B1 The significance level is the probability of rejecting a good batch: the probability that 2 or more of the sample are faulty when the proportion faulty is 0.02. An answer of n = 18 with no working scores 1 mark
An answer of n = 18 with no working scores 1 mark (the final A1). For this B1 accept "the probability of rejecting \(H_0\) when it is true" with context (a good batch, or p = 0.02).
1 mark
M1 Writes the significance level as \(P(X \geqslant 2) = 1 - P(X = 0) - P(X = 1)\) for \(X \sim B(n, 0.02)\)
May be implied by correct values for particular n.
1 mark
A1 \(1 - 0.98^n - 0.02n \times 0.98^{n - 1}\)
Or equivalent, e.g. \(1 - 0.98^{n-1}(0.98 + 0.02n)\). May be implied by correct values of the significance level for two consecutive values of n (e.g. from a calculator's binomial function).
1 mark
M1 Evaluates the significance level for values of n near the answer
At least one value of n between 15 and 22, evaluated correctly to 2 s.f. Accept a calculator's binomial cumulative function.
1 mark
A1 n = 18 gives 0.0495 and n = 19 gives 0.0546 (awrt 0.0495 and 0.0546, i.e. [0.04945, 0.04955) and [0.05455, 0.05465))
Both values are needed (or equivalent values that show 18 is acceptable and 19 is not). Exact values 0.049508... and 0.054615...
1 mark
A1 Largest n = 18
Depends on the second M1 and needs values showing that 18 is acceptable and 19 is not. cao.

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