Chi-Squared Tests

Topic

Includes hypothesis testingFrom an AS paper

Edexcel

Edexcel · Old spec

AS June 2025 Q4

EdexcelAS paperCurrent spec7 marksChi-Squared Tests

4. A 5-question test is taken by 80 students.

Each question was answered either correctly or incorrectly.

Number of questions answered correctly	0	1	2	3	4	5
Observed frequency	15	23	31	9	2	0

(a) Show that the proportion of questions answered correctly by these 80 students is 0.3 (2)

Anisa believes that the number of questions answered correctly can be modelled using a binomial distribution.
In order to test her belief, Anisa first calculates some of the expected frequencies.

Number of questions answered correctly	0	1	2	3	4	5
Expected frequency	13.45	\(r\)	\(s\)	10.58	2.27	0.19

(b) Find the value of \(r\) and the value of \(s\) (3)

Mark scheme (a)

Scheme	Marks	AO
\(\text{mean} = \dfrac{[15 \times 0 +]\, 23 \times 1 + 31 \times 2 + 9 \times 3 + 4 \times 2\, [+ 5 \times 0]}{80}\ [= 1.5]\)	M1	1.1b
\(p = \dfrac{1.5}{5} = 0.3\)*	A1*	1.1b
	(2)

Notes

M1: Attempt to find the mean with at least 3 correct products

A1*: cso fully correct solution leading to given answer 0.3
May be done in a single calculation for M1A1
\(\dfrac{[15 \times 0 +]\, 23 \times 1 + 31 \times 2 + 9 \times 3 + 4 \times 2\, [+ 5 \times 0]}{400} = 0.3\)

Mark scheme (b)

Scheme	Marks	AO
[\(X \sim \mathrm{B}(5, 0.3)\)] \(\mathrm{P}(X = 1) = 0.36015\) or \(\mathrm{P}(X = 2) = 0.3087\)	M1	1.1b
\(r = 80 \times 0.36015 = 28.8\ldots\) awrt 28.8	A1	1.1b
\(s = 80 \times 0.3087 = 24.69\ldots\) awrt 24.7	A1	1.1b
	(3)

Notes

M1: Either correct probability awrt 0.36 or awrt 0.31
(may be implied by a correct value of \(r\) or a correct value of \(s\))

A1: awrt 28.8

A1: awrt 24.7

Mark scheme (c)

Scheme	Marks	AO
Need to combine last 3 columns since \(E_i\) are \(\lt 5\)	B1	2.4
Therefore 4 – 2 degrees of freedom since proportion for Binomial estimated from \(O_i\)	B1	2.4
	(2)
	(7 marks)

Notes

B1: For explaining need to pool last 3 columns/cells and \(E_i \lt 5\)
allow 4 cells remaining after pooling for need to pool last 3 columns/cells
must be referencing expected frequencies (not observed frequencies)

B1: For proportion/parameter/\(p\)/mean/0.3 calculated/estimated (from data/\(O_i\))
and 2 constraints or 4 – 2 or 4 – 1 – 1 which must come from correct reasoning

A2 June 2025 Q3

EdexcelCurrent spec12 marksIncludes hypothesis testingChi-Squared Tests

3. A journalist uses a gender equality test on some randomly chosen films.
She believes that whether or not a film passes the test is associated with the period in which the film is first released. Her data is summarised in Table 1.
The journalist decides to test whether the data supports her belief.

Table 1: Observed frequencies

Period → Test result ↓	1990–1999	2000–2004	2005–2013	Total
Pass	45	70	65	180
Fail	75	90	55	220
Total	120	160	120	400

Some of the expected frequencies she calculates are shown in Table 2 below.

Table 2: Expected frequencies

Period → Test result ↓	1990–1999	2000–2004	2005–2013
Pass			54
Fail			66

(a) Complete Table 2 showing expected frequencies. (2)

(b) Write down hypotheses for a suitable test to assess the journalist’s belief. (1)

(c) Showing your working clearly, test the journalist’s belief using a 5% level of significance.
You should state your critical value and conclusion clearly. (6)

The journalist wants to publish her data and findings.
She decides to publish the data as percentages of pass and fail for each period, as shown in Table 3, rather than as frequencies.

Table 3: Percentages

Period → Test result ↓	1990–1999	2000–2004	2005–2013
Pass (%)	37.5	43.75	54.17
Fail (%)	62.5	56.25	45.83
Total (%)	100	100	100

The editor suggests carrying out the hypothesis test of the journalist’s belief using the percentages in Table 3.

(d)

(i) Describe the effect this would have on the test statistic, justifying your answer.

(ii) Explain whether or not the journalist should follow her editor’s suggestion. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\left[E_1 = \dfrac{180 \times 120}{400} \quad E_2 = \dfrac{180 \times 160}{400} \quad E_3 = \dfrac{220 \times 120}{400} \quad E_4 = \dfrac{220 \times 160}{400}\right]\)
Two of: \(E_1 = 54 \quad E_2 = 72 \quad E_3 = 66 \quad E_4 = 88\)	B1	2.2a
All four of: \(E_1 = 54 \quad E_2 = 72 \quad E_3 = 66 \quad E_4 = 88\)	B1	1.1b
	(2)

Notes

B1: Two of \(E_1, E_2, E_3, E_4\) correct

B1: All four of \(E_1, E_2, E_3, E_4\) correct

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0\): Whether a film passes the (Gender Equality) Test is not associated with the period in which the film is first released \(\mathrm{H}_1\): Whether a film passes the (Gender Equality) Test or not is associated with the period in which the film is first released	B1	2.5
	(1)

Notes

B1: Correct hypotheses mentioning test, and period/date
and association/independence mentioned in at least one hypothesis
Use of correlation/correlated is B0
NB: Relationship/related is condoned if association/independence mentioned at all

Mark scheme (c)

Scheme	Marks	AO
Correct method for finding at least two terms of \(\chi^2\): \(\begin{array}{\|c\|c\|c\|c\|}\hline \dfrac{(O_i - E_i)^2}{E_i} & 1.5 & 0.06 & 2.24 \\ & 1.23 & 0.05 & 1.83 \\ \hline\end{array}\) \(\begin{array}{\|c\|c\|c\|c\|}\hline \dfrac{O_i^2}{E_i} & 37.5 & 68.06 & 78.24 \\ & 85.23 & 92.05 & 45.83 \\ \hline\end{array}\)	M1	1.1b
\(\chi^2 =\) awrt 6.9	A1	1.1b
\(\nu = (3 - 1)(2 - 1) = \mathbf{2}\)	B1	1.1b
\(\chi^2_2(5\%) = 5.991\)	B1ft	1.1b
‘6.9’ \(\gt\) ‘5.991’ Sufficient evidence to reject \(\mathrm{H}_0\)	M1	3.4
Sufficient evidence to support journalist’s belief or Sufficient evidence that whether a film passes the gender equality test is associated with the period in which the film is first released	A1ft	2.2b
	(6)

Notes

M1: Correct method (terms may be unsimplified in an expression, and hence may need to be checked) shown for finding two \(\chi^2\) terms, accept either method
May be implied by a test statistic of awrt 6.9

A1: awrt 6.9

B1: 2 degrees of freedom (can be implied by cv of 5.991)

B1ft: Correct critical value (ft their df)
Must be consistent with their df.

M1: For a correct non-contextual conclusion based on their CV and their test statistic
May be implied by a correct contextual conclusion if no contradictions

A1ft: For a correct contextual conclusion referring to the journalist’s belief or mentioning (gender equality) test and period/date (ft their cv)
Dependent on 1^st A1
Do not allow contradictory statements
B0 for correlation/correlated
Condone ‘related/relationship’ in part (c)
NB: We ignore their hypotheses when marking (c)

Mark scheme (d)

Scheme	Marks	AO
(i) The observed frequencies for each period are more than 100 or All the \(O_i\) and \(E_i\) are lower as percentages or The difference between all the Oi and Ei is lower	M1	3.5a
Therefore, the test statistic would decrease (5.7…)	A1	2.2a
(ii) The journalist should not follow her editor’s suggestion (with an appropriate reason). e.g. As it may/will change the result of the test As they should use the real data rather than percentages	B1	2.4
	(3)
	(12 marks)

Notes

(i) M1: Mentioning that observed totals for each period are more than 100
or
Mentioning that all the \(O_i\) and \(E_i\) have reduced
Or
Mentioning that all the differences between the \(O_i\) and \(E_i\) have reduced
(must state or imply that this is true for all)
or
Calculating the new test statistic correctly (awrt 5.7)

A1: The test statistic would decrease (dependent on the previous M1)

(ii) B1: See scheme.
B0 for vague reasoning such as ‘should not use percentages as it would be inaccurate’
NB: Reasoning for (d)(ii) may be seen in (d)(i)
Do not award final B1 if contradictions between their reasoning in (i) and (ii)

AS June 2025 Q1

EdexcelAS paperCurrent spec10 marksIncludes hypothesis testingChi-Squared Tests

1. A researcher is investigating the relationship between a person’s age and their preferred method of shopping.

A random sample of 300 people is taken and the results are summarised in the table below.

		Preferred method of shopping
		Online	In-store	Total
Age	18 – 30	27	23	50
	31 – 50	31	59	90
	51 – 64	17	43	60
	65 and older	30	70	100
	Total	105	195	300

(a) Write down suitable hypotheses for a test to determine whether or not there is evidence of an association between age and preferred method of shopping. (1)

The researcher assumes that the null hypothesis is true and uses the data in the table to find expected values.

(b)

(i) Identify the cell which would have the smallest expected value.

(ii) Calculate the expected value for this cell. (2)

The value of \(\displaystyle\sum \frac{(O-E)^2}{E}\) for the other 7 cells is 5.060

(c) Test, at the 5% level of significance, whether or not there is evidence of an association between age and preferred method of shopping.
You should state the test statistic, degrees of freedom, critical value and conclusion clearly. (5)

(d) Explain whether or not your conclusion to part (c) would be the same if the test was carried out at a 1% level of significance. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): There is no association between age and preferred method of shopping. \(\mathrm{H}_1\): There is an association between age and preferred method of shopping.	B1	3.4
	(1)

Notes

B1: For both hypotheses in terms of "association" or “independence"
Must mention age and shopping in at least one and be connected correctly to \(\mathrm{H}_0\) and \(\mathrm{H}_1\)
Use of link, relationship, correlation or connection is B0 here.
Hypotheses must be given in part (a).

Scheme	Marks	AO
(i) Online 18 – 30 or 27	B1	2.2a
(ii) [\((50 \times 105) / 300 =\)] 17.5	B1	1.1b
	(2)

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{(\text{‘}27\text{’}-\text{‘}17.5\text{’})^2}{\text{‘}17.5\text{’}} + 5.060\)	M1	1.1b
\(= 10.217\ldots\) awrt 10.2	A1	1.1b
df [\(= (4 - 1)(2 - 1)\)] \(= 3\)	B1	1.1b
[\(10.2 \gt\)] \(\chi^2_{3,(0.05)} = 7.815\)	B1ft	3.1b
[Reject \(\mathrm{H}_0\)] There is evidence of an association between age and preferred method of shopping.	B1ft	2.2b
	(5)

Notes

M1: For use of \(\dfrac{(O-E)^2}{E}\) with their cell + 5.060 (this mark may be implied by awrt 10.2)
Watch out for \(\dfrac{(O-E)^2}{O}\) + 5.060 which is M0

A1: awrt 10.2 (ignore p-value if given)

B1: 3 cao

B1ft: Using the degrees of freedom to find the \(\chi^2\) CV for the appropriate model 7.815 or better
ft their df 1→3.841, 2→5.991, 4→9.488, 5→11.070, 6→12.592, 7→14.067, 8→15.507

B1ft: Correct conclusion or ft conclusion based on their values in (c), in context (age and shopping).
Must be consistent with their CV and test statistic. Independent of their hypotheses.
i.e. test statistic > CV → there is association… test statistic < CV → there is no association…
Allow relationship, link, connection BUT do not accept correlation or contradictory statements

Mark scheme (d)

Scheme	Marks	AO
\(\chi^2\) CV is now 11.345 [\(\gt 10.2\)]…	M1	1.1b
… \(\mathrm{H}_0\) is not rejected/conclusion is reversed	A1ft	2.2b
	(2)
	(10 marks)

Notes

M1: Obtaining new \(\chi^2\) CV 11.345 or awrt 11.3 Condone \(p\) = awrt 0.017 [\(\gt 0.01\)] for this mark
ft their df (all awrt 3sf) 1→6.64, 2→9.21, 4→13.3, 5→15.1, 6→16.8, 7→18.5, 8→20.1

A1ft: ft deduction which must be consistent with their 1% CV and their test statistic (need not be in context). Do not allow contradictory statements.

AS June 2024 Q4

EdexcelAS paperCurrent spec15 marksIncludes hypothesis testingChi-Squared Tests

4. Robin shoots 8 arrows at a target each day for 100 days.

The number of times he hits the target each day is summarised in the table below.

Number of hits	0	1	2	3	4	5	6	7	8
Frequency	1	10	30	34	17	4	2	0	2

Misha believes that these data can be modelled by a binomial distribution.

(a) State, in context, two assumptions that are implied by the use of this model. (2)

(b) Find an estimate for the proportion of arrows Robin shoots that hit the target. (2)

Misha calculates expected frequencies, to 2 decimal places, as follows.

Number of hits	0	1	2	3	4	5	6	7	8
Expected frequency	2.81	12.67	\(r\)	28.05	19.73	\(s\)	2.50	0.40	0.03

Misha correctly used a suitable test to assess her belief.

