Edexcel International A Level Maths · S2 · Statistics 2
Sampling and hypothesis testing
28 free, high-quality past-paper-style questions on Sampling and hypothesis testing for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 243 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-S2-4-B-01S2 · A2Easy6 marks
A university has 9600 enrolled students. The students' union wants to find out how students travel to the campus. The university registry holds an up-to-date list of all enrolled students.
(a)
Identify the population.
(1)
(b)
Suggest a suitable sampling frame and state the sampling unit.
(2)
(c)
Give one advantage and one disadvantage of the students' union taking a census rather than a sample.
(2)
(d)
The university's engineering laboratory receives a delivery of 500 concrete test cubes. The crushing strength of a cube is measured by loading it until it breaks. Explain why the laboratory must use a sample rather than a census to check the crushing strength of the delivery.
(1)
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Answer
Mark
Notes
Part (a)
the 9600 students enrolled at the university
B1
Part (b)
sampling frame: the registry's list of enrolled students
B1
sampling unit: an enrolled student
B1
Part (c)
advantage: every student is included, so the result is exact for the university (no sampling variation)
B1
disadvantage: contacting 9600 students and processing their replies would take a long time / cost a lot
B1
Part (d)
the test destroys each cube, so testing all 500 would leave none to use
B1
Common error: The sampling frame is the list you choose from; the sampling unit is one member of it. Name both in the context of the question, not just “a list” or “a person”.
At a charity stall, each prize envelope contains £0, £2 or £10. A large number of envelopes are used, and the proportions containing £0, £2 and £10 are 0.7, 0.2 and 0.1 respectively. A random sample of two envelopes is taken; the two amounts may be treated as independent. A is the mean amount, in pounds, of the two envelopes in the sample.
(a)
List all the possible samples of two envelopes, ignoring order.
(1)
(b)
Find the sampling distribution of A.
(4)
(c)
Find P(A > 3).
(1)
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Answer
Mark
Notes
Part (a)
{0, 0}, {0, 2}, {0, 10}, {2, 2}, {2, 10}, {10, 10}: all six and no others
B1
Part (b)
values of A: 0, 1, 5, 2, 6, 10 for the six samples
B1
probabilities with the factor 2 for two different amounts, e.g. P({0, 2}) = 2(0.7)(0.2)
M1
at least four of P(A = 0) = 0.49, P(A = 1) = 0.28, P(A = 2) = 0.04, P(A = 5) = 0.14, P(A = 6) = 0.04, P(A = 10) = 0.01 correct
A1
all six correct, presented as a distribution
A1
Part (c)
0.14 + 0.04 + 0.01 = 0.19 (ft their distribution)
B1
Exam tip: When the two values in a sample are different, there are two orders in which they can occur, so multiply by 2. Check that your probabilities add up to 1.
A seed company states that 25% of its sweet pea seeds produce white flowers. A gardener believes that the proportion is higher than this. She plants 40 of the seeds, all of which flower, and records the number, X, that produce white flowers. She tests the company's statement at the 5% level of significance.
(a)
Write down suitable hypotheses for the test.
(1)
(b)
Find the critical region for X. The probability of a value in the critical region should be as close as possible to 5% without exceeding 5%.
(2)
(c)
State the actual significance level of the test.
(1)
(d)
15 of the plants produce white flowers. State the conclusion of the test in context.
The server of a website fails at random at a mean rate of 0.5 times per week. After new hardware is installed, the server fails once in a period of 10 weeks. The owner wants to test, at the 5% level of significance, whether the mean rate of failures has decreased.
(a)
Write down suitable hypotheses.
(1)
(b)
State the distribution of the number of failures, X, in 10 weeks if H0 is true.
(1)
(c)
Find the probability needed for the test.
(2)
(d)
Carry out the test, stating your conclusion in context.
there is sufficient evidence to suggest that the mean rate of server failures has decreased ft
A1
Common error: Using P(X = 1) instead of P(X ≤ 1). The test needs the probability of a result at least as extreme as the one observed, in the direction of H1.
An allotment association manages the site shown in Figure 1. It wants to find out how much water its plot-holders use. The association keeps a register of plot-holders, but the register has not been updated recently. Each plot that is let has exactly one plot-holder.
(a)
Identify the population and use Figure 1 to find its size.
(2)
(b)
The association uses its register as the sampling frame. Use Figure 1 to state how many plot-holders could never be chosen, and explain why this could matter.
(2)
(c)
State the sampling unit.
