Practice questions covering statistical hypothesis testing from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
Questions to attempt
6 questions on Statistical hypothesis testing, each with its mark scheme and worked method. Reveal the answer when you have had a go.
S5.1 · Apply the language of hypothesis testing via a binomial model: null/alternative hypothesis, significance level, test statistic, 1- and 2-tail tests, critical value/region, acceptance region, p-value; extend to correlation coefficients.
Q1
ORIGINAL
Year 1
Year 1: this point starts in the AS subject content, which is what most schools teach in Year 12. The full A-level adds in Year 2: Extending hypothesis testing to correlation coefficients, interpreted using a given p-value or critical value.Year 2 adds: Extending hypothesis testing to correlation coefficients, interpreted using a given p-value or critical value.
1.
A hypothesis test produces a p-value of 0.037. State the conclusion at the 5% significance level and at the 1% significance level. Explain what the significance level represents.
(4)
(Total for Question 1 is 4 marks)
S5.1 · Apply the language of hypothesis testing via a binomial model: null/alternative hypothesis, significance level, test statistic, 1- and 2-tail tests, critical value/region, acceptance region, p-value; extend to correlation coefficients.
Q2
ORIGINAL
Year 1
Year 1: this point starts in the AS subject content, which is what most schools teach in Year 12. The full A-level adds in Year 2: Extending hypothesis testing to correlation coefficients, interpreted using a given p-value or critical value.Year 2 adds: Extending hypothesis testing to correlation coefficients, interpreted using a given p-value or critical value.
2.
A company claims that the probability p of a customer choosing its premium plan has increased from 0.40. State suitable hypotheses and identify the number of tails.
(2)
(Total for Question 2 is 2 marks)
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.
Q3
ORIGINAL
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
3.
A seed supplier claims that the germination probability is 0.8. In an independent sample of 30 seeds, 19 germinate. Test H0:p=0.8 against H1:p<0.8 at the 5% level, given that P(X≤19)=0.0256 for X∼B(30,0.8).
(5)
(Total for Question 3 is 5 marks)
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.
Q4
ORIGINAL
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
4.
A coin is tested with H0:p=0.5 against H1:p>0.5. It lands heads 10 times in 12 tosses. Given P(X≥10)=0.0193 for X∼B(12,0.5), conduct the test at the 5% level.
(4)
(Total for Question 4 is 4 marks)
S5.3 · Conduct a statistical hypothesis test for the mean of a Normal distribution with known, given or assumed variance and interpret the results in context.
Q5
ORIGINAL
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
5.
A Normal population has known standard deviation 10. A random sample of 25 has mean 54.4. Test H0:μ=50 against H1:μ=50 at the 2% significance level. Use the critical values z=±2.326.
(5)
(Total for Question 5 is 5 marks)
S5.3 · Conduct a statistical hypothesis test for the mean of a Normal distribution with known, given or assumed variance and interpret the results in context.
Q6
ORIGINAL
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
6.
A Normal population has known standard deviation 12. Test H0:μ=100 against H1:μ>100 using a random sample of 36 with mean 104, at the 5% level.
(5)
(Total for Question 6 is 5 marks)
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Edexcel 9MA0 · June 2024 · Paper 31 · Q2 · 6 marks
6 marks
2
Show mark scheme
Examiner report
Whilst most students showed they had a good grasp of the concepts of correlation and
regression some failed to give sufficient detail in their answers. In part (a) the requirement to
“interpret” meant, as usual, that we needed a fully contextual response. Some simply stated
“negative correlation” or gave the value of the gradient as – 1.28 whereas a full interpretation of
the rate expected to see mention of the height decreasing by 1.28 m(etres) for each second of
flight. In part (b) most students interpreted this situation as a one-tailed test and stated the
hypotheses correctly in terms of but a few equated this equal to – 0.510 rather than 0 for the
null hypothesis. Almost all the students gave a critical value that matched their alternative
hypothesis, and an encouraging number gave a correct contextual conclusion to the test
mentioning both height and time. Some however lost the final accuracy mark for having
contradictory or incorrect statements such as “reject H0” followed by a correct contextual
conclusion such as “there is insufficient evidence of a negative correlation between h and t”.