(d)

(i) Explain why she used a test with 3 degrees of freedom. (2)

(ii) Complete the test using a 5% level of significance.
You should clearly state your hypotheses, test statistic, critical value and conclusion. (6)

Mark scheme (a)

Scheme	Marks	AO
e.g. The probability of an arrow hitting the target is constant	B1	1.2/3.5b
e.g. The arrows are shot independently	B1	1.2/3.5b
	(2)

Notes

1^st B1 for one suitable comment which must mention “arrows”/ “shots” / “hit” / “target”
or for both comments not in context

2^nd B1 for both suitable comments which must mention “arrows”/ “shots”/“hit” / “target” at least once

Mark scheme (b)

Scheme	Marks	AO
\(\hat{p} = \dfrac{0 \times 1 + 1 \times 10 + 2 \times 30 + 3 \times 34 + 4 \times 17 + 5 \times 4 + 6 \times 2 + 7 \times 0 + 8 \times 2}{100 \times 8}\)	M1	1.1b
\(= \left[\dfrac{2.88}{8}\right] = \underline{\mathbf{0.36}}\)	A1	1.1b
	(2)

Notes

M1 for correct attempt at mean or \(\hat{p}\) (allow at least 3 correct non-zero products seen)

A1 for 0.36 (sight of 2.88 scores M1 and the A mark when divided by 8) (may be seen in (c))

Mark scheme (c)

Scheme	Marks	AO
[Let \(X \sim \mathrm{B}(8, \text{awrt } \text{“}0.36\text{”})\)] \(\mathrm{P}(X = 2) = 0.24936\ldots\) or \(\mathrm{P}(X = 5) = 0.08876\ldots\)	M1	3.4
(E(2)) = awrt 24.94 (24.93~24.95) or (E(5)) = awrt 8.88 (8.87~8.89)	A1	1.1b
For using Sum of Expected frequencies = 100 (accept awrt 100.01)	B1ft	1.1b
	(3)

Notes

M1 for sight or use of a correct binomial model to find either prob (ft their 0.36)

A1 for either correct expected frequency

B1ft for two positive values of \(r\) and \(s\) such that \(r + s = 33.82\) or 33.81

Mark scheme (d)

Scheme

Marks

(i) Need to pool some values to make all \(E_i \gt 5\) [Needn’t specify]

1.1b

Two constraints AND \(p\) was estimated. [“2 constraints” \(\Leftarrow\ 5 - 2 = 3\)]

2.4

(2)

(ii) \(\mathrm{H}_0\): Binomial is a good model for these data
\(\mathrm{H}_1\): Binomial is not a suitable model for these data

2.5

Hits	Ex freq	\(\dfrac{(O_i - E_i)^2}{E_i}\)	\(\dfrac{O_i^2}{E_i}\)
0 or 1	15.48	\(\dfrac{(11 - 15.48)^2}{15.48} = 1.30\)	\(\dfrac{11^2}{15.48} = 7.82\)
2	“24.94”	\(\dfrac{(30 - \text{‘}24.94\text{’})^2}{24.94} = 1.03\)	\(\dfrac{30^2}{24.94} = 36.09\)
3	28.05	\(\dfrac{(34 - 28.05)^2}{28.05} = 1.26\)	\(\dfrac{34^2}{28.05} = 41.21\)
4	19.73	\(\dfrac{(17 - 19.73)^2}{19.73} = 0.38\)	\(\dfrac{17^2}{19.73} = 14.65\)
…5	“11.80”	\(\dfrac{(8 - \text{‘}11.80\text{’})^2}{11.80} = 1.22\)	\(\dfrac{8^2}{11.80} = 5.42\)

M1

M1

1.1b/(2.1)
1.1b

Test statistic \(= 5.188692\ldots\) awrt 5.19 (allow awrt 5.18)

1.1b

CV \(\chi_3^2(5\%) = 7.815\)

1.1b

[Do not rej \(\mathrm{H}_0\)] Data is compatible with Misha’s belief /Binomial model is suitable

2.2b

(6)

(15 marks)

Notes

(i) 1^st B1 for mention of the need to pool since \(E_i \lt 5\) or to get \(E_i \gt 5\)

2^nd B1 for mention of 2 constraints because \(p\) is estimated/calculated (from data)

(ii) 1^st B1 for both hypotheses correct

1^st M1 for correct pooling (0&1) and (…5) (ft their \(r\) and \(s\)) (may be implied by awrt 5.18)

2^nd M1 for correct use of test statistic (at least one correct calc to at least 2 sf from 3^rd or 4^th row)

1^st A1 for awrt 5.19 (accept awrt 5.18, awrt 5.20)

2^nd B1 for 7.815 (or better) [NB p-value = 0.1586 … and is B0]

2^nd A1 dep on both M marks for a correct conclusion with “belief” or “binomial”
e.g \(\mathrm{B}(8, 0.36)\) is a suitable model is A1

A2 June 2024 Q3

EdexcelCurrent spec6 marksIncludes hypothesis testingChi-Squared Tests

3. Tisam took a survey of students’ favourite colours. The results are summarised in the table below.

		Colour
		Red	Blue	Green	Yellow	Black	Total
Year group	1–5	34	15	14	22	3	88
	6–9	23	32	12	9	8	84
	10–12	5	28	19	8	8	68
	Total	62	75	45	39	19	240

Tisam carries out a suitable test to see if there is any association between favourite colour and year group.

(a) Write down the hypotheses for a suitable test. (1)

For her table, Tisam only needs to check one cell to show that none of the expected frequencies are less than 5

(b)

(i) Identify this cell, giving your reason.

(ii) Calculate the expected frequency for this cell. (2)

The test statistic for Tisam’s test is 38.449

(c) Using a 1% level of significance, complete the test.
You should state your critical value and conclusion clearly. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): There is no association between the colour chosen and year group (allow use of “independence” instead of association) \(\mathrm{H}_1\): There is some association between colour and year group	B1	2.5
	(1)

Notes

B1 for both hypotheses in context. Must mention colour and year group at least once

NB: condone use of related/correlated/linked etc if recovered with independent/associated

Mark scheme (b)

Scheme	Marks	AO
(i) Need lowest row total and column total so “Black” and “10-12”	B1	2.2a
(ii) With expected frequency \(\dfrac{68 \times 19}{240} = 5.3833\ldots\)	B1	1.1b
	(2)

Notes

(i) B1 for a choosing “Black” and “10-12” with some correct reasoning mentioning row and column totals (allow clear equivalent, e.g. ‘lowest total frequencies’)

(ii) B1 for a correct expression or awrt 5.38

Mark scheme (c)

Scheme	Marks	AO
\(\upsilon = (5 - 1) \times (3 - 1) = \underline{\mathbf{8}}\)	B1	3.4
cv of \(\chi_8^2(1\%) = \underline{\mathbf{20.090}}\)	B1ft	1.1b
(significant) evidence of an association between colour chosen and year group	B1ft	2.2b
	(3)
	(6 marks)

Notes

1^st B1 for a correct calculation or answer of 8. May be implied by sight of cv of 20.09

2^nd B1ft for 20.090 (accept 20.09) or a correct ft 1% cv using their df and correct to 2 d.p.

3^rd B1ft for a correct conclusion in context (ft their cv)
E.g. “students in different years (tend to) have different favourite colours”
Do not allow contradictory statements. Condone ‘related/relationship’ in part (c)
NB: We ignore their hypotheses when marking (c)

AS June 2024 Q1

EdexcelAS paperCurrent spec6 marksIncludes hypothesis testingChi-Squared Tests

1. Sharma believes that each computer game he sells appeals equally to all age ranges.
To investigate this, he takes a random sample of 100 people who play these games and asks them which of the games \(A\), \(B\) or \(C\) they prefer.
The results are summarised in the table below.

Computer game		\(A\)	\(B\)	\(C\)
Age range	\(\lt 20\)	8	15	6
	20 – 30	21	12	9
	\(\gt 30\)	6	10	13

(a) Write down hypotheses for a suitable test to assess Sharma’s belief. (1)

(b) For the test, calculate the expected frequency for

(i) those players aged under 20 who prefer game \(C\)

(ii) those players aged between 20 and 30 who prefer game \(A\) (2)

Sharma correctly calculates the test statistic for this test to be 11.542 (to 3 decimal places).

(d) Using a 5% significance level, and stating your critical value, comment on Sharma’s belief. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): There is no association between age range and preferred game \(\mathrm{H}_1\): There is an association between age range and preferred game	B1	1.2
	(1)

Notes

B1 for both hypotheses in context
must mention independence/dependence or no association/association
must mention age and game at least once
Do not allow e.g. ‘\(\mathrm{H}_0\): games sold appeal equally to all age ranges’
Use of ‘correlation’, ‘link’, ‘relationship’, ‘connection’ is B0 but allow for 2^nd B1 in (d)
ignore comments about sales

Mark scheme (b)

Scheme	Marks	AO
(i) \(\dfrac{29 \times 28}{100} = \underline{\mathbf{8.12}}\) (ii) \(\dfrac{42 \times 35}{100} = \underline{\mathbf{14.7}}\)	M1 A1	1.1b 1.1b
	(2)

Scheme

Marks

(i) \(\dfrac{29 \times 28}{100} = \underline{\mathbf{8.12}}\)

(ii) \(\dfrac{42 \times 35}{100} = \underline{\mathbf{14.7}}\)

M1
A1

1.1b
1.1b

(2)

Notes

M1 for one correct expression or correct value.

A1 for both correct values (correct to 1 dp). Accept 8.1

Scheme	Marks	AO
[Since (b)(i) \(\gt 5\) there is no pooling so df \(= (3-1) \times (3-1) =\)] 4	B1	1.1b
	(1)

Mark scheme (d)

Scheme	Marks	AO
\(\chi_4^2(0.05) = \underline{\mathbf{9.488}}\)	B1ft	3.4/ 1.1b
[The test is significant so reject \(\mathrm{H}_0\):] Sharma’s belief is not supported / there is significant evidence of an association between age range and computer game preference.	B1	2.2b
	(2)
	(6 marks)

Notes

1^st B1ft for awrt 9.488 or ft their df for a 5% critical value
may see \(\chi_3^2(0.05) = 7.815\) \(\chi_2^2(0.05) = 5.991\) \(\chi_7^2(0.05) = 14.067\) \(\chi_8^2(0.05) = 15.507\)

2^nd B1 Indep of hypotheses but dep on cv < 11.542
for correct statement in context about Sharma’s belief
condone e.g ‘games sold do not appeal equally to all age ranges’
for a contextual statement must mention age and game
B0 for contradictory statements e.g. “test is not significant so belief is not supported”

AS June 2023 Q4

EdexcelAS paperCurrent spec12 marksIncludes hypothesis testingChi-Squared Tests

4. Table 1 below shows the number of car breakdowns in the Snoreap district in each of 60 months.

Number of car breakdowns	0	1	2	3	4	5
Frequency	12	11	19	14	3	1

Table 1

Anja believes that the number of car breakdowns per month in Snoreap can be modelled by a Poisson distribution. Table 2 below shows the results of some of her calculations.

Number of car breakdowns	0	1	2	3	4	\(\geqslant 5\)
Observed frequency \((O_i)\)	12	11	19	14	3	1
Expected frequency \((E_i)\)	9.92			9.64	4.34

Table 2

(a) State suitable hypotheses for a test to investigate Anja’s belief. (1)

(b) Explain why Anja has changed the label of the final column to \(\geqslant 5\) (1)

(d) Find the value of \(\dfrac{(O_i - E_i)^2}{E_i}\) when the number of car breakdowns is

(i) 1

(ii) 3 (2)

(e) Explain why Anja used 3 degrees of freedom for her test. (2)

The test statistic for Anja’s test is 6.54 to 2 decimal places.

(f) Stating the critical value and using a 5% level of significance, complete Anja’s test. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): No. of car breakdowns per month follows a Poisson distribution \(\mathrm{H}_1\): No. of car breakdowns per month does not follow a Poisson distribution	B1	2.5
	(1)

Notes

B1 correct hypotheses mentioning “breakdowns” and “Poisson”. \(\mathrm{Po}(1.8)\) o.e. is B0

Mark scheme (b)

Scheme	Marks	AO
A Poisson distribution will assign some probability to (all) values greater than 5 (o.e.)	B1	2.4
	(1)

Notes

B1 for a suitable reason mentioning the Poisson distribution taking values beyond 5
e.g. sample space for Poisson being \([0, \infty)\)

Mark scheme (c)

Scheme	Marks	AO
Need \(\lambda\): \(\left[\hat{\lambda} =\right] \dfrac{0 \times 12 + 1 \times 11 + 2 \times 19 + 3 \times 14 + 4 \times 3 + 5 \times 1}{(12 + 11 + 19 + 14 + 3 + 1)}\) or 1.8	M1	1.1b
[Under \(\mathrm{H}_0\) \(X \sim \mathrm{Po}(1.8)\)] \(E_1 = 60 \times \mathrm{P}(X = 1) = 17.85(227\ldots)\)	M1	3.4 1.1b
\(E_2 = 60 \times \mathrm{P}(X = 2) = 16.06(705\ldots)\)	A1	1.1b
\(E_{\geqslant 5} = 60 - \displaystyle\sum_0^4 E_i = 2.18(43996\ldots)\)	B1ft
	(4)

Notes

1^st M1 for an expression for the mean with least 3 correct products and correct denominator
M0 if they have found \(\lambda\) by working backwards from \(\mathrm{P}(X = 0) \Rightarrow 60\mathrm{e}^{-\lambda} = 9.92\)

2^nd M1 for a correct method for finding \(E_1\) or \(E_2\) (implied by one correct value)

1^st A1 for awrt 17.85 and awrt 16.07

B1ft for awrt 2.18, or using 60 – (sum of their \(E_i\)) or \(60 \times (1 - \mathrm{P}(X \leqslant 4))\)

Mark scheme (d)

Scheme	Marks	AO
\(\dfrac{(11 - \text{“}17.85\ldots\text{”})^2}{\text{“}17.85\ldots\text{”}} = 2.6287\ldots\) or \(\dfrac{(14 - 9.64)^2}{9.64} = 1.97195\ldots\)	M1	1.1b
awrt 2.63 and awrt 1.97	A1	1.1b
	(2)

Notes

M1 for either correct expression (ft their \(E_1\))

A1 for awrt 2.63 and awrt 1.97

Mark scheme (e)

Scheme	Marks	AO
Need to combine last two columns since \(E_i\) are \(\lt 5\)	B1	1.1a
Degrees of freedom therefore 5 – 2 since mean for Poisson estimated from \(O_i\)	B1	1.1a
	(2)

Notes

1^st B1 for explaining need to pool columns since \(E_i \lt 5\)
allow candidates describing combining last three columns as long as they say \(E_i \lt 5\)

2^nd B1 for mentioning mean/rate/parameter/\(\lambda\) estimated from \(O_i\) and 2 constraints

Mark scheme (f)

Scheme	Marks	AO
\(\chi_3^2(5\%) = 7.815\)	B1	1.1b
(Not significant) insufficient evidence to reject Anja’s belief	B1	2.2b
	(2)
	(12)

Notes

1^st B1 for correct cv of 7.815 (or better)

2^nd B1 for correct conclusion in context mentioning “breakdowns” and “Poisson”
Allow equivalent words for belief, such as “theory” etc.
Must be consistent with their cv
Allow \(\mathrm{Po}(1.8)\) in conclusion instead of Poisson

A2 June 2023 Q3

EdexcelCurrent spec15 marksIncludes hypothesis testingChi-Squared Tests

3. In a class experiment, each day for 170 days, a child is chosen at random and spins a large cardboard coin 5 times and the number of heads is recorded.
The results are summarised in the following table.