(1)
(d)
Give one reason why a census would be practical in this situation.
(1)
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Answer
Mark
Notes
Part (a)
the people who currently rent a plot on the site (the plot-holders)
B1
40 − 3 vacant plots = 37 plot-holders
B1
Part (b)
4 (the hatched plots)
B1
the newest plot-holders cannot be chosen, so their water use is not represented and the results may be biased
B1
Part (c)
one plot-holder
B1
Part (d)
the population is small (37 people on one site), so it is quick and cheap to ask every plot-holder
B1
Common error: The sampling frame is only as good as the list. Anyone missing from the list can never be in the sample, however randomly the sample is chosen.
Figure 1 shows the probability distribution of N, the number of people in a car arriving at a car park. A random sample of two cars is taken, and the numbers of people in the two cars are independent. R is the range (largest value minus smallest value) of the two numbers in the sample.
(a)
Use Figure 1 to find μ = E(N), and explain why μ is not a statistic.
(2)
(b)
List all the possible samples, ignoring order.
(1)
(c)
Find the sampling distribution of R.
(3)
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Answer
Mark
Notes
Part (a)
μ = 1(0.5) + 2(0.3) + 3(0.2) = 1.7
B1
μ is a parameter of the population; it is not calculated from the sample values
B1
Part (b)
{1, 1}, {1, 2}, {1, 3}, {2, 2}, {2, 3}, {3, 3}: all six and no others
B1
Part (c)
R takes the values 0, 1, 2
B1
e.g. P(R = 1) = 2(0.5)(0.3) + 2(0.3)(0.2), using the factor 2 for two different values
M1
P(R = 0) = 0.38, P(R = 1) = 0.42, P(R = 2) = 0.20
A1
Exam tip: A range of 0 means both values are equal, so add the three ‘matching pair’ probabilities: 0.5² + 0.3² + 0.2². Check the final probabilities add up to 1.
Over a long period, 30% of the customers at a juice bar have chosen a smoothie. After a new smoothie menu is introduced, the owner believes that this proportion has increased. She records the number, X, of customers who choose a smoothie in a random sample of 12 customers, and tests her belief at the 5% level of significance. Figure 1 shows the distribution of X if the proportion is still 30%; the probabilities for x ≥ 5 are written above the bars, to 4 decimal places.
(a)
Write down suitable hypotheses.
(1)
(b)
Using Figure 1, find the critical region for X. The probability of a value in the critical region should be as close as possible to 5% without exceeding 5%.
(2)
(c)
State the actual significance level of the test.
(1)
(d)
In the sample, 6 customers choose a smoothie. State the conclusion of the test in context.
6 is not in the critical region, so do not reject H0
M1
there is insufficient evidence to suggest that the proportion of customers choosing a smoothie has increased
A1
Exam tip: Build an upper tail from the far end: add the bar probabilities from the top down until the total would pass 5%. The last total that stays at or below 5% gives the critical region.
A machine on a bottling line stops at random at a mean rate of 2 times per 8-hour shift. After the machine is serviced, the stoppages during the next three shifts are recorded, as shown in Figure 1. The engineer wants to test, at the 10% level of significance, whether the mean rate of stoppages has decreased.
(a)
Write down suitable hypotheses.
(1)
(b)
Use Figure 1 to state the number of stoppages observed, and state the distribution of the number of stoppages, X, in three shifts if H0 is true.
(2)
(c)
Find the probability needed for the test.
(2)
(d)
Carry out the test, stating your conclusion in context. State also whether the conclusion would be the same at the 5% level.
(2)
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Answer
Mark
Notes
Part (a)
H0: λ = 2, H1: λ < 2 (per shift) (or λ = 6 and λ < 6 per three shifts)
B1
Part (b)
2 stoppages
B1
X ~ Po(6)
B1
Part (c)
P(X ≤ 2)
M1
= 0.0620
A1
Part (d)
0.0620 < 0.10, so reject H0: there is sufficient evidence to suggest that the mean rate of stoppages has decreased ft
A1
at 5%: 0.0620 > 0.05, so H0 would not be rejected; the conclusion would be different
B1
Common error: Scaling the rate to the wrong period. Figure 1 covers three shifts, so the mean under H0 is 3 × 2 = 6, not 2.
Records at a clinic show that the number of nights a patient stays after a minor operation is 1, 2 or 3, with probabilities ½, ⅓ and ⅙ respectively. A random sample of the records of three patients is taken, and M is the largest number of nights in the sample. The three values may be treated as independent.
