Statistical sampling - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 18 exam-style questions on Statistical sampling 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.

18 questions
2–8 marks each
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18 questions, marked as you go

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

A secondary school has 1140 students, listed in alphabetical order of surname. A teacher wants a systematic sample of 60 of these students for a survey about how students travel to school.

Explain how the teacher could take this sample.

Give one disadvantage of systematic sampling compared with simple random sampling.

Model answer

Interval: \(1140 \div 60 = 19\)
Number the students on the list from 1 to 1140.
Choose a number at random from 1 to 19 to give the first student, then take every 19th student after that (for example 7, 26, 45, ...) until 60 students have been chosen.
Disadvantage: once the first student is chosen the rest are fixed, so the sample is not completely random; it could also be biased if the list had a pattern repeating every 19 names.

Explanation

The interval is the population size divided by the sample size, here 19, and the first student must be chosen at random from the first 19 so that every student has the same chance of selection. After the random start the selection is fixed, so not every possible sample of 60 can occur. Students often forget the random start, which turns the method into a non-random one.

Mark scheme

1 mark
B1 Finds the interval \(1140 \div 60 = 19\): every 19th student on the list is chosen
Accept "one student in every 19".
1 mark
B1 The first student is chosen at random from the first 19 on the list (numbers 1 to 19), then every 19th student after that is taken (start, start + 19, start + 38, ...) until 60 students are chosen
Must have a random start within the first 19. Starting at the first student on the list (not random) scores B0.
1 mark
B1 A disadvantage, e.g. once the start is chosen the rest of the sample is fixed, so it is not completely random (not every possible sample of 60 can be chosen); or it can be biased if the list has a pattern that repeats every 19 names
Do not accept comments about time or cost, or "not representative" with no reason.

Question 2

4 marks

A sixth-form college has 683 students. The student council wants to choose a simple random sample of 40 of these students to answer questions about the college canteen.

Describe how the student council could select the sample.

Give one disadvantage of using a simple random sample in this situation.

Model answer

Get a list of all 683 students and number them from 001 to 683.
Use a random number generator to produce 3-digit random numbers.
Ignore 000, any number above 683 and any number that has already been chosen.
Keep going until 40 different students have been chosen: these students form the sample.
One disadvantage: a full list of all 683 students is needed, and by chance the sample might contain very few Year 13 students, so the year groups might not be fairly represented.

Explanation

Every student must be on a numbered list so that random numbers can pick them, and repeats are ignored so that 40 different students are chosen. Each possible sample of 40 is then equally likely, so the method is free of selection bias. Its weaknesses are that it needs a full list of the population and that, by chance, it may under-represent a group such as one year.

Mark scheme

1 mark
B1 Obtain a list of all 683 students (a sampling frame) and number them 001 to 683 (or 1 to 683)
Must have both the list (or the idea of every student being included) and the numbering. For a "names in a hat" method, writing each of the 683 names on a separate slip replaces the numbering.
1 mark
B1 Use a random number generator (on a calculator or computer) or random number tables to produce random numbers
Accept "use a computer to choose 40 numbers from 1 to 683 at random". "Pick names out of a hat" (all 683 names) is also acceptable.
1 mark
B1 Ignore any repeats (and, with 3-digit random numbers, 000 and numbers above 683); select the students with the numbers obtained until 40 different students have been chosen
Must mention ignoring repeats (or choosing without replacement) and stopping when 40 have been chosen.
1 mark
B1 A disadvantage in context, e.g. a complete list of all 683 students is needed; or by chance the sample may contain very few students from one year group (or one course), so groups may not be represented in proportion
Do not accept "it is biased" (a simple random sample is free from selection bias), or comments about time or cost with no reason. "Not representative" on its own scores B0: it needs the idea that a particular group may be under- or over-represented by chance.

Question 3

5 marks

A hospital employs 1060 staff in four groups: 486 nurses, 312 support staff, 164 doctors and 98 administrative staff. The hospital manager wants to ask a stratified sample of 50 staff about changes to the staff car park, with each group represented in proportion to its size.

Find the number of staff from each group in the sample, explaining how you have made sure that the total is 50.

Describe how the staff in each group would then be chosen.

Give one reason why a stratified sample is suitable here.

Model answer

Nurses: \(\dfrac{486}{1060} \times 50 = 22.92\)
Support staff: \(\dfrac{312}{1060} \times 50 = 14.72\)
Doctors: \(\dfrac{164}{1060} \times 50 = 7.74\)
Administrative staff: \(\dfrac{98}{1060} \times 50 = 4.62\)
Rounding each to the nearest whole number gives \(23 + 15 + 8 + 5 = 51\), one too many, so the administrative share 4.62 (the share closest to rounding down) is rounded down instead.
Sample: 23 nurses, 15 support staff, 8 doctors and 4 administrative staff (total 50).
Within each group, the staff are listed and numbered, and random numbers are used to choose a simple random sample of the required size.
The groups work different shifts and hours, so they are likely to have different views about the car park, and stratifying makes sure each group is represented in proportion to its size.