Number of heads	0	1	2	3	4	5
Frequency	3	10	45	62	38	12

Marcus believes that a \(\mathrm{B}(5, 0.5)\) distribution can be used to model these data and he calculates expected frequencies, to 2 decimal places, as follows

Number of heads	0	1	2	3	4	5
Expected frequency	\(r\)	26.56	\(s\)	\(s\)	26.56	\(r\)

(a) Find the value of \(r\) and the value of \(s\) (3)

(b) Carry out a suitable test, at the 5% level of significance, to determine whether or not the \(\mathrm{B}(5, 0.5)\) distribution is a good model for these data.
You should state clearly your hypotheses, the test statistic and the critical value used. (6)

Nima believes that a better model for these data would be \(\mathrm{B}(5, p)\)

To test her model, Nima uses this value of \(p\), to calculate expected frequencies as follows

Number of heads	0	1	2	3	4	5
Expected frequency	2.07	14.65	41.44	58.63	41.47	11.74

The test statistic for Nima’s test is 1.62 (to 3 significant figures)

(d) State,

(i) giving your reasons, the degrees of freedom

(ii) the critical value

that Nima should use for a test at the 5% significance level. (3)

(e) With reference to Marcus’ and Nima’s test results, comment on

(i) the probability of the coin landing on heads,

(ii) the independence of the spins of the coin.

Give reasons for your answers. (2)

Mark scheme (a)

Scheme	Marks	AO
[\(X \sim \mathrm{B}(5, 0.5)\)] \(\mathrm{P}(X = 0) = \mathrm{P}(X = 5) = 0.03125\) or \(\mathrm{P}(X = 2)\) or \(\mathrm{P}(X = 3) = 0.3125\)	M1	1.1b
[multiply by 170 to get] \(r = \underline{\mathbf{5.31}}(25)\); \(\ s = \underline{\mathbf{53.1}}(25)\)	A1;A1	1.1b(x2)
	(3)

Notes

M1 for 1 correct probability which may be embedded (0.03125 or 0.3125 or \(0.5^5\) or \({}^5C_2\, 0.5^2 \times 0.5^3\))

1^st A1 for \(r\) = awrt 5.31 (condone \(\frac{85}{16}\))

2^nd A1 for \(s\) = awrt 53.1 (condone \(\frac{425}{8}\))

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0: \mathrm{B}(5, 0.5)\) is a suitable model \(\quad \mathrm{H}_1: \mathrm{B}(5, 0.5)\) is NOT a …	B1	2.5
\(\dfrac{(O_i - E_i)^2}{E_i}\): \(\dfrac{(3-\text{‘}5.31\text{’})^2}{\text{‘}5.31\text{’}} = 1.00\ldots\), \(\dfrac{(10-26.56)^2}{26.56} = 10.3\ldots\), \(\dfrac{(45-\text{‘}53.1\text{’})^2}{\text{‘}53.1\text{’}} = 1.23\ldots\), \(\dfrac{(62-\text{‘}53.1\text{’})^2}{\text{‘}53.1\text{’}} = 1.48\ldots\), \(\dfrac{(38-26.56)^2}{26.56} = 4.92\ldots\), \(\dfrac{(12-\text{‘}5.31\text{’})^2}{\text{‘}5.31\text{’}} = 8.41\ldots\) or \(\dfrac{O_i^2}{E_i}\): \(\dfrac{3^2}{\text{‘}5.31\text{’}} = 1.69\ldots\), \(\dfrac{10^2}{26.56} = 3.76\ldots\), \(\dfrac{45^2}{\text{‘}53.1\text{’}} = 38.1\ldots\), \(\dfrac{62^2}{\text{‘}53.1\text{’}} = 72.3\ldots\), \(\dfrac{38^2}{26.56} = 54.3\ldots\), \(\dfrac{12^2}{\text{‘}5.31\text{’}} = 27.1\ldots\)	M1	1.1b
\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i}\) or \(\displaystyle\sum\frac{O_i^2}{E_i} - 170 = 27.4\ldots\) awrt 27.4 or awrt 27.5	A1	1.1b
Degrees of freedom is \(6 - 1 = \underline{\mathbf{5}}\), and critical value is 11.07(0)	B1ft B1ft	1.1b(x2)
[Significant result] Marcus’ model/\(\mathrm{B}(5, 0.5)\) is not a good fit. (o.e.)	A1	2.2b
	(6)

Notes

1^st B1 for both hypotheses mentioning \(\mathrm{B}(5, 0.5)\) or Marcus’ distribution at least once

M1 for at least one correct (ft) term or expression of the test statistic (accept 2sf)

1^st A1 for awrt 27.4 or awrt 27.5 (correct value here scores M1A1)

2^nd B1 for 5 or ft if ‘their \(r\)’ < 5, then df (= 4 – 1) = 3

3^rd B1 for 11.07(0) (or better) for ft df = 4 \(\rightarrow\) 9.488 or df = 3 \(\rightarrow\) 7.815

A1 dep on 1^st M1 for a suitable conclusion in context rejecting \(\mathrm{B}(5, 0.5)\)/Marcus’ model
Must be compatible with their test statistic and their CV. Just ‘Bin is not a good fit’ is A0
A0 if inconsistent comments are seen e.g. “do not reject \(\mathrm{H}_0\), \(\mathrm{B}(5, 0.5)\) is not a good fit”

Mark scheme (c)

Scheme	Marks	AO
\(\hat{p} = \left[\dfrac{0 \times 3 + 1 \times 10 + \ldots + 5 \times 12}{170 \times 5}\right] = 0.58588\ldots\) awrt 0.586	B1	1.1b
	(1)

Notes

B1 for awrt 0.586 allow \(\frac{498}{850}\) o.e.

Mark scheme (d)

Scheme	Marks	AO
(i) Need to pool (first 2) cells (0 and 1 since \(E(0) \lt 5\)) and use of \(\hat{p}\)	M1	2.4
Degrees of freedom: 5 groups – 2 constraints = 3	A1	1.1b
(ii) Critical value is 7.815	B1ft	1.1b
	(3)

Notes

(i) M1 for both reasons, must mention pooling or show pooling or mention exp. value < 5 and use of estimated parameter

A1 for df = 3 (must have scored the M1 for this mark)

(ii) B1 for 7.815 (or better) allow this independent of the M1 only allow ft on df=4 \(\rightarrow\) 9.488

Mark scheme (e)

Scheme	Marks	AO
(i) Nima’s model is a good fit (since 1.62 < ‘7.815’)/Marcus’ is not and this suggests coin is biased/probability of head approx. 0.6	B1	2.4
(ii) Nima’s test suggests binomial is a good model and therefore independence of spins is a reasonable assumption	B1	2.2b
	(2)
	(15 marks)

Notes

(i) 1^st B1 for stating Nima’s (binomial) model is a good fit/do not reject \(\mathrm{H}_0\) for Nima’s model/Marcus’ model is not a good fit and suggest that coin is probably biased/ \(p \gt 0.5\) (\(p\) closer to ‘their (c)’)
Only comparing 1.62 with ‘27.4’ to reach \(p \gt 0.5\) is incorrect and scores B0

(ii) 2^nd B1 for mention of Nima’s test suggests Binomial distribution is suitable and that spins are independent (ignore reference to Marcus’ test for 2^nd B1)

AS June 2023 Q2

EdexcelAS paperCurrent spec6 marksIncludes hypothesis testingChi-Squared Tests

2. A bag contains a large number of balls, all of the same size and weight. The balls are coloured Red, Blue or Yellow.

Jasmine asks each child in a group of 150 children to close their eyes, select a ball from the bag and show it to her. The child then replaces the ball and repeats the process a second time.

If both balls are the same colour the child receives a prize.

The results are given in the table below.

1st colour → 2nd colour ↓	Red	Blue	Yellow	Total
Red	31	11	18	60
Blue	8	10	9	27
Yellow	21	9	33	63
Total	60	30	60	150

Jasmine carries out a test, at the 5% level of significance, to see whether or not the colour of the 2nd ball is independent of the colour of the 1st ball.

(a) Calculate the expected frequencies for the cases where both balls are the same colour. (2)

The test statistic Jasmine obtained was 12.712 to three decimal places.

(b) Use this value to complete the test, stating the critical value and conclusion clearly. (3)

With reference to your calculations in part (a) and the nature of the experiment,

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{E(RR)} = \dfrac{60 \times 60}{150}\) or \(\mathrm{E(BB)} = \dfrac{30 \times 27}{150}\) or \(\mathrm{E(YY)} = \dfrac{60 \times 63}{150}\)	M1	1.1b
\(\mathrm{E(RR)} = \underline{\mathbf{24}}\) and \(\mathrm{E(BB)} = \underline{\mathbf{5.4}}\) and \(\mathrm{E(YY)} = \underline{\mathbf{25.2}}\)	A1	1.1b
	(2)

Notes

M1 for at least one correct calculation seen

A1 for all 3 expected frequencies correct, must be exact.

Mark scheme (b)

Scheme	Marks	AO
[\(\mathrm{E(BB)} \gt 5\) so no need for pooling] \(\nu = (3-1)(3-1) = 4\)	B1	3.1b
Critical value: \(\chi_4^2(5\%) = \underline{\mathbf{9.488}}\)	B1ft	1.1b
(significant) evidence that the colour of balls is not independent	B1ft	2.2b
	(3)

Notes

1^st B1 for 4 degrees of freedom

2^nd B1ft for awrt 9.488 or ft their degrees of freedom for 5% cv

3^rd B1ft for correct contextual conclusion mentioning “colour” and “independent”
Must be consistent with their cv
Allow associated instead of not independent

Mark scheme (c)

Scheme	Marks	AO
\(O_i \gt E_i\) for pairings of the same colour and some children may be cheating when selecting the 2^nd ball so choice not independent (o.e.)	B1	3.5a
	(1)
	(6)

Notes

B1 Stating or indicating that \(O_i \gt E_i\) and a suitable reason for the lack of independence that would invalidate the modelling assumption. Such as:

cheating/looking
The first ball being replaced by the child so more likely to be picked again

AS June 2022 Q3

EdexcelAS paperCurrent spec9 marksIncludes hypothesis testingChi-Squared Tests

3. In a game, a coin is spun 5 times and the number of heads obtained is recorded.
Tao suggests playing the game 20 times and carrying out a chi-squared test to investigate whether the coin might be biased.

(a) Explain why playing the game only 20 times may cause problems when carrying out the test. (1)

Chris decides to play the game 500 times. The results are as follows

Number of heads	0	1	2	3	4	5
Observed frequency	2	27	93	181	146	51

Chris decides to test whether or not the data can be modelled by a binomial distribution, with the probability of a head on each spin being 0.6

She calculates the expected frequencies, to 2 decimal places, as follows

Number of heads	0	1	2	3	4	5
Expected frequency	5.12	38.40	115.20	172.80	129.60	38.88

(b) State the number of degrees of freedom in Chris’ test, giving a reason for your answer. (1)

(c) Carry out the test at the 5% level of significance.
You should state your hypotheses, test statistic, critical value and conclusion clearly. (5)

(d) Showing your working, find an alternative model which would better fit Chris’ data. (2)

Mark scheme (a)

Scheme	Marks	AO
Not all the expected frequencies are likely to be over 5 Or the sample size is too small.	B1	3.5b
	(1)

Notes

B1: For recognising the limitations of using a chi squared model on small sample sizes eg 20 is not large, not enough data, sample needs to be larger, you may need to combine cells.

Mark scheme (b)

Scheme	Marks	AO
5 degrees of freedom since the parameter is not estimated from the data [and the totals agree]	B1	2.4
	(1)

Notes

B1: For 5 [dof] and a correct reason indicating parameter(probability) is not estimated.
Condone missing comment about totals

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{H}_0\): \(\mathrm{B}(5, 0.6)\) is a suitable model \(\mathrm{H}_1\): \(\mathrm{B}(5, 0.6)\) is not a suitable model	B1	3.4
\(\displaystyle\sum \frac{(O-E)^2}{E} = \frac{(2 - 5.12)^2}{5.12} + \ldots + \frac{(51 - 38.88)^2}{38.88}\)	M1	2.1
\(= 15.8063\ldots\) awrt 16	A1	1.1b
[\(15.8 \gt\)] \(\chi^2_{5,(0.05)} = 11.070\)	B1ft	1.1b
\(\mathrm{B}(5, 0.6)\) is not a suitable model [for the number of heads spun]	A1ft	3.5a
	(5)

Notes

B1: Both hypotheses correct Must have \(\mathrm{B}(5, 0.6)\) or binomial with number (\(n\)) = 5 and probability(\(p\)) = 0.6 (in at least 1) and be attached to \(\mathrm{H}_0\) and \(\mathrm{H}_1\) the right way round.

M1: Attempting to find the test statistic \(\displaystyle\sum \frac{(O-E)^2}{E}\) (at least two correct expressions, fractions or decimals) or \(\displaystyle\chi^2 = \sum \frac{O^2}{E} = \frac{(2)^2}{\text{“}5.12\text{”}} + \ldots + \frac{51^2}{38.88} - 500\) (at least two correct expressions, fractions or decimals plus the − 500) Implied by awrt 15.8

A1: Awrt16

B1ft: Allow 11.07 or awrt 11.070 For correct CV, ft their answer to (b)
NB dof 3 is 7.815 dof 4 is 9.488

A1ft: Ft "their 11.070" and their CV or \(p\) value. A correct conclusion independent of the hypotheses ie [If they should reject \(\mathrm{H}_0\) then they need "is not a suitable model. If they should accept \(\mathrm{H}_0\) then they need "is suitable"…] Allow Binomial is not a suitable model eg condone \(\mathrm{B}(500, 0.6)\) is not a suitable model. Do not accept contradictory statements

NB If \(p\) value [0.007419] given instead of CV they could get B1M1A1B0A1 unless they give the CV as well

Mark scheme (d)

Scheme	Marks	AO
\(\dfrac{[0 \times 2] + (1 \times 27) + (2 \times 93) + (3 \times 181) + (4 \times 146) + (5 \times 51)}{500}\) [\(= 3.19\)]	M1	3.3
\(\mathrm{B}([5],\ p = \dfrac{3.19}{5} = 0.638)\)	A1	1.1b
	(2)
	(9 marks)

Notes

M1: For a correct method using the data to improve the model. Implied by 3.19

A1: Correct model. Condone use of any value of \(n\) Accept Binomial with \(p = 0.638\)

A2 June 2022 Q1

EdexcelCurrent spec9 marksIncludes hypothesis testingChi-Squared Tests

1. A researcher is investigating the number of female cubs present in litters of size 4
He believes that the number of female cubs in a litter can be modelled by \(\mathrm{B}(4, 0.5)\)
He randomly selects 100 litters each of size 4 and records the number of female cubs.
The results are recorded in the table below.