(a)
Explain why M is a statistic.
(1)
(b)
Show that P(M = 1) = ⅛.
(1)
(c)
Find the sampling distribution of M.
(3)
(d)
Find E(M).
(2)
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Answer
Mark
Notes
Part (a)
M is calculated from the sample values only and involves no unknown parameters
B1
Part (b)
all three stays are 1 night: (½)3 = ⅛
B1
Part (c)
P(M ≤ 2) = (⅚)3 = 125/216, or a full listing of samples
Exam tip: For the largest value, P(M ≤ k) = P(all three ≤ k) is quick to find; subtract consecutive values to get P(M = k). Listing every sample also works but takes longer.
Over a long period, 35% of the customers at a farm shop have paid in cash. The manager wants to test whether this proportion has changed. A random sample of 25 customers is taken and the number, X, who pay in cash is recorded.
(a)
Write down suitable hypotheses.
(1)
(b)
Using a 5% level of significance, find the critical region for X. The probability in each tail should be as close as possible to 2.5%.
(4)
(c)
Find the actual significance level of the test.
(1)
(d)
In the sample, 4 customers pay in cash. State, with a reason, the conclusion of the test in context.
(2)
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Answer
Mark
Notes
Part (a)
H0: p = 0.35, H1: p ≠ 0.35
B1
Part (b)
X ~ B(25, 0.35) and lower tail: P(X ≤ 3) = 0.0097, P(X ≤ 4) = 0.0320
there is sufficient evidence to suggest that the proportion of customers who pay in cash has changed ft
A1
Model sketch
Figure: distribution of X under H₀, X ~ B(25, 0.35), with the critical region X ≤ 4 or X ≥ 14 in red
Exam tip: “As close as possible to 2.5%” allows a tail just above 2.5%: compare how far each possible boundary probability is from 0.025. Here P(X ≤ 4) = 0.0320 is closer than 0.0097, so the actual significance level (0.0575) ends up slightly above 5%.
Along a stretch of river, sightings of kingfishers occur at random at a mean rate of 1.5 per hour. After building work starts nearby, a volunteer suspects that the rate of sightings has fallen. She watches the river for 3 hours and counts the number, X, of sightings. She tests her suspicion at the 10% level of significance.
(a)
State the distribution of X if the rate has not changed.
(1)
(b)
Write down suitable hypotheses.
(1)
(c)
Find the critical region for X. The probability of a value in the critical region should be as close as possible to 10%.
(3)
(d)
State the actual significance level of the test.
(1)
(e)
She sees 1 kingfisher in the 3 hours. State the conclusion of the test in context.
(2)
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Answer
Mark
Notes
Part (a)
X ~ Po(4.5)
B1
Part (b)
H0: λ = 1.5, H1: λ < 1.5 (per hour)
B1
Part (c)
uses the lower tail of Po(4.5)
M1
P(X ≤ 1) = 0.0611 and P(X ≤ 2) = 0.1736
A1
critical region X ≤ 1
A1
Part (d)
0.0611
B1
Part (e)
1 is in the critical region, so reject H0
M1
there is sufficient evidence to suggest that the rate of kingfisher sightings has fallen
A1
Model sketch
Figure: distribution of X under H₀, X ~ Po(4.5), with the critical region X ≤ 1 in red
Common error: Using Po(1.5) for a 3-hour count. Scale the mean to the length of time observed: 1.5 × 3 = 4.5.
A ferry company claims that 85% of its crossings arrive on time. A regular passenger believes the true proportion is lower. She records a random sample of 20 crossings and finds the number, X, that arrive on time. She tests the claim at the 5% level of significance.
(a)
Write down suitable hypotheses.
(1)
(b)
Let Y = 20 − X be the number of crossings that do not arrive on time. State the distribution of Y if H0 is true.
(1)
(c)
Find the critical region for X. The probability of a value in the critical region should be as close as possible to 5% without exceeding 5%.
(3)
(d)
State the actual significance level.
(1)
(e)
13 of the 20 crossings arrive on time. State the conclusion of the test in context.