Although we did not require a comparison of the test statistic with the critical value if one was
seen it had to be correct and some students lost the final mark for statements such as
– 0.510 < – 0.5494 or for comparing 0.510 with the significance level of 0.05 In part (c) most
students realised that the given equation was not an appropriate model but did not give a strong
enough explanation as to why this was so. Those who noted that the scatter diagram showed a
curved set of data scored the mark whilst those who suggested that the points seemed to form a
(negative) quadratic probably also gave themselves lead in to part (d). Of those who did not
score the mark in part (c) some simply discussed correlation rather than stating that the line had
negative gradient throughout whereas the scatter diagram had a positive gradient initially and
then a negative gradient. Others just compared one or two points rather than concentrating on
the overall shape of the scatter diagram and some simply stated “non-linear” without developing
their answer and drawing out a difference between the scatter diagram and Amar’s linear model.
Part (d) proved challenging for most students and many simply moved on to the next question.
There were many convoluted attempts involving calculus or multiplying out the bracket or
substituting a value such as (1, 28). Students should be encouraged to consider the number of
marks available as a guide to the amount of work required to answer the question. Those who
had identified the scatter diagram as representing a parabola and saw the connection with the
quadratic nature of Jane’s model could simply estimate the equation of the axis of symmetry
and usually chose a suitable value of k.
Text version
Amar measures a bird's height h metres at time t seconds for 10 values of t. The regression line is h = 38.6 - 1.28t and the product moment correlation coefficient is -0.510. The scatter diagram shows points at approximately (1, 28), (1.5, 34), (2, 36), (2.5, 38), (3.5, 40), (4.5, 39), (5, 37), (6.5, 33), (7, 27) and (8.5, 20), so the height peaks at about t = 3.5 seconds before falling.
Edexcel 9MA0 · June 2024 · Paper 31 · Q4 · 6 marks
6 marks
4
Show mark scheme
Examiner report
In part (a) most gave the hypotheses correctly in terms of p though some failed to have the H0
and H1 labels and some had p = 0 for the null hypothesis perhaps confusing this with a test for
correlation. Most used the correct distribution to try and find the critical region but notation
was often poor with P(X = 8) being written when P(X ⩽ 8) was meant or calculated and a
number struggled to find P(X ⩾ 9) correctly leading to upper tails of the critical region of
X ⩾ 8 or X ⩾ 10. There were many errors with the probabilities in each tail with
P(X = 0) = 0.0147 (a truncated answer) frequently seen instead of 0.0148 the correct rounded
value. Students don’t seem to be aware that the table function on their calculator truncates
values rather than rounding them. Some students gave their critical regions inside probability
statements and of course failed to secure the accuracy mark, and some didn’t think that X = 0
was a valid part of the critical region and omitted this from their answer. Many students who
had identified a critical region were able to add their tail probabilities to give the significance
level, but some merely wrote down 0.025 or doubled the probability of 0.0155 found in part (a).
In part (c) some students didn’t have a critical region of the correct form and were unable to
access this mark. Others simply said that they would “accept H0” without any attempt to give a
contextual conclusion and some thought that because the test was not significant Freya’s belief
was therefore correct. A few students incorrectly used the probability 7
40
and then compared
this against their probabilities of 0.0148 or 0.0155 from part (a).
Text version
The proportion of left-handed adults in a country is 10%. Freya believes the proportion among adults under 25 is different from 10% and takes a random sample of 40 adults under 25.
Edexcel 9MA0 · June 2023 · Paper 31 · Q2 · 9 marks
9 marks
2
Show mark scheme
Examiner report
This question was also answered well.
In part (a) students were often aware of the conditions needed for a binomial distribution
but did not always present their answers in context with the precision required. Some
focussed on the need for a fixed number of trials or that packets either did or did not
contain a prize but these features were clearly identified in the question. Those scoring
the mark either identified the need for prizes to be placed in packets at random or that the
probability of a packet containing a prize needed to be constant. There was some
confusion between “events”, such as a prize being placed in a packet, being independent
and the “probability” of such an event being constant. We sometimes saw a statement
such as “the probability of a packet containing a prize is independent” which didn’t score
this first mark.
Part (b) was generally answered well though some gave answers to 2dp or 2sf rather than
the 3sf as instructed on the front of the paper. Most answered (i) correctly but in (ii)
common incorrect answers were from those who found P(T = 3) or P(T ⩽ 3). In part (c)
many were able to identify the correct distribution and those who stated this were
guaranteed the method mark even if their answer to (b)(ii) was incorrect. Whilst many
did obtain the correct answer here those with an incorrect answer to (b)(ii) who simply
wrote down a probability could not be given either mark. Some students in part (c) didn’t
identify the binomial model but wrote an expression for the probability of the form
5
2
3
2
C
(1
)
p
p
−
but a few forgot the 5
2
C term.