Number of female cubs	0	1	2	3	4
Observed number of litters	10	33	33	15	9

He calculated the expected frequencies as follows

Number of female cubs	0	1	2	3	4
Expected number of litters	6.25	\(r\)	\(s\)	\(r\)	6.25

(a) Find the value of \(r\) and the value of \(s\) (3)

(b) Carry out a suitable test, at the 5% level of significance, to determine whether or not the number of female cubs in a litter can be modelled by \(\mathrm{B}(4, 0.5)\)
You should clearly state your hypotheses and the critical value used. (6)

Mark scheme (a)

Scheme	Marks	AO
\(r = \mathrm{P}(X = 3) \times 100\) or \(r = \mathrm{P}(X = 1) \times 100\) or \(s = \mathrm{P}(X = 2) \times 100\)	M1	3.4
\(r = \underline{25}\) (value may be in table)	A1	1.1b
\(s = \underline{37.5}\) (value may be in table)	A1	1.1b
	(3)

Notes

M1 Using the Binomial model to expected value. Allow if both probs 0.25 and 0.375 seen
May be implied by a correct value of \(r\) or \(s\). Alternatives \(r = 6.25 \times 4\) or \(s = 6.25 \times 6\)

1^st A1 for \(r = 25\)

2^nd A1 for \(s = 37.5\)

SC “B1” If M0 scored but their values of \(r\) and \(s\) satisfy \(2r + s = 87.5\) score as M0A0A1

Mark scheme (b)

Scheme

Marks

\(\mathrm{H}_0\): \(\mathrm{B}(4, 0.5)\) is a suitable model (o.e.) Condone \(\mathrm{B}(0.5, 4)\)
\(\mathrm{H}_1\): \(\mathrm{B}(4, 0.5)\) is not a suitable model (o.e.)

2.5

\(\dfrac{(O_i - E_i)^2}{E_i}\)	2.25	2.56	0.54	4	1.21
\(\dfrac{O_i^2}{E_i}\)	16	43.56	29.04	9	12.96

1.1b

\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i} = 10.56\) or \(\displaystyle\sum\frac{O_i^2}{E_i} - N = 110.56 - 100 = 10.56 \ \left(= \frac{264}{25}\right)\)

1.1b

\(\nu = 5 - 1 = 4\)

1.1b

CV = 9.488 (Calc 9.487729035…)

B1ft

1.1b

Significant so there is evidence that the researcher's model is not suitable

2.2b

(6)

Total 9

Notes

1^st B1 Both hypotheses correct using the correct notation in at least one or written in full e.g. binomial with \(n = 4\) and \(p = 0.5\)

M1 Calculating either \(\dfrac{(O_i - E_i)^2}{E_i}\) or \(\dfrac{O_i^2}{E_i}\) at least 4 correct. Implied by sight of awrt 10.6

1^st A1 Allow 10.6 (from correct working)

2^nd B1 Correct dof May be implied by CV of 9.48 or 9.49 or better

3^rd B1ft For 9.488 or better. Can ft their dof NB \(\chi^2_3(5\%) = 7.815\) (allow awrt 7.815)

2^nd A1 Indep’ of hypotheses but dep on 1^st A1
Evaluating the outcome by drawing a correct inference.
Compatible with comparison of 10.56 or 10.6 with their CV (which must be > 1)
They must say model not suitable (o.e.)
They do not need to state the comparison or say reject \(\mathrm{H}_0\) etc
No need to explicitly see \(\mathrm{B}(4, 0.5)\) mentioned here

AS June 2022 Q1

EdexcelAS paperCurrent spec7 marksIncludes hypothesis testingChi-Squared Tests

1. Stuart is investigating a treatment for a disease that affects fruit trees. He has 400 fruit trees and applies the treatment to a random sample of these trees. The remainder of the trees have no treatment. He records the number of years, \(y\), that each fruit tree remains free from this disease.

The results are summarised in the table below.

		Treatment
		Applied	Not applied
Number of years free from this disease	\(y \lt 1\)	15	25
	\(1 \leqslant y \lt 2\)	35	61
	\(2 \leqslant y\)	124	140

The data are to be used to determine whether or not there is an association between the application of the treatment and the number of years that a fruit tree remains free from this disease.

(a) Calculate the expected frequencies for

(i) Applied and \(y \lt 1\)

(ii) Not applied and \(1 \leqslant y \lt 2\) (2)

The value of \(\displaystyle\sum \frac{(O-E)^2}{E}\) for the other four classes is 2.642 to 3 decimal places.

(b) Test, at the 5% level of significance, whether or not there is an association between the application of the treatment and the number of years a fruit tree remains free from this disease.
You should state your hypotheses, test statistic, critical value and conclusion clearly. (5)

Mark scheme (a)

Scheme	Marks	AO
(i) \(\dfrac{40 \times 174}{400}\) (ii) \(\dfrac{96 \times 226}{400}\)	M1	1.1b
\(= 17.4\) \(= 54.24\)	A1	1.1b
	(2)

Notes

M1 A correct method to work out either expected frequencies – or 1 correct

A1 17.4 and 54.24 (accept 54.2)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0\): There is no association between the application of the treatment and the number of years that a fruit tree remains free from this disease. \(\mathrm{H}_1\): There is an association between the application of the treatment and the number of years that a fruit tree remains free from this disease.	B1	3.4
\(\displaystyle\sum \frac{(O-E)^2}{E} = \frac{(15 - \text{“}17.4\text{”})^2}{\text{“}17.4\text{”}} + \frac{(61 - \text{“}54.24\text{”})^2}{\text{“}54.24\text{”}} + 2.642\)	M1	1.1b
\(= 3.815\ldots\) awrt 3.82	A1	1.1b
[\(3.82 \lt\)] \(\chi^2_{2,(0.05)} = 5.991\)	B1	3.1b
There is no evidence of association between the application of the treatment and the number of years that a fruit tree remains free from this disease.	A1ft	2.2b
	(5)
	(7 marks)

Notes

B1: For both hypotheses in terms of "association" or independence" Must mention application/treatment and years in at least one and be connected correctly to \(\mathrm{H}_0\) and \(\mathrm{H}_1\)
[Use of link, relationship or connection. is B0 but allow for last A1ft]

M1: A correct method to find the total \(\chi^2\) value. ft their values from (a)
If no method shown at least 1 of the two missing \(\chi^2\) contributions must be correct
(\(0.331\ldots \left(\dfrac{48}{145}\right)\) and 0.8425… allow 2sf). Implied by awrt 3.82

A1: awrt 3.82 or awrt 3.83

B1: Using the degrees of freedom to find the \(\chi^2\) CV for the appropriate model. awrt 5.991 allow 5.9915

A1ft: Ft "their 3.82" and their CV or p-value. Correct conclusion in context. (application or treatment and years) This is independent of hypotheses ie if they should accept \(\mathrm{H}_0\) then they need eg there is no association between …. If they should reject \(\mathrm{H}_0\) then they need there is an association"… Allow relationship, link, connection for association BUT do not accept correlation or contradictory statements

NB If p-value [0.148388] given instead of CV could get B1M1A1B0A1 unless they give the CV as well

A2 October 2021 Q1

EdexcelCurrent spec7 marksIncludes hypothesis testingChi-Squared Tests

1. Kelly throws a tetrahedral die \(n\) times and records the number on which it lands for each throw.

She calculates the expected frequency for each number to be 43 if the die was unbiased.

The table below shows three of the frequencies Kelly records but the fourth one is missing.

Number	1	2	3	4
Frequency	47	34	36	\(x\)

(a) Show that \(x = 55\) (1)

Kelly wishes to test, at the 5% level of significance, whether or not there is evidence that the tetrahedral die is unbiased.

(b) Explain why there are 3 degrees of freedom for this test. (1)

Mark scheme (a)

Scheme	Marks	AO
\(x = 4 \times 43 - 47 - 34 - 36 = 55\)*	B1*	3.4
	(1)

Notes

B1*: Using the uniform model to show the missing observed value eg \(x = \dfrac{43 - 0.25 \times (47 + 34 + 36)}{0.25} = 55\)

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{H}_0\): The die is unbiased \(\mathrm{H}_1\): The die is biased	B1	2.1
Test Statistic \(= \dfrac{(47 - 43)^2}{43} + \dfrac{(34 - 43)^2}{43} + \dfrac{(36 - 43)^2}{43} + \dfrac{(55 - 43)^2}{43}\)	M1	1.1b
\(= 6.744\ldots\)	A1	1.1b
\(\chi^2_{(3, 0.05)} = 7.815\)	B1	1.1b
Not in the critical region since \(7.815 \gt \text{“}6.74\ldots\text{”}\) therefore insufficient evidence to reject \(\mathrm{H}_0\) Inconclusive test - consistent with the die being unbiased.	A1	3.5a
	(5)
	(7 marks)

Notes

B1: Both hypotheses correct. eg The data fits a discrete uniform distribution

M1: Attempting to find \(\sum\dfrac{(O - E)^2}{E}\) or \(\sum\dfrac{O^2}{E} - N\) May be implied by awrt 6.74 or \(p\) value of 0.0805…

A1: awrt 6.74 or \(\dfrac{290}{43}\) oe May be implied by \(p\) value of 0.0805…

B1: awrt 7.82 (Calc 7.8147…)

A1: Drawing correct inference in context. Need the word die or tetrahedral

A2 October 2020 Q5

EdexcelCurrent spec13 marksIncludes hypothesis testingChi-Squared Tests

5. A factory produces pins.

An engineer selects 40 independent random samples of 6 pins produced at the factory and records the number of defective pins in each sample.

Number of defective pins	0	1	2	3	4	5	6
Observed frequency	19	11	7	2	0	1	0

(a) Show that the proportion of defective pins in the 40 samples is 0.15 (2)

The engineer suggests that the number of defective pins in a sample of 6 can be modelled using a binomial distribution. Using the information from the sample above, a test is to be carried out at the 10% significance level, to see whether the data are consistent with the engineer’s suggested model.

The value of the test statistic for this test is 2.689

(b) Justifying the degrees of freedom used, carry out the test, at the 10% significance level, to see whether the data are consistent with the engineer’s suggested model.
State your hypotheses clearly. (8)

The engineer later discovers that the previously recorded information was incorrect.
The data should have been as follows.

Number of defective pins	0	1	2	3	4	5	6
Observed frequency	19	11	6	3	1	0	0

(c) Describe the effect this would have on the value of the test statistic that should be used for the hypothesis test.
Give reasons for your answer. (3)

Mark scheme (a)

Scheme	Marks	AO
\(p = \dfrac{(0) + 11 + 14 + 6 + (0) + 5 + (0)}{6 \times 40}\)	M1	2.1
\(p = \underline{\mathbf{0.15}}\)*	A1*cso	1.1b
	(2)

Notes

M1: Correct expression for \(p\) (may be seen in stages). Allow \(\tfrac{36}{240}\) but not \(\tfrac{6}{40}\) on its own

A1*cso: \(p = 0.15\) stated and no incorrect working seen

Mark scheme (b)

Scheme

Marks

\(X \sim \mathrm{B}(6, 0.15)\)

\(x\)	2	3	4	5	6
\(40 \times \mathrm{P}(X = x)\)	7.04…	1.65…	0.219…	0.015…	0.00…

3.4

Require \(40 \times \mathrm{P}(X \geqslant k) \gt 5\)
Exp. frequency for \(X \geqslant 2 = 8.94\ldots\) / \(X \geqslant 3 = 1.89\ldots\)

1.1b

Combine last 5 cells / only 3 cells in total

2.2a

2 is subtracted (as there are 2 restrictions) and the proportion used from data (and 1 equal totals)

2.4

\(3 - 2 = 1\) degree of freedom

1.1b

\(\mathrm{H}_0\): Binomial distribution is a suitable model
\(\mathrm{H}_1\): Binomial distribution is not a suitable model

3.4

Critical value \(\chi^2_{(1, 0.10)} = 2.705\) or 2.706

B1ft

1.1b

Test statistic is not in the critical region, insufficient evidence to reject \(\mathrm{H}_0\) (\(2.689 \lt 2.705/6\))
Data are consistent with binomial/engineer’s/suggested model.

B1ft

3.5a

(8)

Notes

M1: Attempting to find expected frequencies, at least 2 correct trunc. or rounded 1dp

M1: Recognising need to combine cells (Sight of awrt 8.94 implies M1M1)

A1: Combining cells for \(X \geqslant 2\) (to make 3 cells)

B1: Justifying why 2 is subtracted with \(p\) being calculated from data

A1: 1 degree of freedom

B1: Correct hypotheses (0.15 must not be included) Allow engineer’s model.

B1ft: Correct critical value (ft their df) May see \(\chi^2_{(2, 0.10)} = 4.605\) or \(\chi^2_{(3, 0.10)} = 6.251\)

B1ft: Correct inference (ft comparison of their CV with 2.689).
Condone \(p = 0.15\) included here. Do not allow contradictory statements to score here.
Hypotheses must be correct way round.

Mark scheme (c)

Scheme	Marks	AO
The total amount/proportion of defective pins remains the same.	M1	2.4
The cells for \(X \geqslant 2\) are still combined in the test.	M1	1.1b
So there is no change to the value of the test statistic.	A1	2.2a
	(3)
	(13 marks)

Notes

M1: Determining the number (\(N = 36\))/proportion (\(p = 0.15\)) of defective pins has not changed. e.g. \(11 + 12 + 9 + 4 = 36\). But not \(7 + 2 + 1 = 6 + 3 + 1\)

M1: Understanding the cells for \(X \geqslant 2\) are still combined in the test

A1: (dep on both M1s) Concluding that there is no change to the value of the test statistic.