(2)
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Answer
Mark
Notes
Part (a)
H0: p = 0.85, H1: p < 0.85
B1
Part (b)
Y ~ B(20, 0.15)
B1
Part (c)
P(X ≤ 13) = P(Y ≥ 7) = 1 − 0.9781 = 0.0219
M1
P(X ≤ 14) = P(Y ≥ 6) = 1 − 0.9327 = 0.0673 > 0.05
A1
critical region X ≤ 13
A1
Part (d)
0.0219
B1
Part (e)
13 is in the critical region, so reject H0
M1
there is sufficient evidence to suggest that fewer than 85% of crossings arrive on time
A1
Model sketch
Figure: distribution of X under H₀, X ~ B(20, 0.85), with the critical region X ≤ 13 in red
Exam tip: The tables stop at p = 0.5. Count the other outcome instead: X ≤ 13 on time is the same event as Y ≥ 7 late.
A national survey found that 35% of households in a country own a dog. A researcher believes that the proportion in one particular city is different. She takes a random sample of 200 households in the city and finds that 84 of them own a dog. Let X be the number of households in a sample of 200 that own a dog.
(a)
Write down suitable hypotheses.
(1)
(b)
Explain why a Normal distribution may be used to approximate the number of households owning a dog, and state its mean and variance under the null hypothesis.
(3)
(c)
Using this approximation, test at the 5% level of significance whether the proportion of households owning a dog in the city differs from 35%.
(5)
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Answer
Mark
Notes
Part (a)
H0: p = 0.35, H1: p ≠ 0.35
B1
Part (b)
n is large and p is not too far from 0.5 (np = 70 and n(1 − p) = 130 are both large)
B1
mean 70
B1
variance 45.5, so Y ~ N(70, 45.5)
B1
Part (c)
P(X ≥ 84) ≈ P(Y > 83.5), continuity correction
M1
standardises: z = (83.5 − 70)/√45.5
M1
z = 2.00, probability = 1 − 0.9772 = 0.0228
A1
compares with 0.025 (two-tailed): 0.0228 < 0.025, reject H0
M1
there is sufficient evidence to suggest that the proportion of households in the city owning a dog is different from 35%
A1
Exam tip: In a two-tailed test, compare the tail probability with half the significance level, here 0.025, not 0.05.
During a meteor shower, an astronomer has previously observed meteors at random at a mean rate of 12 per hour. On a night with thin cloud she counts 25 meteors in 3 hours. She wants to test, at the 5% level of significance, whether the cloud reduces the mean rate of meteors she observes.
(a)
Write down suitable hypotheses.
(1)
(b)
State the distribution of the number of meteors, X, observed in 3 hours under H0, and a suitable Normal approximation to it.
(2)
(c)
Carry out the test, stating your conclusion in context.
(5)
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Answer
Mark
Notes
Part (a)
H0: λ = 12, H1: λ < 12 (per hour)
B1
Part (b)
X ~ Po(36)
B1
Y ~ N(36, 36)
B1
Part (c)
P(X ≤ 25) ≈ P(Y < 25.5)
M1
z = (25.5 − 36)/6 = −1.75
M1
probability = 1 − 0.9599 = 0.0401
A1
0.0401 < 0.05, reject H0
M1
there is sufficient evidence to suggest that the mean rate of meteors observed is lower on a cloudy night
A1
Exam tip: For a lower tail the continuity correction goes up: P(X ≤ 25) ≈ P(Y < 25.5), because the bar for 25 runs from 24.5 to 25.5.
Figure 1 shows a spinner with three sectors, marked with scores 2, 5 and 10. The spinner is spun three times; the three scores are independent. M is the median of the three scores.
(a)
Use Figure 1 to write down the probability distribution of the score on one spin.
Exam tip: The median of three values equals a score k at the two ends of the scale only when at least two of the three values are k. Find the end values this way and get the middle one by subtracting from 1.
Returns to an online clothing shop occur at random at a mean rate of 3 per day. After the shop adds a new sizing guide to its website, the manager wants to test, at the 10% level of significance, whether the mean rate of returns has changed. The number of returns, X, in a 3-day period is recorded. Figure 1 shows the cumulative probabilities P(X ≤ x) for X ~ Po(9), with four values labelled.
(a)
Explain why Po(9) is used for X if the rate has not changed, and write down suitable hypotheses.
(2)
(b)
Using Figure 1, find the critical region for X. The probability in each tail should be as close as possible to 5% without exceeding 5%.
(4)
(c)
Find the actual significance level of the test.
(1)
(d)
There are 4 returns in the 3-day period. State the conclusion of the test in context.
4 is not in the critical region, so do not reject H0
M1
there is insufficient evidence to suggest that the mean rate of returns has changed
A1
Exam tip: A cumulative graph gives the lower tail directly. For the upper tail, P(X ≥ c) = 1 − P(X ≤ c − 1): read the point one step to the left of the boundary.
