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Testing a claim about a population proportion \(p\) with the binomial distribution: writing hypotheses, finding a p-value or a critical region (one-tailed), and writing a conclusion in context. A complete test is typically worth around 4 to 8 marks.
Key facts
\(H_0: p = p_0\);\(H_1: p \gt p_0\), \(p \lt p_0\) or \(p \ne p_0\)
Under \(H_0\): \(X \sim B(n, p_0)\)
Upper tail: \(P(X \ge x) = 1 - P(X \le x - 1)\)
Lower tail: \(P(X \le x)\)
Critical region: the largest tail whose probability is no more than the significance level
Actual significance level = probability of the critical region under \(H_0\)
Notes
Setting up
Define \(p\) in context as the probability of success (the population proportion). \(H_0: p = p_0\), the claimed value.
Choose \(H_1\) from the wording: ‘increased’ or ‘more’ gives \(p \gt p_0\); ‘decreased’ or ‘less’ gives \(p \lt p_0\); ‘changed’ or ‘different’ gives \(p \ne p_0\)
Under \(H_0\), the number of successes in a sample of size \(n\) is \(X \sim B(n, p_0)\)
Method 1: p-value
For \(H_1: p \gt p_0\), find \(P(X \ge x) = 1 - P(X \le x - 1)\), where \(x\) is the observed value; for \(H_1: p \lt p_0\), find \(P(X \le x)\)
Method 2: critical region
Upper tail: the smallest \(c\) with \(P(X \ge c) \le\) the significance level gives the critical region \(X \ge c\). Lower tail: the largest \(c\) with \(P(X \le c) \le\) the significance level gives \(X \le c\)
Use a table of cumulative probabilities on your calculator and quote the probabilities either side of each boundary.
If a question asks for tails ‘as close as possible to’ a value instead, follow that wording.
If the observed value is in the critical region, reject \(H_0\)
Conclusion
Give the decision, then a conclusion in context that does not claim certainty, e.g. ‘there is evidence to suggest that the proportion of customers who use the app has increased’.
How to answer each type of question
M = method mark, A = answer mark, B = a mark on its own.
One-tailed test using a p-value
4 to 5 marksGrade D
Write the hypotheses, with \(p\) defined.
State the distribution under \(H_0\)
Find the probability of the observed value or a more extreme one.
Compare, decide and conclude in context.
Example. A company claims that 30% of its customers use its app. The manager believes that the proportion is higher. In a random sample of 25 customers, 12 use the app. Test the manager's belief at the 5% level of significance.
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\(H_0: p = 0.3\),\(H_1: p \gt 0.3\), where \(p\) is the proportion of customers who use the app
(B1)
Under \(H_0\), \(X \sim B(25, 0.3)\)
(M1)
\(P(X \ge 12) = 1 - P(X \le 11)\)
\(= 1 - 0.9558 = 0.0442\)
(A1)
\(0.0442 \lt 0.05\), so reject \(H_0\)
(M1)
There is sufficient evidence to suggest that more than 30% of customers use the app
(A1)
One-tailed critical region (lower tail)
3 to 4 marksGrade C
Use the distribution under \(H_0\)
Find the largest \(c\) with \(P(X \le c)\) no more than the significance level.
Quote the probabilities either side and write the region as \(X \le c\)
Example. A treatment works for 70% of patients. A researcher suspects that a new batch is less effective. It is given to 20 patients, and \(X\) is the number for whom it works. Find the critical region for a test at the 5% level, and state the actual significance level.
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\(H_0: p = 0.7\),\(H_1: p \lt 0.7\); \(X \sim B(20, 0.7)\)
Critical region \(X \le 10\) (A1); actual significance level 0.0480
(B1)
Smallest sample size for a significant result
3 marksGrade A*
Write the probability of the most extreme outcome in terms of \(n\)
Set it to be no more than the significance level.
Solve using logarithms and round up.
Example. A test of \(H_0: p = 0.2\) against \(H_1: p \lt 0.2\) is carried out at the 5% level using a sample of size \(n\). Find the smallest value of \(n\) for which observing no successes would lead to rejecting \(H_0\)
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\(P(X = 0) = 0.8^n \le 0.05\)
(M1)
\(n \ge \dfrac{\log 0.05}{\log 0.8} = 13.4\)
(M1)
\(n = 14\)
(A1)
(check: \(0.8^{13} = 0.0550\) and \(0.8^{14} = 0.0440\))
Don’t lose marks
Using \(P(X = x)\) instead of \(P(X \ge x)\) or \(P(X \le x)\) for the p-value.
Using \(1 - P(X \le x)\) for \(P(X \ge x)\)
Writing the critical region as a probability instead of as values of \(X\)
Conclusions that claim certainty, or that don't mention the context.
More tips
Memory tricks
Use your calculator's table mode for binomial CD to list \(P(X \le x)\) for many values at once and spot the boundary.
Write \(P(X \ge c) = 1 - P(X \le c - 1)\) every time: it avoids the most common slip.
Exam technique
For critical regions, show the probabilities either side of each boundary: they earn the method mark.
Write critical regions in terms of \(X\), e.g. \(X \le 3\) or \(X \ge 13\)
Read the tail rule in the question: ‘at most 2.5%’ and ‘as close as possible to 2.5%’ can give different answers.