The hypothesis test in part (d) was answered well with correct notation used for the
hypotheses and the appropriate one-tailed test identified. Most selected the correct
binomial model but it was disappointing to see a number finding P(X = 9) rather than
P(X ⩽ 9). A correct conclusion in context was often given: some using the simple
response that there is evidence to support Kamil’s claim and others stating that there was
evidence that the proportion of packets containing a prize is less than
1
7. It was
encouraging to see that most of those using the second approach were referring to
“proportion” rather than “number of” this time.
Text version
A machine fills packets with sweets, and 1/7 of the packets also contain a prize. The packets are placed in boxes containing 40 packets each. The random variable T is the number of packets containing a prize in a box.
Edexcel 9MA0 · June 2023 · Paper 31 · Q4 · 6 marks
6 marks
4
Show mark scheme
Examiner report
Whilst it was rare to see an incorrect answer to part (a) a minority of students were unable
to use their calculator to evaluate this simple probability using a normal distribution.
In part (b) most students stated their hypotheses correctly in terms of with very few
using a one-tailed test. To score the method mark in this type of question the students
simply need to select and write down the correct model (distribution). A number seemed
to be listing numbers they needed for their calculator, but we expect to see the appropriate
mathematical notation used. Fortunately, many obtained a correct probability and would
therefore receive the method mark by implication. Some chose to standardise and find a
z value (1.9088… which should be compared with 1.96) or a critical region
(𝑆̄ ⩾ 177.248. . . ) but given that a p-value was required in part (c) this was not the most
efficient route. A common error in stating the model is to use an incorrect value for the
mean. On this occasion we condoned
2
6.8
~ N 177.2,
52
S
for the M1 and first A1 but
penalised the final accuracy mark. In a hypothesis test the probability (or critical region)
is always calculated assuming H0 is true and so the mean should be 175.4 and the
probability calculated is (
)
P
177.2 .
S
Some students stopped after stating that the result
was not significant or not rejecting H0 but a good number gave a correct conclusion in
context.
Part (c) was not answered well with only a few scoring this mark. The common error was
to use 0.028 or 0.025 but others used values like 175.4
Text version
The heights of adult men from region A are normally distributed with mean 175.4 cm and standard deviation 6.8 cm. A student claims that the mean height in region B is different from the mean in region A. A random sample of 52 adult men from region B has mean height 177.2 cm. The student assumes that the standard deviation is 6.8 cm in both regions.
Edexcel 9MA0 · June 2022 · Paper 31 · Q4 · 6 marks
6 marks
4
Show mark scheme
Examiner report
Part (a) was well answered with the majority of students writing their hypotheses in
terms of p.
Part (b) proved to be a challenge for many of the students with a reasonable proportion
unable to even begin, which is unusual for a question of this nature – as it is quite
standard.
Most students were able to state the correct distribution and write some probabilities,
but many were not able to identify correctly the CR from their probabilities.
Responses scoring M1A1A1 were rare. This was often due students giving probability
statements rather than giving a CR
Other issues included:
• giving probabilities to only 2 s.f.
• selecting the region between the two Critical values as their CR
Students who were aware of what a two-tailed CR looks like almost always got part (c)
correct. The students who did not really understand the idea of the correct CR seemed to
make a fairly random guess at the actual significance level.
For those students who stated both tails of the critical region in (b), the majority
correctly compared 15 with their critical region and then usually supported the manager.
Others used the longer version of the contextual statement and most did so correctly.
Text version
Past records show that 10% of a dentist's customers arrive late. A new manager believes that the proportion has changed. A random sample of 50 customers is taken.
Edexcel 9MA0 · June 2022 · Paper 31 · Q6 · 9 marks
9 marks
6
Show mark scheme
Examiner report
In part (a) a large number of students simply wrote ‘negative correlation’, with no
context. The word Interpret in the question hints that context is required. Of the
students that did try to truly interpret the correlation in a contextual statement, the vast
majority were successful.
Whilst many students realised that one- tailed hypotheses were needed in part (b) few
used ρ for the parameter. Most students had ± the correct CV but the final A1 was lost
by many students, as they compared the positive value 0.3887 to the negative test
statistic. This showed these students were unfamiliar with the meaning of the CV/CR
and how it related to their test statistic.
Part (d) differentiated well between students with basic knowledge and those with
deeper understanding.
The majority of students scored 5/5 on this part. Almost all used the LH approach on the
MS. This was usually well done, though a good number made errors with the laws of
indices and maintained a + between the two terms after making the equation into powers
of 10. The students knew their log rules, but many did only so to a basic degree, and
came unstuck during their working. Sometimes this was to such an extent that they lost
the marks.