A supplier of solar garden lights states that 10% of its lights fail a waterproofing test. A retailer suspects that the proportion is higher. A random sample of 30 lights is tested, and Figure 1 shows the results. The retailer tests the supplier's statement at the 5% level of significance.
(a)
Write down suitable hypotheses.
(1)
(b)
Find the critical region for the number of lights, Y, that fail the test. The probability of a value in the critical region should be as close as possible to 5% without exceeding 5%.
(3)
(c)
State the actual significance level of the test.
(1)
(d)
Use Figure 1 to complete the test, stating your conclusion in context.
(2)
(e)
State one assumption about the lights needed for the binomial model to be valid.
7 lights failed; 7 is in the critical region, so reject H0
M1
there is sufficient evidence to suggest that more than 10% of the lights fail the waterproofing test
A1
Part (e)
the lights fail independently of each other / each light has the same probability of failing
B1
Common error: Stopping at the first tail probability below 5% without checking the next value. Quote both P(Y ≥ 7) and P(Y ≥ 6) to show that 7 is the boundary.
On a hillside, orchids grow at random at a mean rate of 1.25 per square metre. After sheep are removed from the hillside, an ecologist believes that the rate has increased. She marks out the plot shown in Figure 1 and counts the orchids in it. She tests her belief at the 5% level of significance.
(a)
Give two conditions needed for the number of orchids in a plot to be modelled by a Poisson distribution.
(2)
(b)
Write down suitable hypotheses.
(1)
(c)
Use Figure 1 to carry out the test, stating your conclusion in context.
(5)
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Answer
Mark
Notes
Part (a)
orchids occur independently of each other
B1
orchids occur singly and at a constant mean rate per square metre
B1
Part (b)
H0: λ = 1.25, H1: λ > 1.25 (per m2)
B1
Part (c)
area 3 × 2 = 6 m2, so X ~ Po(7.5) under H0
B1
12 orchids in the plot
B1
P(X ≥ 12) = 1 − P(X ≤ 11) = 1 − 0.9208
M1
= 0.0792 > 0.05, so do not reject H0
A1
there is insufficient evidence to suggest that the rate of orchids has increased ft
A1
Exam tip: Count carefully from a figure and say what you counted. Then scale the rate to the area: 1.25 per m2 over 6 m2 gives a mean of 7.5.
A city's transport plan assumed that 25% of people who work in the city centre walk to work. A researcher believes that this proportion is now higher. She asks a random sample of 120 people who work in the city centre how they mainly travel to work, and Figure 1 shows the results.
(a)
Write down suitable hypotheses.
(1)
(b)
Use Figure 1 to write down the number of people in the sample who walk to work.
(1)
(c)
X is the number of people in a sample of 120 who walk to work. Explain why a Normal distribution may be used to approximate the distribution of X under H0, and state its mean and variance.
(3)
(d)
Using this approximation, test the researcher's belief at the 5% level of significance.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
H0: p = 0.25, H1: p > 0.25
B1
Part (b)
39
B1
Part (c)
n is large and np = 30 and n(1 − p) = 90 are both greater than 5
B1
mean 120 × 0.25 = 30
B1
variance 120 × 0.25 × 0.75 = 22.5
B1
Part (d)
P(X ≥ 39) ≈ P(Y > 38.5), Y ~ N(30, 22.5)
M1
z = (38.5 − 30)/√22.5 = 1.79
M1
P = 1 − 0.9633 = 0.0367
A1
0.0367 < 0.05: reject H0; there is sufficient evidence to suggest that more than 25% of people who work in the city centre walk to work
A1
Exam tip: For an upper tail the continuity correction goes down: P(X ≥ 39) ≈ P(Y > 38.5). Read whole-number frequencies off the chart using the gridlines before you approximate.
A water company receives complaints at random at a mean rate of 15 per day. After a pipe-replacement programme, the company wants to test, at the 5% level of significance, whether the mean rate of complaints has decreased. Figure 1 shows the number of complaints on each of five days after the programme.
(a)
Write down suitable hypotheses.
(1)
(b)
Use Figure 1 to find the total number of complaints in the five days, and state the distribution of the total, X, under H0.
(2)
(c)
Explain why a Normal approximation is suitable here.