What each grade needs
What you need to be able to do, from the first marks up to the top grade.
Grade E
Write hypotheses for a binomial test\(H_0: p = p_0\), with \(H_1\) from the wording and \(p\) defined in context.
Grade D
Carry out a one-tailed test with a p-valueFind \(P(X \ge x)\) or \(P(X \le x)\) under \(H_0\) and compare it with the significance level.
Grade C
Find a one-tailed critical regionFind the boundary where the tail probability is no more than the significance level.
Grade A
Carry out a complete test in contextHypotheses, distribution, probability or critical region, decision and conclusion in context.
Grade A*
Find a sample size that gives significanceE.g. find the smallest \(n\) for which \(X = 0\) is in the critical region.
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
With the standard treatment, 40% of patients with an illness recover within a week. A doctor is concerned that a cheaper treatment reduces the proportion of patients who recover within a week. A test of \(H_0: p = 0.4, \ H_1: p \lt 0.4\) is carried out at the 1% level of significance, where \(p\) is the proportion of patients who recover within a week. In a random sample of 26 patients, 7 recover within a week.
Using \(X \sim B(26, 0.4)\), find \(P(X \le 7)\) and state, with a reason, whether \(H_0\) is rejected.[3]
Show the answer and mark scheme
Answer: \(P(X \le 7) = 0.1216\). Since \(0.1216 \gt 0.01\), do not reject \(H_0\).
M1 for \(P(X \le 7)\)
A1 for awrt 0.1216
A1ft for a correct decision from a comparison with 0.01
Question 2Medium8 marks
A game designer claims that a spinner lands on red with probability \(\frac{1}{4}\). A player believes that it lands on red less often than this. In 40 spins, the spinner lands on red 4 times. The player carries out a test at the 5% level of significance and writes: \(H_0: p = 0.1\), \(H_1: p \lt 0.1\) \(P(X = 4) = 0.0113\), which is less than 0.05, so reject \(H_0\).
(a) Identify two errors in the player’s working.[2]
(b) Carry out the test correctly.[4]
(c) Find the critical region for this test.[2]
Show the answer and mark scheme
(a)Answer: The hypotheses should use the claimed value \(p = 0.25\), not the sample proportion 0.1; the player should find \(P(X \le 4)\), not \(P(X = 4)\).
B1 for the hypotheses should be about the claimed population proportion: \(H_0: p = 0.25\), \(H_1: p \lt 0.25\) (0.1 is the sample proportion)
B1 for the probability should be \(P(X \le 4)\) (the observed value or a more extreme one), not \(P(X = 4)\)
Worked solution: The hypotheses must be about the population parameter under the designer’s claim, \(p = 0.25\); 0.1 is just the sample proportion. And the test needs the probability of a result at least as extreme as the one observed, \(P(X \le 4)\), not the probability of exactly 4.
(b)Answer: \(H_0: p = 0.25\), \(H_1: p \lt 0.25\); \(X \sim \mathrm{B}(40, 0.25)\), \(P(X \le 4) = 0.0160 \lt 0.05\): reject \(H_0\). There is evidence that the spinner lands on red less often than \(\frac{1}{4}\).
B1 for \(H_0: p = 0.25\), \(H_1: p \lt 0.25\)
M1 for \(X \sim \mathrm{B}(40, 0.25)\) and \(P(X \le 4)\)
A1 for 0.0160 (awrt 0.016)
A1 for reject \(H_0\), with a conclusion in context: evidence that the spinner lands on red less often than the designer claims
Worked solution: \(H_0: p = 0.25\), \(H_1: p \lt 0.25\). Under \(H_0\), \(X \sim \mathrm{B}(40, 0.25)\). \(P(X \le 4) = 0.0160 \lt 0.05\), so reject \(H_0\): there is evidence that the spinner lands on red less often than \(\frac{1}{4}\) of the time.
(c)Answer: \(X \le 5\)
M1 for \(P(X \le 5) = 0.0433\) and \(P(X \le 6) = 0.0962\)
A1 for \(X \le 5\)
Worked solution: \(P(X \le 5) = 0.0433 \le 0.05\) and \(P(X \le 6) = 0.0962 \gt 0.05\), so the critical region is \(X \le 5\).
Question 3Hard6 marks
At a restaurant, 35% of customers buy a dessert. The owner is concerned that higher prices have reduced the proportion of customers who buy a dessert. A random sample of 18 customers is taken, and \(X\) is the number who buy a dessert.
(a) Find the critical region for a test of \(H_0: p = 0.35, \ H_1: p \lt 0.35\) at the 1% level of significance, and state the actual significance level.[4]
(b) In the sample, 0 buy a dessert. State, with a reason, the conclusion of the test in context.[2]
Show the answer and mark scheme
(a)Answer: \(X \le 1\); actual significance level 0.0046.
M1 for \(X \sim B(18, 0.35)\)
A1 for \(P(X \le 1) = 0.0046\) and \(P(X \le 2) = 0.0236\)
A1 for \(X \le 1\)
A1 for 0.0046
(b)Answer: 0 is in the critical region, so reject \(H_0\). There is sufficient evidence to suggest that the proportion of customers who buy a dessert has decreased.
M1 for a correct decision from their critical region