A surprisingly high number of students used natural logarithms and hence the exponential
function in this question showing a less secure grip of this topic.
Text version
Anna records resting heart rate h beats per minute and weekly exercise time m minutes for a random sample of 19 students. The scatter diagram shows a clear curved downward pattern: h generally falls from roughly 70-76 at low m towards roughly 62-65 by m = 150-400, with the curve becoming flatter as m increases. She codes x = log_10(m) and y = log_10(h). The product moment correlation coefficient between x and y is -0.897, and the regression line of y on x is y = -0.05x + 1.92.
Edexcel 9MA0 · October 2021 · Paper 31 · Q2 · 6 marks
6 marks
2
Show mark scheme
Examiner report
Most scored the mark in part (a) by identifying that the correlation was negative. When we ask students to
“describe” we are simply requiring a non-contextual description. We will ask them to “interpret” when we
require a contextual description. In this case we allowed students to describe the correlation in terms of the
variables so answers such as “as x increases y decreases” were accepted. Some thought the correlation was
strong and others weak but at this stage all we wanted was the negative description. Part (b) caused some
problems with some stating that the scatter diagram showed negative correlation but failing to say whether
or not this was compatible with Marc’s suggestion. Some students felt that there was no correlation shown
by the scatter diagram and did not reconsider this even after correctly answering part (d).
Most answered part (c) correctly though a number “lost” the minus sign between their calculator and
writing down the answer on the page. Usually there were sufficient figures given for us to award the mark
when the minus sign was included. In part (d) most attempted the hypotheses and the majority used ρ. Some
didn’t have “= 0” for their null hypothesis with alternative values or inequalities being used and of course
some used r or simply wrote the hypotheses in words. We allowed (+) for the critical value provided the
actual value was compatible with their alternative hypothesis and many achieved this mark but few scored
the final mark as the conclusion was not related to the context of the question.
Text version
Marc records, for a random sample of 16 students, x = the number of letters in the student's last name and y = the number of letters in the student's first name. The scatter diagram plots the points (3,7), (6,7), (8,4), (7,4), (5,6), (3,8), (11,5), (3,5), (4,8), (5,4), (4,7), (9,4), (7,5), (10,5), (6,6), (6,3), on axes 0 <= x <= 12 and 0 <= y <= 10.
Edexcel 9MA0 · October 2021 · Paper 31 · Q5 · 11 marks
11 marks
5
Show mark scheme
Examiner report
Most students used their calculators in part (a) with reasonable success but a frequent error was to use 0.1
instead of 0.01 and without any working being shown there was no opportunity to award the method mark.
Part (b) was a straightforward probability calculation designed to be completed on a calculator as the
specification requires and so it was disappointing that nearly 25% of the candidates scored no marks at all on
this question. Identifying the conditional probability in part (c) defeated many students and some of those
who seemed to be attempting to use a conditional probability were unable to use the correct notation to write
their expression. Those who did start with a correct conditional probability and moved onto a ratio of
probabilities often had a correct denominator but a common error was to have P(F > 160) as the numerator
rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something
here. The hypotheses were usually stated in terms of µ and most of the time 166.5 rather than 164.6 was
used. Selecting the correct model caused a number of problems: some used 164.6 as the mean and others
used 7.4 (or occasionally 6.1) as the standard deviation, simply stating the correct model will score the
method mark here but a number of students didn’t show us the model they were using and we had to award
this mark by implication from correct working. Most students successfully used the probability approach (as
we would recommend) with a few attempting a critical region or using a z value and comparing with 1.6449
from the tables. The final mark required a correct interpretation of their calculation in context and some
students simply stopped after stating that the result was significant and failed to score this final mark.
Text version
The heights F cm of females from a country are normally distributed with mean 166.5 cm and standard deviation 6.1 cm. For a different country, female heights are normally distributed with standard deviation 7.4 cm. Mia believes the mean height in the different country is less than 166.5 cm. She takes a random sample of 50 females there and obtains a sample mean of 164.6 cm.