AS October 2020 Q2

EdexcelAS paperCurrent spec15 marksIncludes hypothesis testingChi-Squared Tests

2. In an experiment, James flips a coin 3 times and records the number of heads. He carries out the experiment 100 times with his left hand and 100 times with his right hand.

	Number of heads
	0	1	2	3
Left hand	7	29	42	22
Right hand	13	35	36	16

(a) Test, at the 5% level of significance, whether or not there is an association between the hand he flips the coin with and the number of heads.
You should state your hypotheses, the degrees of freedom and the critical value used for this test. (7)

(b) Assuming the coin is unbiased, write down the distribution of the number of heads in 3 flips. (1)

(c) Carry out a \(\chi^2\) test, at the 10% level of significance, to test whether or not the distribution you wrote down in part (b) is a suitable model for the number of heads obtained in the 200 trials of James’ experiment.
You should state your hypotheses, the degrees of freedom and the critical value used for this test. (7)

Mark scheme (a)

Scheme

Marks

\(\mathrm{H}_0\): There is no association between the hand and the number of heads
\(\mathrm{H}_1\): There is an association between the hand and the number of heads

2.5

	0	1	2	3
E	\(\dfrac{100 \times 20}{200}\) \(= 10\)	\(\dfrac{100 \times 64}{200}\) \(= 32\)	\(\dfrac{100 \times 78}{200}\) \(= 39\)	\(\dfrac{100 \times 38}{200}\) \(= 19\)

M1
A1

1.1b
1.1b

\(\displaystyle\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(7-10)^2}{10} + \frac{(13-10)^2}{10} + \ldots + \frac{(16-19)^2}{19}\)

1.1b

\(= 3.7714\ldots\) awrt 3.77

1.1b

Degrees of freedom [\(= (4-1) \times (2-1)\)] \(= 3\)
\(\chi^2_{3,0.05} = 7.815\)

3.1b

(Do not reject \(\mathrm{H}_0\).) There is not enough evidence to suggest an association between the hand flipping the coin and the number of heads.

2.2b

(7)

Notes

B1: For both hypotheses correct with at least one in context.

M1: For attempt at \(\dfrac{\text{row total} \times \text{column total}}{\text{grand total}}\) (may be implied by one correct expected frequency). Working may be seen in table.

A1: All correct expected frequencies

M1: For applying \(\displaystyle\sum \frac{(O-E)^2}{E}\) ft their values

A1: awrt 3.77

M1: For using degrees of freedom to set up \(\chi^2\) model

A1: Correct conclusion in context with all other marks scored.

Mark scheme (c)

Scheme

Marks

\(\mathrm{H}_0\): \(\mathrm{B}(3, 0.5)\) is a suitable model
\(\mathrm{H}_1\): \(\mathrm{B}(3, 0.5)\) is not a suitable model

B1ft

3.4

	0	1	2	3
E	\(200 \times \mathrm{P}(X = 0)\) \(= 25\)	\(200 \times \mathrm{P}(X = 1)\) \(= 75\)	\(200 \times \mathrm{P}(X = 2)\) \(= 75\)	\(200 \times \mathrm{P}(X = 3)\) \(= 25\)

M1
A1

2.1
1.1b

\(\displaystyle\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(20-25)^2}{25} + \frac{(64-75)^2}{75} + \frac{(78-75)^2}{75} + \frac{(38-25)^2}{25}\)

1.1b

\(= 9.493\ldots\) awrt 9.49

1.1b

[df \(= 3\)] \(\chi^2_{3,0.1} = 6.251\)

3.1b

(Reject \(\mathrm{H}_0\)) \(\mathrm{B}(3, 0.5)\) is not a suitable model for the number of heads.

3.5a

(7)

(15 marks)

Notes

B1ft: For both hypotheses correct. Must have binomial and (3, 0.5) or ft their distribution in part (b)

M1: For attempt at expected frequencies using their distribution from part (b) (may be implied by one correct or correct ft expected frequency)

A1: All correct expected frequencies

M1: For applying \(\displaystyle\sum \frac{(O-E)^2}{E}\) ft their values

A1: awrt 9.49

M1: For using degrees of freedom to set up \(\chi^2\) model

A1: Correct conclusion in context with all other marks scored.

A2 June 2019 Q4

EdexcelCurrent spec19 marksIncludes hypothesis testingChi-Squared Tests

4. Liam and Simone are studying the distribution of oak trees in some woodland. They divided the woodland into 80 equal squares and recorded the number of oak trees in each square. The results are summarised in Table 1 below.

Number of oak trees in a square	0	1	2	3	4	5	6	7 or more
Frequency	1	4	21	23	13	11	7	0

Table 1

Liam believes that the oak trees were deliberately planted, with 6 oak trees per square and that a constant proportion \(p\) of the oak trees survived.

(a) Suggest the model Liam should use to describe the number of oak trees per square. (2)

Liam decides to test whether or not his model is suitable and calculates the expected frequencies given in Table 2.

Number of oak trees in a square	0 or 1	2	3	4	5	6
Expected frequency	5.53	14.89	24.26	22.24	10.87	2.21

Table 2

(b) Showing your working clearly, complete the test using a 5% level of significance. You should state your critical value and conclusion clearly. (7)

Simone believes that a Poisson distribution could be used to model the number of oak trees per square. She calculates the expected frequencies given in Table 3.

Number of oak trees in a square	0 or 1	2	3	4	5	6 or more
Expected frequency	12.69	16.07	\(s\)	14.58	\(t\)	9.37

Table 3

(d) Write down hypotheses to test the suitability of Simone’s model. (1)

The test statistic for this test is 8.749

(e) Complete the test. Use a 5% level of significance and state your critical value and conclusion clearly. (3)

(f) Using the results of these tests, explain whether the origin of this woodland is likely to be cultivated or wild. (2)

Mark scheme (a)

Scheme	Marks	AO
[\(T\) = no. of oak trees in a square] \(T \sim\) Binomial	M1	3.3
\(T \sim \mathrm{B}(6, p)\)	A1	1.1b
	(2)

Notes

M1 for choosing binomial A1 for \(\mathrm{B}(6, p)\) can be in words and allow \(\mathrm{B}(6, 0.55)\)

Mark scheme (b)

Scheme

Marks

Expected frequency for 6 is less than 5 so pool: new \(E_i = 13.08\)

2.1

\(\frac{(O_i - E_i)^2}{E_i}\)	0.051	2.51	0.0654	3.84	1.85
\(\frac{O_i^2}{E_i}\)	4.521	29.617	21.805	7.599	24.771

\(\displaystyle\sum \frac{(O_i - E_i)^2}{E_i} = 8.313\)

M1, A1

1.1b ×2

\(p\) needed estimating (\(\hat{p} = 0.55\)) so \(\nu = 5 - 2 = 3\); cv 7.815

B1, B1ft

1.1b ×2

Significant result, so Liam’s model is not suitable

M1, A1

1.1b
2.2b

(7)

Notes

1^st M1 for pooling last 2 classes (\(E_i = 13.08\) but accept 13.1)

2^nd M1 for at least 3 correct values or expressions. Either row to at least 2 sf

1^st A1 for awrt 8.31 (8.31 gets 3/3) [NB no pooling gives awrt 16.8458.. and implies M0M1A0]

1^st B1 for 3 degrees of freedom 2^nd B1ft for critical value of 7.815 (e.g. \(\nu = 4\) use 9.488)

3^rd M1 for a correct conclusion (non-contextual ignore any contradictory contextual comments for this mark) based on their cv and their test statistic
This mark can be implied by a fully correct solution ending with correct contextual conclusion

2^nd A1 for correct conclusion in context with all other marks scored

Mark scheme (c)

Scheme	Marks	AO
[\(R\) = no. of oak trees in a square for Simone’s model] \(R \sim \mathrm{Po}(3.3)\)	M1	3.3
Correct expression for \(s\) or \(t\) using Poisson	M1	3.4
\(s = \underline{\mathbf{17.67}}\) and \(t = \underline{\mathbf{9.62}}\)	A1, A1	1.1b ×2
	(4)

Notes

1^st M1 for selecting a correct model \(\mathrm{Po}(3.3)\) [Allow \(\mathrm{Po}(\text{awrt } 3.3)\)]

2^nd M1 for use of the model with an expression or correct value for \(s\) or \(t\)

1^st A1 for one correct 2^nd A1 for both correct (allow awrt 2dp)

Mark scheme (d)

Scheme	Marks	AO
\(\mathrm{H}_0\): Poisson is a good fit (for no. of oak trees per square) \(\mathrm{H}_1\): Poisson is not a good fit (for no. of oak trees per square)	B1	2.5
	(1)

Notes

B1 for correct hypotheses must mention Poisson: use of \(\mathrm{Po}(3.3)\) is B0

Mark scheme (e)

Scheme	Marks	AO
No pooling needed so degrees of freedom is \(6 - 2 = 4\)	B1	1.1b
Critical value is 9.488 (accept 9.49)	B1	1.1a
Not significant so Poisson (or Simone’s) model is suitable	B1	2.2b
	(3)

Notes

1^st B1 for correct degrees of freedom \(\nu = 4\) only

2^nd B1 for selecting correct critical value (9.488 only)

3^rd B1 for not significant conclusion based on 8.749 vs their cv (condone use of \(\mathrm{Po}(3.3)\) here)

Mark scheme (f)

Scheme	Marks	AO
Poisson model has better fit so suggests that oak trees occur at random Or binomial suggests deliberately planted or cultivated	B1	2.2b
Therefore the forest is likely to be wild not cultivated	B1	3.5a
	(2)
	(19 marks)

Notes

1^st B1 for choosing Poisson as better or stating Poisson implies wild or bino’l implies cultivated

2^nd B1 (dep on rejecting bin and accepting Poisson) for clearly stating woodland is wild
If the tests give the same results then 2^nd B0 automatically

AS June 2019 Q2

EdexcelAS paperCurrent spec7 marksIncludes hypothesis testingChi-Squared Tests

2. A spinner used for a game is designed to give scores with the following probabilities

Score	1	2	3	4	6
Probability	\(\dfrac{3}{10}\)	\(\dfrac{1}{10}\)	\(\dfrac{1}{10}\)	\(\dfrac{2}{5}\)	\(\dfrac{1}{10}\)

The spinner is spun 80 times and the results are as follows

Score	1	2	3	4	6
Frequency	15	4	12	41	8

Test, at the 10% level of significance, whether or not the spinner is giving scores as it is designed to do. Show your working and state your hypotheses clearly. (7)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\): Spinner is working as designed (o.e.)
\(\mathrm{H}_1\): Spinner is not working as designed (o.e.)

1.2

\(E_i\)	24	8	8	32	8
\(O_i\)	15	4	12	41	8
\(\dfrac{(O_i - E_i)^2}{E_i}\)	\(\dfrac{81}{24}\)	\(\dfrac{16}{8}\)	\(\dfrac{16}{8}\)	\(\dfrac{81}{32}\)	0
\(\dfrac{O_i^2}{E_i}\)	\(\dfrac{225}{24}\)	\(\dfrac{16}{8}\)	\(\dfrac{144}{8}\)	\(\dfrac{1681}{32}\)	\(\dfrac{64}{8}\)

M1
A1

M1

3.4
1.1b

1.1b

\(\displaystyle\sum \frac{(O_i - E_i)^2}{E_i} = 3.375 + 2 + 2 + 2.53125 + 0 = 9.90625\)
or \(\displaystyle\sum \frac{O_i^2}{E_i} - N = 9.375 + 2 + 18 + 52.53125 + 8 - 80 = 9.90625\)

1.1b

\(\nu = 5 - 1 = 4\) so \(\chi^2_4(10\%)\) cv \(= 7.779\) or better

3.4

Result is significant so there is evidence that the spinner is not operating as designed

A1cso

3.5a

(7 marks)

Notes

1^st B1 for both hypotheses given in suitable context

1^st M1 for using the model to find at least 2 correct expected frequencies

1^st A1 for all correct \(E_i\)

2^nd M1 for attempt to find test statistic (at least two correct expressions, fractions or decimals)

2^nd A1 for a correct test statistic (awrt 9.91) [ accept \(\tfrac{317}{32}\) ]

2^nd B1 for correct critical value (allow 7.78)
NB \(p\)- value \(= 0.042036\ldots\) so allow awrt 0.042

3^rd A1cso dep on all previous marks for a correct conclusion in context
(can be in terms of model or spinner’s design) Must mention spinner and scores or design
Accept “spinner is not accurate”

AS June 2019 Q1

EdexcelAS paperCurrent spec6 marksIncludes hypothesis testingChi-Squared Tests

1. A leisure club offers a choice of one of three activities to its 150 members on a Tuesday evening. The manager believes that there may be an association between the choice of activity and the age of the member and collected the following data.

Activity Age \(a\) years	Badminton	Bowls	Snooker
\(a \lt 20\)	9	3	3
\(20 \leqslant a \lt 40\)	10	10	14
\(40 \leqslant a \lt 50\)	16	15	5
\(50 \leqslant a \lt 60\)	15	13	11
\(a \geqslant 60\)	4	19	3

(a) Write down suitable hypotheses for a test of the manager’s belief. (1)

The manager calculated expected frequencies to use in the test.

(b) Calculate the expected frequency of members aged 60 or over who choose snooker, used by the manager. (1)

The test statistic used to test the manager’s belief is 19.583

(d) Using a 5% level of significance, complete the test of the manager’s belief. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): There is no association between age and activity \(\mathrm{H}_1\): There is an association between age and activity	B1	1.2
	(1)

Notes

B1 for both hypotheses in terms of “association” or “independence”. Must mention age and activity (or sport). [Use of “relationship” or “link” here is B0 but allow for last B1ft]

Mark scheme (c)

Scheme	Marks	AO
Since expected value in \(a \lt 20\) and snooker \(= 3.6 \lt 5\) we amalgamate two rows	B1	2.4
Table is now \(4 \times 3\) so degrees of freedom is \((4-1) \times (3-1) = 6\) (*)	B1*	1.1b
	(2)

Notes

1^st B1 for a reason to get a 4 x 3 table based on amalgamation of rows
Must mention \(a \lt 20\) and snooker and see 3.6 and be combining rows (not columns)

2^nd B1* for 6 degrees of freedom clearly coming from a 4 x 3 table formed from amalgamation of rows. [ 8 – 2 = 6 is B0 ]

Mark scheme (d)

Scheme	Marks	AO
Critical value \(\chi^2_6(5\%) = 12.592\)	B1	1.1b
[Significant result]: so there is evidence to support manager’s belief	B1ft	2.2b
	(2)
	(6 marks)

Notes

1^st B1 for correct critical value (allow 12.6 or 12.59 or awrt 12.592)
NB \(p\)-value \(= 0.0032839\ldots\) so allow awrt 0.00328

2^nd B1ft for a correct comparison and conclusion (ft their cv) [Independent of hypotheses]
e.g. there is an “association” or “relationship” or “link” between age and activity is OK
BUT there is a “correlation” between age and activity is B0

Do not accept contradictory contextual statements e.g. “manager’s belief supported , there is no association between age and activity”

AS June 2018 Q4

EdexcelAS paperCurrent spec7 marksIncludes hypothesis testingChi-Squared Tests

4. Abram carried out a survey of two treatments for a plant fungus. The contingency table below shows the results of a survey of a random sample of 125 plants with the fungus.