Explanation

Each group gets its share of the 50 places in proportion to its size, \(\dfrac{\text{size}}{1060} \times 50\), and the staff in each group are then chosen at random. Rounding all four shares to the nearest whole number gives 51, so one share has to be rounded the other way; any whole numbers either side of the exact shares that total 50 are acceptable. Students often give 51 staff or forget that the selection within each stratum must be random.

Mark scheme

1 mark
M1 Finds at least two of the exact shares \(\dfrac{486}{1060} \times 50\), \(\dfrac{312}{1060} \times 50\), \(\dfrac{164}{1060} \times 50\), \(\dfrac{98}{1060} \times 50\); or a correct final answer of 23, 15, 8 and 4 with no working
An answer of 23, 15, 8 and 4 with no working scores 2 marks (M1 A1); the explanation of the total, the description and the reason are marked separately.
1 mark
A1 All four shares correct: 22.9, 14.7, 7.7 and 4.6 (to 1 d.p. or better: 22.924..., 14.716..., 7.735..., 4.622...)
May be implied by a correct final set of numbers.
1 mark
B1 Explains that rounding every share to the nearest whole number gives 23 + 15 + 8 + 5 = 51, so one number must be rounded down instead, and gives a set of whole numbers totalling 50, e.g. 23, 15, 8, 4
Accept any set totalling 50 in which each number is one of the two whole numbers either side of its exact share: 23, 15, 8, 4 (the usual choice: 4.62 has the smallest fractional part of the shares that round up), or 23, 15, 7, 5, or 23, 14, 8, 5, or 22, 15, 8, 5 (in the order nurses, support staff, doctors, administrative staff). A set that does not total 50, or a set with no explanation of the adjustment, scores B0.
1 mark
B1 Within each group, the staff are chosen by simple random sampling (e.g. a numbered list of the staff in that group and random numbers)
Must say the selection within each group is random. "Ask staff in each group until the numbers are reached" is quota sampling and scores B0.
1 mark
B1 A reason, e.g. the different groups of staff may work different shifts or hours, so they may have different views about the car park, and stratifying makes sure each group is represented in proportion to its size
Must relate to the groups of staff. Do not accept "it is quicker" or "it is cheaper".

Question 4

6 marks

A cinema wants to estimate the mean age of the people who watch films there on Saturdays. Three methods of choosing 50 people are suggested.

Method A: ask the first 50 people who arrive on one Saturday morning.
Method B: on one Saturday, choose one of the first 20 people to buy a ticket at random, and then ask every 20th person to buy a ticket after that, until 50 people have been asked.
Method C: an interviewer in the foyer asks people until 25 males and 25 females have been asked.

Name the type of sampling used in each method.

For Method A and for Method C, give a reason why the method may give a biased estimate of the mean age.

State, with a reason, which of the three methods you would recommend.

Model answer

Method A: opportunity sampling. Method B: systematic sampling. Method C: quota sampling.
Method A only reaches people at the morning showings, who may be mostly families with young children, so their ages are not typical of all Saturday customers.
Method C lets the interviewer choose whom to ask, so the sample is not random and may favour particular ages; the quotas control only gender, not age.
I would recommend Method B: it has a random start, it spreads the sample over the whole day's ticket sales and the interviewer cannot choose who is asked.

Explanation

The three methods are opportunity, systematic and quota sampling. Method A is biased towards the morning audience, and Method C depends on the interviewer's choices, so neither is random. Method B, with its random start and selection spread over the whole day, is the best of the three; a reason in context is needed for each mark.

Mark scheme

1 mark
B1 Method A: opportunity (convenience) sampling
Condone misspelling.
1 mark
B1 Method B: systematic sampling
1 mark
B1 Method C: quota sampling
Do not accept stratified sampling: the people are not chosen at random within each group.
1 mark
B1 Method A: only people at the morning showings are asked, and the ages of morning audiences (e.g. families with children at children's films) are unlikely to be typical of the whole day
Must be in context: refer to the morning, or to the first people to arrive, being of different ages from the rest.
1 mark
B1 Method C: the interviewer chooses whom to ask, so the sample is not random (e.g. the interviewer may approach people of a particular age), and quotas for gender do nothing to control age
Accept "not random, so interviewer bias". Do not accept time or cost comments.
1 mark
B1 Method B, because it has a random start and spreads the sample across the whole day's ticket sales, with no choice by an interviewer
Must choose B with a reason. Choosing A or C scores B0.

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