Edexcel 9MA0 · October 2020 · Paper 31 · Q2 · 7 marks
7 marks
2
Show mark scheme
Examiner report
Very few students were unable to identify the negative correlation in part (a) with many
attempting to describe the strength as well (split fairly evenly between weak and strong)
though this was not required here. In part (b) it was clear that more students were
familiar with the large data set. Many identified “rainfall” as the variable on the y-axis
and they usually gave the correct units although some thought it should be mm3. The
most common incorrect answer here was “cloud cover” but, of course, this variable is
not available on the large data set for Perth. Part (c) was a standard hypothesis test for
correlation. It was encouraging to see a good number using ρ for the hypotheses but
fewer realised that the test was two-tailed. The phrase “…believes there is a correlation
between…” shows that a two tailed test should be used, however many were swayed by
the negative value of r and opted for a one tailed test. It was good to see that most
chose the critical value from the tables that was compatible with their alternative
hypothesis and this meant that they could also access the final mark provided their
comment was in context, “support for Stav’s belief” was perhaps the simplest way of
including the context. There were some good answers to part (d) but many others were
unable to pull together the various parts of the question to make a correct inference.
Some stated that the humidity was high and other assumed this but interpreted the
negative value of r correctly and were then able to state that they would expect the
number of hours of sunshine to be low. Some gave a suggested value for the number of
hours of sunshine but this needed to be realistic and values more than 5 were not
accepted. A few students knew (from the large data set) that humidity greater than 95%
implies foggy conditions and argued correctly that this would suggest a low amount of
sunshine.
Text version
A random sample of 15 days is taken from the large data set for Perth in June and July 1987. A scatter diagram compares two variables. The x-axis is Daily Mean Temperature in degC, running from 0 to 20. The y-axis is unlabelled and has a scale break near the origin. The 15 plotted points show a clear negative trend, with the higher y-values mainly at x about 11 to 15 and the lower y-values mainly at x about 15 to 19; two observations lie on the horizontal axis. No numerical y-scale is shown.
Edexcel 9MA0 · October 2020 · Paper 31 · Q5 · 15 marks
15 marks
5
Show mark scheme
Examiner report
It was disappointing that over 20% of candidates scored no marks here as there were
some straightforward marks available to those who had studied this part of the
specification. Part (a) was a very straightforward opening part and the vast majority of
students could use their calculator efficiently to score this mark. Part (b) was a fairly
standard hypothesis test and a good number made progress here. Some still did not use
µ for the hypotheses and a null hypothesis of
10
T =
was all too common. Most
attempts used a normal distribution with a correct variance but a significant number of
students used a mean of 11.5 rather than 10. This is not a correct method for a
hypothesis test (the probabilities should be calculated on the assumption that H0 is true)
but, on this occasion, for those who still managed to reach a correct conclusion a special
case enabled them to secure one mark. A variety of different approaches were used in
part (b), some finding the critical value for X and others calculating the test statistic of
1.677… but some of these candidates compared this value to the significance level
rather than the critical value of 1.6669 from the tables. The first part of (c) was another
fairly straightforward calculation which many scoring this mark, sometimes including
those who failed to score the mark in part (a). Students need to be aware of which “tail”
their calculator gives them. Beyond (c) (i) the question was more challenging but it is
encouraging that over 50% of students scored some of these first 6 marks.
In (c)(ii) although a good number of students recognised that a conditional probability
was required, few could find a correct expression for the numerator and a common error
was to see
(c)(i)
P(
0)
T >
used rather than a correct numerator of P(0 < T < 2). In part (iii)
many realised that the model would assign some probability to negative values of T but
this will always be true when a normal distribution is used as a model. The point here
was that a significant proportion of the values for T were negative and so this means that
the normal model is not very good. Most responses seemed to believe that a model
must match reality exactly but a few did comment that P(T > 0) should be closer to 1 or
that the difference between (i) and (ii) was too big and therefore the model was not a
good one to use.
Fully correct answers to part (d) were rare and this part gave the A* students an
opportunity to shine. Successful students realised that they needed to look at P(T > 2)
and often found this by taking 1 – (c)(i) but they rarely realised that the median splits
this probability into 2 so they were seeking the value of t such that P(T > t) =
0.40…Some who reached this stage tried, incorrectly to use their calculator, failing to
realise that their calculator gave t for probabilities of the form P(T < t), a few used the
tables provided to find z = 0.2533 and some did prepare correctly to use their calculator
and found P(T < t ) = 0.5978… (the 0.402… + (c)(i)) from whence the correct value of t
= 5.9 could be found. It was encouraging to see some good use of diagrams in this
question being used to help students identify which probabilities they needed.
Text version
A health centre claims that the time X minutes a doctor spends with a patient is modelled by N(10, 4^2). A receptionist samples 20 patients and obtains a mean time of 11.5 minutes. The centre also claims that the time T minutes a dentist spends on a routine appointment is modelled by T ~ N(5, 3.5^2).
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