		Treatment
		No action	Plant sprayed once	Plant sprayed every day
Outcome	Plant died within a month	15	16	25
	Plant survived for 1 – 6 months	8	25	10
	Plant survived beyond 6 months	7	14	5

Abram calculates expected frequencies to carry out a suitable test. Seven of these are given in the partly-completed table below.

		Treatment
		No action	Plant sprayed once	Plant sprayed every day
Outcome	Plant died within a month			17.92
	Plant survived for 1 – 6 months	10.32	18.92	13.76
	Plant survived beyond 6 months	6.24	11.44	8.32

The value of \(\displaystyle\sum \frac{(O-E)^2}{E}\) for the 7 given values is 8.29

Test at the 2.5% level of significance, whether or not there is an association between the treatment of the plants and their survival. State your hypotheses and conclusion clearly. (7)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\): There is no association between the treatment of the plants and their survival/outcome.
\(\mathrm{H}_1\): There is an association between the treatment of the plants and their survival/outcome

3.4

	No action	Plant sprayed once	Plant sprayed every day
Plant died within a month	13.44	24.64	17.92
Plant survived for 1 – 6 months	10.32	18.92	13.76
Plant survived beyond 6 months	6.24	11.44	8.32

M1
A1

1.1b
1.1b

\(\displaystyle\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(15 - \text{“}13.44\text{”})^2}{\text{“}13.44\text{”}} + \frac{(16 - \text{“}24.64\text{”})^2}{\text{“}24.64\text{”}} + 8.29\)

1.1b

awrt 11.5

1.1b

Degrees of freedom \((3-1)(3-1) = 4\)
\(\chi^2_{4,0.025} = 11.143\)

3.1b

Reject \(\mathrm{H}_0\) There is an association between the treatment of the plants and their survival/outcome

dA1ft

2.2b

(7 marks)

Notes

B1: For correct hypotheses at least one in context. Allow independent and not independent. Do not accept correlation.

M1: For attempt at \(\dfrac{(\text{Row Total})(\text{Column Total})}{(\text{Grand Total})}\) to find expected frequencies. ( they may put numbers in table)

A1: awrt 13.44 and 24.64 This may be implied by a correct value of \(\chi^2\)

M1: For applying \(\displaystyle\sum \frac{(O-E)^2}{E}\) ft their expected values. If no method shown at least 1 of the two missing \(\chi^2\) contributions must be correct – you may need to check this (correct ones are \(0.181\ldots\) and \(3.0296\ldots\) allow 2sf) (condone missing 8.29)

A1: awrt 11.5

M1: For using degrees of freedom to set up \(\chi^2\) model critical value, implied by CV 11.143 or better

dA1ft: dependent on the 2^nd and 3^rd M marks. Correct conclusion ft their \(\displaystyle\sum \frac{(O-E)^2}{E}\) there is an association between the treatment of the plants and their survival/outcome: - do not allow contradicting statements. Do not award if hypotheses are the wrong way round or there are no hypotheses.

AS June 2018 Q1

EdexcelAS paperCurrent spec10 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

1. A researcher is investigating the distribution of orchids in a field. He believes that the Poisson distribution with a mean of 1.75 may be a good model for the number of orchids in each square metre. He randomly selects 150 non-overlapping areas, each of one square metre, and counts the number of orchids present in each square.

The results are recorded in the table below.

Number of orchids in each square metre	0	1	2	3	4	5	6
Number of squares	30	42	35	26	11	6	0

He calculates the expected frequencies as follows

Number of orchids in each square metre	0	1	2	3	4	5	More than 5
Number of squares	26.07	45.62	39.91	23.28	10.19	3.57	\(r\)

(a) Find the value of \(r\) giving your answer to 2 decimal places. (1)

The researcher will test, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution with mean 1.75

(b) State clearly the hypotheses required to test whether or not this Poisson distribution is a suitable model for these data. (1)

The test statistic for this test is 2.0 and the number of degrees of freedom to be used is 4

(d) Stating your critical value clearly, determine whether or not these data support the researcher’s belief. (2)

The researcher works in another field where the number of orchids in each square metre is known to have a Poisson distribution with mean 1.5

He randomly selects 200 non-overlapping areas, each of one square metre, in this second field, and counts the number of orchids present in each square.

(e) Using a Poisson approximation, show that the probability that he finds at least one square with exactly 6 orchids in it is 0.506 to 3 decimal places. (4)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0\): \(\mathrm{Po}(1.75)\) is a suitable model \(\mathrm{H}_1\): \(\mathrm{Po}(1.75)\) is not a suitable model	B1	3.4
	(1)

Notes

B1: For both hypotheses correct. Must have \(\mathrm{Po}(1.75)\) or Poisson with mean 1.75 and be attached to \(\mathrm{H}_0\) and \(\mathrm{H}_1\) the right way round.

Mark scheme (c)

Scheme	Marks	AO
Cells are combined for expected frequencies < 5 so combine the last 3 cells	B1	2.4
subtract 1 since totals agree	B1	2.4
	(2)

Notes

B1: Explaining why there are 5 classes. Must mention combine the 3 cells when frequencies < 5 or to combine the 3 cells to make frequency > 5

B1: Explaining why 1 is subtracted. Must say/show 1 is subtracted and Totals agree or Total frequency must be 150 or only need 4 pieces of data to find the other or \(\lambda\) is known or 1.75 is given.

NB B0 for “only 1 constraint” on its own.

Mark scheme (d)

Scheme	Marks	AO
\(\chi^2_4 = 9.488\)	B1	1.1b
therefore, the researcher’s belief is supported or evidence that Po(1.75) is a good model for the number of orchids in each square metre	B1ft	3.5a
	(2)

Notes

B1: awrt 9.49

B1ft: ft their critical value only. For drawing the correct conclusion – condone missing 1.75.
If hypotheses are the wrong way round or there are no hypotheses in (b) award B0

Mark scheme (e)

Scheme	Marks	AO
\(\mathrm{P}(\text{exactly 6 orchids}) =\) awrt 0.00353	B1	1.1b
\(X \sim \mathrm{B}(200, \text{“}0.00353\text{”})\) mean \(= 200 \times \text{“}0.00353\text{”} =\) awrt 0.706	M1	3.3
\(Y \sim \mathrm{Po}(\text{“}0.706\text{”})\) \(1 - \mathrm{P}(Y = 0) = 1 - \mathrm{e}^{-\text{“}0.706\text{”}}\)	M1	3.4
\(= 0.506\)*	A1*	2.1
	(4)
	(10 marks)

Notes

B1: awrt 0.00353. May be implied by awrt 0.706 for mean.

M1: Selecting the model \(\mathrm{B}(200, \text{“their P(exactly 6 orchids)”})\) and using \(np\) (\(0 \lt p \lt 1\)) to find the mean. May be implied by awrt 0.706

M1: Using the model Po(their \(np\)) and using or writing \(1 - \mathrm{P}(Y = 0)\) or \(1 - \mathrm{P}(Y \leqslant 0)\) or \(1 - \mathrm{e}^{-\text{“}0.706\text{”}}\)

A1*: only award if the previous 3 marks have been awarded. and 0.506 stated.

S3 June 2018 Q6

EdexcelOld spec18 marksIncludes hypothesis testingChi-Squared Tests

6. David carries out an experiment with 4 identical dice, each with faces numbered 1 to 6. He rolls the 4 dice and counts the number of dice showing an even number on the uppermost face. He repeats this 150 times. The results are summarised in the table below.

No. of dice showing an even number	0	1	2	3	4
Frequency	12	45	36	39	18

David defines the random variable \(C\) as the number of dice showing an even number on the uppermost face when the four dice are thrown.

David claims that \(C \sim \mathrm{B}(4, 0.5)\)

(a) Stating your hypotheses clearly and using a 1% level of significance, test David’s claim. Show your working clearly. (9)

John claims that \(C \sim \mathrm{B}(4, p)\)

(b) Calculate an estimate of the value of \(p\) from the summary of the results of David’s experiment. Show your working clearly. (2)

John decides to test his claim. He calculates expected frequencies using the results of David’s experiment and obtains the following table.

No. of dice showing an even number	0	1	2	3	4
Expected frequency	8.65	36.00	\(d\)	39.00	\(e\)

(d) State suitable hypotheses to test John’s claim. (1)

John obtained a test statistic of 16.9 and carries out a test at the 1% level of significance.

(e) State what conclusion John should make about his claim. (3)

Mark scheme (a)

Scheme

Marks

\(\mathrm{H}_0 :\) B(4, 0.5) is a suitable model \(\mathrm{H}_1 :\) B(4, 0.5) is not a suitable model

Even number count	\(O_i\)	\(E_i\)	\(\frac{(O_i - E_i)^2}{E_i}\)	\(\frac{O_i^2}{E_i}\)
0	12	9.375	0.735	15.36
1	45	37.5	1.5	54
2	36	56.25	7.29	23.04
3	39	37.5	0.06	40.56
4	18	9.375	7.935	34.56

\(E_i = 150 \times \mathrm{P}(X = i)\)

M1A1

\(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\) or \(\chi^2 = \sum \frac{O_i^2}{E_i} - N\)

M1A1

\(\chi^2 = 17.52\) or \(\chi^2 = 167.52 - 150 = 17.52\) awrt 17.5

\(\nu = 4, \quad \chi^2_4(1\%) = 13.277\)

B1, B1ft

(Reject \(\mathrm{H}_0\),) B(4,0.5) is not a suitable model or David’s claim incorrect.

(9)

Notes

(corrected from the printed mark scheme: the second statistic is printed as \(\chi = 167.52 - 150 = 17.52\))

1st B1 Accept ‘Binomial with \(p = 0.5\)’ replacing ‘B(4, 0.5)’

1st M1 for attempt at \(E_i = 150 \times \mathrm{P}(X = i)\) with at least 2 values correct.

1st A1 at least 4 \(E_i\) correct to 3sf cao. Condone truncation.

2nd M1 for at least 2 correct calculations from 4^th or 5^th column.
2nd A1 at least 4 correct to 3sf from 4^th or 5^th column. Condone truncation.

3rd A1 for a test statistic of awrt 17.5 Answer only implies 2ndM1 2ndA1 3rdA1

4th A1 for correct conclusion rejecting binomial model. Condone missing parameters here.
Award provided their test statistic >11.345

Mark scheme (b)

Scheme	Marks
\(\hat{p} = \dfrac{0 \times 12 + 1 \times 45 + 2 \times 36 + 3 \times 39 + 4 \times 18}{4 \times 150} = 0.51\)	M1 A1
	(2)

Notes

1st M1 for attempting \(\hat{p} = \frac{\sum fx}{600}\) with at least 2 values on the numerator correct

1st A1 for 0.51 cao

Mark scheme (c)

Scheme	Marks
\(d = 150 \times 6 \times 0.51^2 \times 0.49^2 = 56.205009\) awrt 56.2	M1, A1
\(e = 150 - (8.65 + 36.00 + 39.00 + \text{"}d\text{"}) = 10.144991\) awrt 10.1 or 10.2 or \(e = 150 \times 0.51^4 = 10.1478015\)	B1ft
	(3)

Notes

1st M1 \(d = 150 \times 6 \times (\text{their } \hat{p})^2 \times (1 - \text{their } \hat{p})^2\)

1st A1 awrt 56.2

1st B1ft awrt 10.1 or follow from "\(d\)"

Mark scheme (d)

Scheme	Marks
\(\mathrm{H}_0 :\) B(4, \(p\)) is a suitable model \(\mathrm{H}_1 :\) B(4, \(p\)) is not a suitable model	B1
	(1)

Notes

1st B1 accept \(\mathrm{H}_0 :\) Binomial is a suitable model \(\mathrm{H}_1 :\) Binomial is not a suitable model

Mark scheme (e)

Scheme	Marks
\(\nu = 3, \quad \chi^2_3(1\%) = 11.345\)	B1B1ft
(16.9>11.345) Reject \(\mathrm{H}_0\) Binomial is not a suitable model or John’s claim incorrect or equivalent contextualised statement that rejects the Binomial model.	B1
	(3)
	(18 marks)

Notes

1st B1 \(\nu = 3\), 2nd B1 11.345, follow through their \(\nu \neq\) their value in part (a)

3rd B1 Correct statement rejecting \(\mathrm{H}_0\)

S3 June 2017 Q4

EdexcelOld spec14 marksIncludes hypothesis testingChi-Squared Tests

4. A psychologist carries out a survey of the perceived body weight of 150 randomly chosen people. He asks them if they think they are underweight, about right or overweight. His results are summarised in the table below.

	Underweight	About right	Overweight
Male	20	22	30
Female	16	28	34

The psychologist calculates two of the expected frequencies, to 2 decimal places, for a test of independence between perceived body weight and gender. These results are shown in the table below.

	Underweight	About right	Overweight
Male	17.28
Female	18.72

(a) Complete the table of expected frequencies shown above. (2)

(b) Test, at the 10% level of significance, whether or not perceived body weight is independent of gender. State your hypotheses clearly. (7)

The psychologist now combines the male and female data to test whether or not body weight types are chosen equally.

(c) Find the smallest significance level, from the tables in the formula booklet, for which there is evidence of a preference. (5)

Mark scheme (a)

Scheme	Marks
\(\dfrac{72 \times 50}{150} = 24,\ \dfrac{78 \times 50}{150} = 26\)	M1
\(\dfrac{72 \times 64}{150} = 30.72,\ \dfrac{78 \times 64}{150} = 33.28\)	A1
	(2)

Notes

M1 For one correct \(\frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\); can be implied by correct answers.

A1 24, 26, 30.72, 33.28 only.