(1)
(d)
Carry out the test, stating your conclusion in context.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
H0: λ = 15, H1: λ < 15 (per day) (or λ = 75 and λ < 75 for five days)
B1
Part (b)
11 + 14 + 9 + 15 + 12 = 61
B1
X ~ Po(75)
B1
Part (c)
λ = 75 is large (greater than 10), so Po(75) ≈ N(75, 75)
B1
Part (d)
P(X ≤ 61) ≈ P(Y < 61.5), Y ~ N(75, 75)
M1
z = (61.5 − 75)/√75 = −1.56
M1
P = 1 − 0.9406 = 0.0594
A1
0.0594 > 0.05: do not reject H0; there is insufficient evidence to suggest that the mean rate of complaints has decreased
A1
Common error: Testing one day at a time. Combine the five days into one count; its mean under H0 is 5 × 15 = 75, and the Normal approximation then has variance 75 too.
Over a long period, 20% of the mobile phones returned to a repair centre have had a battery fault. After the manufacturer changes its battery supplier, the manager of the centre wants to test whether the proportion of returned phones with a battery fault has changed. A random sample of 50 returned phones is inspected and the number, X, with a battery fault is recorded.
(a)
Write down suitable hypotheses.
(1)
(b)
Using a 5% level of significance, find the critical region for X. The probability in each tail should be as close as possible to 2.5% without exceeding 2.5%.
(4)
(c)
Find the actual significance level of the test.
(1)
(d)
In the sample, 5 phones have a battery fault. State the conclusion of the test in context.
(2)
(e)
The manager now believes that the new supplier has reduced the proportion. A further random sample of 200 returned phones contains 29 with a battery fault. Using a suitable approximation, test this belief at the 5% level of significance.
(5)
Show mark scheme
Answer
Mark
Notes
Part (a)
H0: p = 0.2, H1: p ≠ 0.2
B1
Part (b)
X ~ B(50, 0.2) and lower tail: P(X ≤ 4) = 0.0185, P(X ≤ 5) = 0.0480
5 is not in the critical region, so do not reject H0
M1
there is insufficient evidence to suggest that the proportion of returned phones with a battery fault has changed ft
A1
Part (e)
H0: p = 0.2, H1: p < 0.2, and Y ~ N(40, 32)
B1
P(X ≤ 29) ≈ P(Y < 29.5)
M1
z = (29.5 − 40)/√32 = −1.86
M1
probability = 1 − 0.9686 = 0.0314
A1
0.0314 < 0.05: reject H0; there is sufficient evidence to suggest that the proportion of returned phones with a battery fault has been reduced
A1
Model sketch
Figure: distribution of X under H₀, X ~ B(50, 0.2), with the critical region X ≤ 4 or X ≥ 17 in red
Exam tip: A larger sample can detect a change that a small one cannot. Here 5 out of 50 (10%) was not significant in the two-tailed test, but 29 out of 200 (14.5%) is significant in the one-tailed test.
Calls to a taxi firm's booking line arrive at random at a mean rate of 0.5 per minute. After the firm launches a booking app, the manager wants to test whether the mean rate of calls to the booking line has changed.
(a)
Write down suitable hypotheses.
(1)
(b)
The number of calls, X, in a 15-minute period is recorded. Using a 10% level of significance, find the critical region for X. The probability in each tail should be as close as possible to 5%.
(4)
(c)
Find the actual significance level of the test.
(1)
(d)
In the 15-minute period there are 3 calls. State, with a reason, the conclusion of the test in context.
(2)
(e)
The manager now suspects that the rate of calls has decreased. The number of calls, W, in a 90-minute period is to be recorded. Using a suitable approximation, find the critical region for W for a test at the 5% level of significance, where the probability of a value in the critical region should be as close as possible to 5% without exceeding 5%. State the conclusion if there are 31 calls in the 90 minutes.
(5)
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Answer
Mark
Notes
Part (a)
H0: λ = 0.5, H1: λ ≠ 0.5 (per minute)
B1
Part (b)
X ~ Po(7.5) and lower tail: P(X ≤ 2) = 0.0203, P(X ≤ 3) = 0.0591
there is sufficient evidence to suggest that the mean rate of calls to the booking line has changed from 0.5 per minute ft
A1
Part (e)
under H0W ~ Po(45), approximated by N(45, 45)
B1
(w + 0.5 − 45)/√45 ≤ −1.6449, continuity correction and percentage point used
M1
w ≤ 33.47
A1
critical region W ≤ 33
A1
31 is in the critical region: reject H0; there is sufficient evidence to suggest that the mean rate of calls to the booking line has decreased ft
A1
Exam tip: Read the rule in the question. “As close as possible to 5%” lets a tail go slightly above 5% (here 0.0591); “without exceeding 5%” does not, and would give X ≤ 2 instead. With a Normal approximation, set the standardised boundary equal to the percentage point and round the integer boundary so the probability stays below the level.