Mark scheme (b)

Scheme

Marks

\(\mathrm{H}_0 :\) Perceived (body) weight is independent of gender (no association)
\(\mathrm{H}_1 :\) Perceived (body) weight is not independent of gender (association)

\(O\)	\(E\)	\(\dfrac{(O-E)^2}{E}\)	\(\dfrac{O^2}{E}\)
20	17.28	0.428148	23.14815
22	24	0.166667	20.16667
30	30.72	0.016875	29.29688
16	18.72	0.395214	13.67521
28	26	0.153846	30.15385
34	33.28	0.015577	34.73558
150	150	1.176327	151.1763

M1A1

\(\displaystyle\sum \frac{(O-E)^2}{E}\) or \(\displaystyle\sum \frac{O^2}{E} - 150 = 1.18\)

\(\nu = (3-1)(2-1) = 2,\ \chi^2_2(10\%) = 4.605\)

B1B1ft

(Accept \(\mathrm{H}_0\)) Perceived (body) weight is independent of gender (no association)

A1ft

(7)

Notes

B1 Both hypotheses required. Must mention “Perceived”, “weight” and “gender” at least once.
Use of “relationship” or “correlation” or “connection” or “link” award B0.

M1 for at least 2 correct terms (as in 3^rd or 4^th column) or correct expressions.
A1 for all correct. Accept 2sf accuracy. Allow truncation e.g. 1.17…

A1 Awrt 1.18-1.19

B1B1ft 2 can be implied by 4.605 seen

A1ft A correct comment in context - must mention “weight” and “gender”. Condone “relationship” or “connection” here but not “correlation”.
Follow through from their test stat and cv, but hypotheses must be correct.

Mark scheme (c)

Scheme

Marks

\(O\)	\(E\)	\(\dfrac{(O-E)^2}{E}\)	\(\dfrac{O^2}{E}\)
36	50	3.92	25.92
50	50	0	50
64	50	3.92	81.92
150	150	7.84	157.84

B1
M1A1

\(\displaystyle\sum \frac{(O-E)^2}{E}\) or \(\displaystyle\sum \frac{O^2}{E} - 150 = 7.84\)

\(\nu = 2,\ \chi^2_2(2.5\%) = 7.378\)

(5)

(14 marks)

Notes

B1 for \(E_i = 50\), could be implied.

M1 for combining values and for attempting \(\dfrac{(O-E)^2}{E}\) or \(\dfrac{O^2}{E}\) with at least 2 correct expressions or values.

A1 for all correct, can be implied by correct answer below.

A1 Awrt 7.84

A1 0.025 or 2.5%

S3 June 2017 Q2

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

Figure 1: a circular dial marked from 0 to 345 degrees in steps of 15, with a pointer from the centre — Figure 1

The pointer shown in Figure 1 is spun so that it comes to rest between 0 and 360 degrees.

Linda claims that it is equally likely to come to rest at any point between 0 and 360 degrees. She spins the pointer 100 times and her results are summarised in the table below. She calculates expected frequencies for some of the possible outcomes and these are also given in the table below.

Angle (degrees)	0–45	45–90	90–180	180–315	315–360
Frequency	18	16	18	29	19
Expected frequency	12.5	\(a\)	\(b\)	\(c\)	12.5

(a) Find the values of the missing expected frequencies \(a\), \(b\) and \(c\). (2)

(b) Stating your hypotheses clearly and using a 5% level of significance, test whether or not Linda’s claim is supported by these data. (8)

Mark scheme (a)

Scheme	Marks
\(a = \dfrac{45}{360} \times 100 = 12.5\)	M1
\(b = \dfrac{90}{360} \times 100 = 25\)
\(c = \dfrac{135}{360} \times 100 = 37.5\)	A1
	(2)

Notes

M1 For one correct calculation

A1 12.5, 25, 37.5 only

Mark scheme (b)

Scheme

Marks

\(\mathrm{H}_0 :\) Continuous uniform distribution is a suitable model
\(\mathrm{H}_1 :\) Continuous uniform distribution is not a suitable model

\(O\)	\(E\)	\(\dfrac{(O-E)^2}{E}\)	\(\dfrac{O^2}{E}\)
18	12.5	2.42	25.92
16	12.5	0.98	20.48
18	25	1.96	12.96
29	37.5	1.927	22.43
19	12.5	3.38	28.88
100	100	10.67	110.67

M1A1

\(\displaystyle\sum \frac{(O-E)^2}{E} = 10.67\) or \(\displaystyle\sum \frac{O^2}{E} - 100 = 10.67\)

\(\nu = 5 - 1 = 4,\ \chi^2_4(5\%) = 9.488\)

B1,B1ft

(10.7>9.488) Reject \(\mathrm{H}_0\)

Linda’s claim is not supported.

A1ft

(8)

(10 marks)

Notes

B1 Both. Accept in terms of Linda’s claim. Accept U[0,360).

M1 for attempting \(\dfrac{(O-E)^2}{E}\) or \(\dfrac{O^2}{E}\) with at least 3 correct expressions or values.

A1 for all values correct to 2dp or as fractions. Can be implied by correct answer if all not listed.

A1 awrt 10.7

B1,B1ft 4 can be implied by awrt 9.488 seen or \(p =\) awrt 0.03

M1 Correct for their test stat and cv or thier \(p\) to 5% . Can be implied by correct conclusion from their test stat and cv.

A1ft A correct comment suggesting that continuous uniform model is not suitable or Linda’s claim is not correct. Linda’s claim must be described in full if Linda is not mentioned.
Follow through from their test stat and cv, but hypotheses must be correct.

S3 June 2016 Q6

EdexcelOld spec17 marksIncludes hypothesis testingChi-Squared Tests

6. An airport manager carries out a survey of families and their luggage. Each family is allowed to check in a maximum of 4 suitcases. She observes 50 families at the check-in desk and counts the total number of suitcases each family checks in. The data are summarised in the table below.

Number of suitcases	0	1	2	3	4
Frequency	6	25	12	6	1

The manager claims that the data can be modelled by a binomial distribution with \(p = 0.3\)

(a) Test the manager’s claim at the 5% level of significance. State your hypotheses clearly.
Show your working clearly and give your expected frequencies to 2 decimal places. (8)

The manager also carries out a survey of the time taken by passengers to check in. She records the number of passengers that check in during each of 100 five-minute intervals.

The manager makes a new claim that these data can be modelled by a Poisson distribution. She calculates the expected frequencies given in the table below.

Number of passengers	0	1	2	3	4	5 or more
Observed frequency	5	40	31	18	6	0
Expected frequency	16.53	29.75	\(r\)	\(s\)	7.23	3.64

(b) Find the value of \(r\) and the value of \(s\) giving your answers to 2 decimal places. (3)

Mark scheme (a)

Scheme

Marks

\(\mathrm{H_0}\): Binomial with \(p = 0.3\) is a good fit. \(\mathrm{H_1}\): Binomial with \(p = 0.3\) is not a good fit.

	0	1	2 or more
Observed	6	25	19
Expected	50x0.2401 =12.005 or 12.01 or 12.00	50x0.4116 =20.58	50x0.2646+50x0.0756+50x0.0081 =13.23+3.78+0.405 =17.415 or 17.41 or 17.42
\(\frac{(O - E)^2}{E}\)	3.003751	0.949291	0.144256
\(\frac{O^2}{E}\)	2.998751	30.36929	20.72926

M1A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 4.097\ldots\) or \(\displaystyle\sum \frac{O^2}{E} - N = 54.097\ldots - 50 = 4.097\ldots\) awrt 4.09-4.1(0)

dM1A1

\(\nu = 3 - 1 = 2\)

B1ft

\(\chi^2_2(5\%) = 5.991\ (\gt 4.1(0))\)

B1ft

Insufficient evidence to reject \(\mathrm{H_0}\) (Accept \(\mathrm{H_0}\))
Binomial with \(p = 0.3\) is a good fit.

(8)

Notes

B1 both including \(p = 0.3\)

M1 with some combined columns and at least one \(E\) correct to 2sf

A1 all correct to 2dp and total of expected values is 50.

dM1 either method

A1 awrt 4.09-4.1(0)

B1 ft their columns -1

B1 ft their

A1 cao

Mark scheme (c)

Scheme

Marks

\(\mathrm{H_0}\): Poisson is a good fit. \(\mathrm{H_1}\): Poisson is not a good fit.

	0	1	2	3	4 or more
Observed	5	40	31	18	6
Expected	16.53	29.75	26.78	16.07	10.87
\(\dfrac{(O - E)^2}{E}\)	\(\frac{11.53^2}{16.53} = 8.042\ldots\)	\(\frac{10.25^2}{29.75} = 3.532\ldots\)	\(\frac{4.22^2}{26.78} = 0.665\ldots\)	\(\frac{1.93^2}{16.07} = 0.232\ldots\)	\(\frac{4.87^2}{10.87} = 2.182\ldots\)
\(\dfrac{O^2}{E}\)	\(\frac{5^2}{16.53} = 1.512\ldots\)	\(\frac{40^2}{29.75} = 53.782\ldots\)	\(\frac{31^2}{26.78} = 35.885\ldots\)	\(\frac{18^2}{16.07} = 20.162\ldots\)	\(\frac{6^2}{10.87} = 3.312\ldots\)

\(\displaystyle\sum \frac{(O - E)^2}{E} = 14.65 - 14.66\) or \(\displaystyle\sum \frac{O^2}{E} - N = 114.65 - 100 = 14.65 - 14.66\)

M1A1

\(\nu = 5 - 1 - 1 = 3\)

B1 cao

\(\chi^2_3(1\%) = 11.345\ (\lt 14.65)\)

B1ft

Sufficient evidence to reject \(\mathrm{H_0}\)
Poisson is not a good fit.

A1 cao

(6)

(17 marks)

Notes

B1 no parameters included

M1 either method

B1 ft their \(\nu\)

S3 June 2016 Q2

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

2. A new drug to vaccinate against influenza was given to 110 randomly chosen volunteers. The volunteers were given the drug in one of 3 different concentrations, \(A\), \(B\) and \(C\), and then were monitored to see if they caught influenza. The results are shown in the table below.

	\(A\)	\(B\)	\(C\)
Influenza	12	29	9
No influenza	15	23	22

Test, at the 10% level of significance, whether or not there is an association between catching influenza and the concentration of the new drug. State your hypotheses and show your working clearly. You should state your expected frequencies to 2 decimal places. (10)

Mark scheme

Scheme

Marks

\(\mathrm{H_0}\): Drug concentration and catching influenza are independent / not associated
\(\mathrm{H_1}\): Drug concentration and catching influenza are not independent / associated

	\(A\)	\(B\)	\(C\)
Influenza	\(\frac{50 \times 27}{110} = 12.272\ldots\)	\(\frac{50 \times 52}{110} = 23.636\ldots\)	\(\frac{50 \times 31}{110} = 14.090\ldots\)	50
No influenza	\(\frac{60 \times 27}{110} = 14.727\ldots\)	\(\frac{60 \times 52}{110} = 28.363\ldots\)	\(\frac{60 \times 31}{110} = 16.909\ldots\)	60
	27	52	31	110

M1A1

\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)	\(\dfrac{O^2}{E}\)
12	12.272…	0.0060…	11.7333…
29	23.636…	1.2171…	35.5807…
9	14.090…	1.8392…	5.7483…
15	14.727…	0.0050…	15.2777…
23	28.363…	1.0142…	18.6506…
22	16.909…	1.5327…	28.6236…

M1A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 5.6145\ldots\) or \(\displaystyle\sum \frac{O^2}{E} - N = 115.62\ldots - 110 = 5.6145\ldots\) awrt 5.61-5.62

\(\nu = (3 - 1)(2 - 1) = 2\), \(\chi^2_2(10\%) = 4.605\)

B1B1ft

Reject \(\mathrm{H_0}\)

Drug concentration and catching influenza are not independent / are associated.

A1cao

(10 marks)

Notes

B1 hyps correct way around

M1 for correct expression at least once

A1 all seen and correct 2dp or better. Can be implied by test statistic of awrt 5.61-5.62.

M1 either method at least one correct

A1 at least 3 correct values. Can be implied by test statistic of awrt 5.61-5.62

A1 awrt 5.61-5.62

B1 cao

B1 follow through their \(\nu\)

M1 must correctly reject / not reject the null hypothesis based on their test stat and cv oe

S3 June 2015 Q6

EdexcelOld spec19 marksIncludes hypothesis testingChi-Squared Tests

Figure 1: a target of four concentric circles of radii 4 cm, 7 cm, 9 cm and 10 cm; the regions from the centre outwards are Green, Red, Blue and Yellow — Figure 1

The sketch in Figure 1 represents a target which consists of 4 regions formed from 4 concentric circles of radii 4 cm, 7 cm, 9 cm and 10 cm. The regions are coloured as labelled in Figure 1.
A random sample of 100 children each choose a point on the target and their results are summarised in the table below.

Colour of region	Green	Red	Blue	Yellow
Frequency	22	39	25	14

Caitland is trying to model the distribution of the points chosen by the children. She defines the random variable \(D\) to be the distance, in cm, of a point from the centre of the target and assumes \(D \sim \mathrm{U}[0, 10]\).

(a) Stating your hypotheses clearly and using a 1% level of significance, test whether or not U[0, 10] is a suitable model for these data. (9)

Henry claims that the points are randomly distributed over the target and the probability of a point being in any particular region is proportional to the area of that region.
He calculates expected frequencies and obtains the following table.

Colour of region	Green	Red	Blue	Yellow
Expected frequency	16	33	\(r\)	\(s\)

(b) Find the value of \(r\) and the value of \(s\). (3)

Henry obtained a test statistic of 6.188 and no groups were pooled.

Phoebe believes that the children chose the region of the target according to colour. She believes that boys and girls would favour different colours and splits the original data by gender to obtain the following table.

Observed frequencies

Colour of region	Green	Red	Blue	Yellow	Total
Boys	10	12	10	3	35
Girls	12	27	15	11	65

(d) State suitable hypotheses to test Phoebe’s belief. (1)

Phoebe calculated the following expected frequencies to carry out a suitable test.