A cereal manufacturer states that 10% of its boxes contain a collectable card. A child believes that the proportion is lower than this. Boxes may be treated as independent.
(a)
The child's family buys n boxes and none of them contains a card. Find the smallest value of n for which this result would be significant at the 5% level, using a one-tailed test.
(3)
(b)
The family buys 50 boxes. Write down suitable hypotheses and find the critical region for the number of boxes, X, containing a card, at the 5% level of significance. The probability of a value in the critical region should be as close as possible to 5% without exceeding 5%.
(3)
(c)
State the actual significance level of this test.
(1)
(d)
The 50 boxes contain 2 cards. State the conclusion of the test in context.
(2)
(e)
A shopkeeper believes that the proportion of boxes containing a card is not 10%. In a random sample of 50 boxes, 10 contain a card. Test the shopkeeper's belief at the 5% level of significance.
2 is not in the critical region, so do not reject H0
M1
there is insufficient evidence to suggest that fewer than 10% of boxes contain a card
A1
Part (e)
H1: p ≠ 0.1; P(X ≥ 10) = 1 − 0.9755 = 0.0245
M1
0.0245 < 0.025: reject H0; there is sufficient evidence to suggest that the proportion of boxes containing a card is not 10%
A1
Exam tip: “Smallest n” questions are solved with logarithms: the inequality sign stays the same when you take logs, but dividing by ln 0.9 (negative) reverses it.
A town has 6000 households. The council believes that more than 40% of households recycle their food waste, and plans a survey to test this belief. The council keeps a register of the addresses of all households in the town.
(a)
State the population and suggest a suitable sampling frame.
(2)
(b)
Give one reason why the council might prefer a sample to a census.
(1)
(c)
A random sample of 25 households is taken, and X is the number that recycle their food waste. Write down suitable hypotheses and find the critical region for X at the 5% level of significance. The probability of a value in the critical region should be as close as possible to 5% without exceeding 5%.
(3)
(d)
In the sample, 15 households recycle their food waste. State the actual significance level and the conclusion of the test in context.
(3)
(e)
A councillor says that the survey proves that more than 40% of households recycle their food waste. Comment on this statement.
(1)
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Answer
Mark
Notes
Part (a)
population: the 6000 households in the town
B1
sampling frame: the council's register of household addresses
B1
Part (b)
a sample is quicker / cheaper than surveying all 6000 households
there is sufficient evidence to suggest that more than 40% of households recycle their food waste
A1
Part (e)
it is not proof: if p were 0.4, a result in the critical region would still occur with probability 0.0344, so the test only gives evidence at the 5% level
B1
Common error: Writing “this proves …” or “H0 is false”. A hypothesis test only measures the strength of evidence; conclusions should say “sufficient evidence to suggest”.
A bus company states that 25% of the passengers on one of its routes pay with a travel card. A researcher wants to test whether this proportion is different. In a random sample of 20 passengers, X pay with a travel card. Sam proposes the critical region shown in Figure 1 for a test at the 5% level of significance.
(a)
Write down suitable hypotheses.
(1)
(b)
Use Figure 1 to write down Sam's critical region, and find its actual significance level.
(3)
(c)
Explain why Sam's critical region is not suitable, and find the critical region in which the probability in each tail is as close as possible to 2.5% without exceeding 2.5%.
(3)
(d)
In the sample, 2 passengers pay with a travel card. Compare the conclusions reached using Sam's critical region and the correct one.
(2)
(e)
In a second random sample of 150 passengers, 26 pay with a travel card. Using a suitable approximation, test at the 5% level of significance whether the proportion is different from 25%.
the lower tail 0.0913 is far above 0.025 (overall 10.5% > 5%)
B1
P(X ≤ 1) = 0.0243 ≤ 0.025, so lower tail X ≤ 1
M1
P(X ≥ 9) = 0.0409 > 0.025, so upper tail X ≥ 10; critical region X ≤ 1 or X ≥ 10
A1
Part (d)
with Sam's region, 2 would lead to rejecting H0
B1
with the correct region, 2 is not in it: do not reject H0; insufficient evidence to suggest that the proportion paying with a travel card is different from 25%
0.0192 < 0.025: reject H0; there is sufficient evidence to suggest that the proportion paying with a travel card is different from 25%
A1
Exam tip: In a two-tailed test each tail gets half the significance level. A critical region whose tails add up to well over the stated level is not a valid region at that level.
Flaws occur at random in a woven fabric at a mean rate of 0.8 per metre. After a new loom is installed, the mill manager wants to test whether the mean rate of flaws has changed. Figure 1 shows the flaws in a 5 m length of fabric from the new loom.
(a)
Write down suitable hypotheses and find the critical region for the number of flaws, X, in a 5 m length, using a 10% level of significance. The probability in each tail should be as close as possible to 5% without exceeding 5%.
(5)
(b)
Use Figure 1 to complete the test, stating your conclusion in context.
(2)
(c)
Find the actual significance level of the test.
(1)
(d)
Later, the manager suspects that the new loom has reduced the mean rate of flaws. A 60 m roll from the new loom has 36 flaws. Using a suitable approximation, test the manager's suspicion at the 5% level of significance.
0.0485 < 0.05: reject H0; there is sufficient evidence to suggest that the mean rate of flaws has decreased
A1
Exam tip: A result can be close to the edge without being in the critical region: 8 flaws has P(X ≥ 8) = 0.0511, just above 5%, so it is not significant. Always compare with the region you found, not with how far the value looks from the mean.
Electronic components are delivered in large batches. A buyer uses the two-stage sampling plan shown in Figure 1 to decide whether to accept a batch. A batch is satisfactory if the proportion, p, of faulty components is 0.05. Rejecting a batch can be treated as rejecting H0: p = 0.05 in favour of H1: p > 0.05. Components are faulty independently of each other.
(a)
For a satisfactory batch, find the probability that the batch is accepted after the first sample.
(1)
(b)
For a satisfactory batch, find the probability that a second sample is needed.
(2)
(c)
Show that the probability that a satisfactory batch is rejected is 0.113 to 3 significant figures.
(4)
(d)
For a satisfactory batch, find the expected number of components inspected.
(2)
(e)
The buyer wants the actual significance level of the plan to be at most 10%. State whether the plan in Figure 1 meets this. The buyer then changes the second stage so that the batch is rejected only if 3 or more of the second sample are faulty. Find the new probability of rejecting a satisfactory batch, and comment.
0.0897 ≤ 0.10, so the changed plan meets the requirement ft
A1
Exam tip: Follow every path in a flowchart that ends in ‘reject’ and add their probabilities. A path through two stages multiplies the probabilities at each stage because the two samples are independent.
Figure 1 shows five tokens in a bag. Two tokens are taken at random from the bag, without replacement, and X̄ is the mean of the numbers on the two tokens. The two tokens marked 2 are different tokens, so they are treated as distinct throughout.
(a)
Find μ, the mean of the numbers on the five tokens in the bag, and explain why X̄ is a statistic.
(2)
(b)
List all the possible samples of two tokens, ignoring order, labelling the two tokens marked 2 as 2a and 2b.
(2)
(c)
Find the sampling distribution of X̄.
(3)
(d)
Show that E(X̄) = μ.
(2)
(e)
Find P(|X̄ − μ| < 1).
(1)
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Answer
Mark
Notes
Part (a)
μ = (1 + 2 + 2 + 5 + 8)/5 = 3.6
B1
X̄ is calculated from the sample values only and involves no unknown parameters
B1
Part (b)
writing the 2s as 2a and 2b: {1, 2a}, {1, 2b}, {1, 5}, {1, 8}, {2a, 2b}, {2a, 5}, {2b, 5}, {2a, 8}, {2b, 8}, {5, 8}: at least 8 correct
M1
all 10 and no others, each with probability 1/10
A1
Part (c)
values of X̄: 1.5, 2, 3, 3.5, 4.5, 5, 6.5
B1
probability 2/10 for the values from two samples (1.5, 3.5 and 5)
M1
P = 0.2, 0.1, 0.1, 0.2, 0.1, 0.2, 0.1 respectively, all correct
2.6 < X̄ < 4.6: values 3, 3.5, 4.5, so P = 0.1 + 0.2 + 0.1 = 0.4
B1
Common error: Treating the two tokens marked 2 as one. They are different tokens, so a 1 with a 2 can happen in two ways; without replacement there are 10 equally likely pairs, not 7.
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