Expected frequencies

Colour of region	Green	Red	Blue	Yellow
Boys	7.7	13.65	8.75	4.9
Girls	14.3	25.35	16.25	9.1

(e) Show how the value of 25.35 was obtained. (1)

Phoebe carried out the test using 2 degrees of freedom and a 10% level of significance.
She obtained a test statistic of 1.411

(f) Explain clearly why Phoebe used 2 degrees of freedom. (1)

(g) Stating your critical value clearly, determine whether or not these data support Phoebe’s belief. (2)

Mark scheme (a)

Scheme

Marks

\(\mathrm{H_0}\): U[0, 10] is a suitable model \(\mathrm{H_1}\): U[0, 10] is not a suitable model

\(D\)	\(O_i\)	\(E_i\)	\(\frac{(O_i - E_i)^2}{E_i}\)	\(\frac{O_i^{\,2}}{E_i}\)
0 – 4	22	40	8.1	12.1
4 – 7	39	30	2.7	50.7
7 – 9	25	20	1.25	31.25
9 – 10	14	10	1.6	19.6

Values of \(D\) (B1); Expected Freq (M1A1); 4th or 5th col (M1); \(\chi^2 = 13.65\) (A1)

B1
M1A1
M1
A1

\(\nu = 3\), \(\chi^2_3(1\%) = 11.345\)

B1, B1

[Reject \(\mathrm{H_0}\),] the uniform distribution over [0, 10] is not a suitable model

(9)

Notes

2nd B1 for the correct values for \(D\) (can be implied by 40, 30, 20, and 10.)

1st M1 for at least 2 expected frequencies or clear use of a correct formula e.g. \(0.4N\)

1st A1 for all the correct \(E_i\)

2nd M1 for at least 2 correct calculations from 4th or 5th column

2nd A1 for a test statistic of 13.65 (accept 13.7 to 3 sf)
Awrt 13.7 only scores 2nd B1M1A1M1A1

3rd A1 for a correct conclusion rejecting the uniform model. Award provided their test statistic > 11.345

Mark scheme (d)

Scheme	Marks
\(\mathrm{H_0}\): The colour/region chosen for the points is independent of gender (or no assoc’) \(\mathrm{H_1}\): The colour/region chosen for the points is dependent on gender (or assoc’)	B1
	(1)

Notes

B1 Independence or association mentioned at least once if ditto marks used.
Allow connection but not correlation.

Mark scheme (f)

Scheme	Marks
Expected frequency for Yellow and Boys is 4.9 < 5 so col. must be pooled/combined. [This gives a \(2 \times 3\) table so \(\nu = (2 - 1) \times (3 - 1) = 2\)]	B1
	(1)

Notes

B1 for recognising there is an \(E_i \lt 5\) and need for pooling/combining oe

Mark scheme (g)

Scheme	Marks
cv = 4.605	B1
[Not significant] so the data do not support Phoebe’s belief oe	B1
	(2)
	(19 marks)

Notes

2nd B1 for correctly stating that Phoebe’s belief is not supported by the data oe (depends on their cv being > 1.411)

S3 June 2014 (R) Q6

EdexcelOld spec17 marksIncludes hypothesis testingChi-Squared Tests

6. Bags of £1 coins are paid into a bank. Each bag contains 20 coins.

The bank manager believes that 5% of the £1 coins paid into the bank are fakes. He decides to use the distribution \(X \sim \mathrm{B}(20, 0.05)\) to model the random variable \(X\), the number of fake £1 coins in each bag.

(a) State the assumptions necessary for the binomial distribution to be an appropriate model in this case. (2)

The bank manager checks a random sample of 150 bags of £1 coins and records the number of fake coins found in each bag. His results are summarised in Table 1.

Number of fake coins in each bag	0	1	2	3	4 or more
Observed frequency	43	62	26	13	6
Expected frequency	53.8	56.6	\(r\)	8.9	\(s\)

Table 1

(b) Calculate the values of \(r\) and \(s\), giving your answers to 1 decimal place. (3)

(c) Carry out a hypothesis test, at the 5% significance level, to see if the data supports the bank manager’s statistical model. State your hypotheses clearly. (7)

The assistant manager thinks that a binomial distribution is a good model but suggests that the proportion of fake coins is higher than 5%. She calculates the actual proportion of fake coins in the sample and uses this value to carry out a new hypothesis test on the data. Her expected frequencies are shown in Table 2.

Number of fake coins in each bag	0	1	2	3	4 or more
Observed frequency	43	62	26	13	6
Expected frequency	44.5	55.7	33.2	12.5	4.1

Table 2

(d) Explain why there are 2 degrees of freedom in this case. (2)

(e) Given that she obtains a \(\chi^2\) test statistic of 2.67, test the assistant manager’s hypothesis that the binomial distribution is a good model for the number of fake coins in each bag. Use a 5% level of significance and state your hypotheses clearly. (3)

Mark scheme (b)

Scheme	Marks
\(r = 150 \times \mathrm{P}(X = 2) = 150 \times \dbinom{20}{2} \times 0.05^2 \times 0.95^{18}\)	M1
\(r = 28.3015\ldots\) awrt 28.3	A1
\(s = 150 - (53.8 + 56.6 + 28.3 + 8.9) = 2.4\)	A1ft
	(3)

Notes

M1A1 for one of \(r\) or \(s\) correct

A1ft for other value if using 150 - … and answer must be >0

Mark scheme (c)

Scheme

Marks

\(\mathrm{H_0}\): Bin(20, 0.05) is a suitable model
\(\mathrm{H_1}\): Bin(20, 0.05) is not a suitable model

Combining last two groups

	\(\geqslant 3\)
Observed frequency	19
Expected frequency	11.3

\(\nu = 4 - 1 = 3\)

Critical value, \(\chi^2(0.05) = 7.815\) (accept 9.488 if their \(\nu = 4\))

B1ft

Test statistic, \(\sum \frac{(O - E)^2}{E} = \frac{(43 - 53.8)^2}{53.8} + \frac{(62 - 56.6)^2}{56.6} + \cdots\)

\(= 2.168\ldots + 0.515\ldots + 0.186\ldots + 5.246\ldots\)
\(= 8.117\) (accept 10.16 if groups not combined)

A1ft

In critical region, sufficient evidence to reject \(\mathrm{H_0}\), accept \(\mathrm{H_1}\)
Significant evidence at 5% level to reject the manager’s model

A1ft

(7)

Notes

1st B1 can be in words but must include \(p = 0.05\)

3rd B1 ft on their \(\nu\)

Test statistic alternative method
Test stat \(= \sum \frac{O^2}{E} - 150 = \frac{43^2}{53.8} + \frac{62^2}{56.6} + \cdots - 150 = 8.117\ldots\)

1st A1 ft if their groups not combined

2nd A1 ft their test and critical values but must be comment in context e.g. mention of “manager’s model” or “fake coins”

Mark scheme (d)

Scheme	Marks
\(\nu = 4 - 2 = 2\)
4 classes due to pooling	B1
2 restrictions (equal total and mean/proportion)	B1
	(2)

Notes

1st B1 evidence that pooling is required

2nd B1 must have correct reasons for restrictions.

Mark scheme (e)

Scheme	Marks
\(\mathrm{H_0}\): Binomial distribution is a good model \(\mathrm{H_1}\): Binomial distribution is not a good model	B1
Critical value, \(\chi^2(0.05) = 5.991\)	B1
Test statistic is not in critical region, insufficient evidence to reject \(\mathrm{H_0}\) Accept the assistant manager’s model for the number of fake coins per bag.	B1
	(3)
	(17 marks)

S3 June 2014 (R) Q2

EdexcelOld spec7 marksIncludes hypothesis testingChi-Squared Tests

2. A survey asked a random sample of 200 people their age and the main use of their mobile phone.

The results are shown in Table 1 below.

		Main use of their mobile phone
		Internet	Texts	Phone calls
Age	Under 20	27	14	9
	From 20 to 40	32	34	29
	Over 40	15	19	21

Table 1

The data are to be used to test whether or not age and main use of their mobile phone are independent.

Table 2 shows the expected frequencies for each group, assuming people’s age and main use of their mobile phone are independent.

		Main use of their mobile phone
		Internet	Texts	Phone calls
Age	Under 20	18.5	16.75	14.75
	From 20 to 40	35.15	31.825	28.025
	Over 40	20.35	18.425	16.225

Table 2

(a) For users under 20 choosing the Internet as the main use of their mobile phone,

(i) verify that the expected frequency is 18.5

(ii) show that the contribution to the \(\chi^2\) test statistic is 3.91 to 3 significant figures.

(2)

(b) Given that the \(\chi^2\) test statistic for the data is 9.893 to 3 decimal places, test at the 5% level of significance whether or not age and main use of their mobile phone are independent. State your hypotheses clearly. (5)

Mark scheme (b)

Scheme	Marks
\(\mathrm{H_0}\): users age and main mobile phone use are independent/ no association between users age and main mobile phone use \(\mathrm{H_1}\): users age and main mobile phone use are not independent/ some association between users age and main mobile phone use	B1
\(\nu = 4\)	B1
Critical value \(\chi^2 = 9.488\)	B1ft
Test statistic is in critical region therefore significant evidence to reject \(\mathrm{H_0}\) and accept \(\mathrm{H_1}\).	M1
Evidence at 5% level that age and main phone use are not independent.	A1ft
	(5)
	(7 marks)

Notes

3rd B1 ft on their value of \(\nu\)

M1 for attempt to compare test statistic and their critical value

A1 ft on test statistic and critical value but must be comment in context. (A0 if hypotheses are the wrong way around)

S3 June 2014 Q5

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests

5. A research station is doing some work on the germination of a new variety of genetically modified wheat.

They planted 120 rows containing 7 seeds in each row.

The number of seeds germinating in each row was recorded. The results are as follows

Number of seeds germinating in each row	0	1	2	3	4	5	6	7
Observed number of rows	2	6	11	19	25	32	16	9

(a) Write down two reasons why a binomial distribution may be a suitable model. (2)

(b) Show that the probability of a randomly selected seed from this sample germinating is 0.6 (2)

The research station used a binomial distribution with probability 0.6 of a seed germinating. The expected frequencies were calculated to 2 decimal places. The results are as follows

Number of seeds germinating in each row	0	1	2	3	4	5	6	7
Expected number of rows	0.20	2.06	\(s\)	23.22	\(t\)	31.35	15.68	3.36

(d) Stating your hypotheses clearly, test, at the 1% level of significance, whether or not the data can be modelled by a binomial distribution. (7)

Mark scheme (a)

Scheme	Marks
The seeds are independent / There are a fixed number of seeds in a row / There are only two outcomes to the seed germinating – either it germinates or it does not / The probability of a seed germinating is constant	B1 B1
	(2)

Notes

Any two and at least one must have context. 2 correct, no context B1B0. Do not award B0B1.

Mark scheme (b)

Scheme	Marks
\(\dfrac{(0 \times 2) + (1 \times 6) + (2 \times 11) + (3 \times 19) + (4 \times 25) + (5 \times 32) + (6 \times 16) + (7 \times 9)}{120 \times 7} = \dfrac{504}{840}\)	M1
\(= 0.6\) **	A1cso
	(2)

Notes

M1 require at least two correct terms in numerator and /(120x7) or /120 then /7

A1 cso as given answer

Mark scheme (c)

Scheme	Marks
\(p = 0.6 \quad q = 0.4\) \(s = 120 \times 21q^5p^2 = 120 \times 21 \times 0.4^5 \times 0.6^2 = 9.29\)	B1
\(t = 120 \times 35q^3p^4 = 120 \times 35 \times 0.4^3 \times 0.6^4 = 34.84\)	B1
	(2)

Notes

Cao for each B1

Mark scheme (d)

Scheme

Marks

\(\mathrm{H_0}\): A binomial distribution is a suitable model.
\(\mathrm{H_1}\): A binomial distribution is not a suitable model.

Observed number of rows	19	19	25	32	25
Expected number of rows	11.55	23.22	34.84	31.35	19.04
\(\dfrac{(O - E)^2}{E}\)	4.81	0.77	2.78	0.013	1.87
\(\dfrac{O^2}{E}\)	31.26	15.55	17.94	32.66	32.83

\(\nu = 5 - 2 = 3\)

B1ft

Critical value for \(\chi^2 = 11.345\)

B1ft

\(\displaystyle\sum \frac{(O - E)^2}{E} = 10.23\) or \(\displaystyle\sum \frac{O^2}{E} - N = 130.23 - 120 = 10.23\)

M1A1

10.23 < 11.345 therefore do not reject \(\mathrm{H_0}\)
A binomial is a suitable model.

(7)

(13 marks)

Notes

1st B1 for both hypotheses. B0 if they include 0.6 Condone \(X \sim \mathrm{B}(n, p)\) etc

1st M1 for using some combined columns (<8)

2nd B1ft follows from ‘their no of columns’ -2

3rd B1ft follows from the degrees of freedom

2nd M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\) with at least 2nd (3 seeds) and 4th (5 seeds) accurate to 2sf
Contradictory statements score M0 e.g. “significant” do not reject \(\mathrm{H_0}\)

1st A1 for awrt 10.2

2nd A1 dependent on 2nd M for a correct comment suggesting that binomial model is suitable. No follow through.
Condone mention of 0.6 here. Hypotheses wrong way round scores A0

S3 June 2014 Q3

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

3. A number of males and females were asked to rate their happiness under the headings “not happy”, “fairly happy” and “very happy”.

The results are shown in the table below

		Happiness			Total
		Not happy	Fairly happy	Very happy	Total
Gender	Female	9	43	34	86
Gender	Male	13	25	16	54
Total		22	68	50	140

Stating your hypotheses, test at the 5% level of significance, whether or not there is evidence of an association between happiness and gender. Show your working clearly. (10)

Mark scheme

Scheme

Marks

		Happiness
		Not happy	Fairly happy	Very happy
Gender	Female	13.51	41.77	30.71
Gender	Male	8.49	26.23	19.29

M1
A1

\(\mathrm{H_0}\): Happiness and gender are independent/ not associated
\(\mathrm{H_1}\): Happiness and gender are not independent/ associated

\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)	\(\dfrac{O^2}{E}\)
9	13.51	1.508	5.996
43	41.77	0.0361	44.264
34	30.71	0.351	37.637
13	8.49	2.402	19.915
25	26.23	0.0575	23.829
16	19.29	0.560	13.274

dM1
A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 4.91\) or \(\displaystyle\sum \frac{O^2}{E} - N = 144.91 - 140 = 4.91\)

\(\nu = (3 - 1)(2 - 1) = 2\)

\(\displaystyle\sum \frac{(O - E)^2}{E} \lt 5.991\)

B1ft

4.91 < 5.991 so ‘insufficient evidence to reject \(\mathrm{H_0}\)’ or ‘Accept \(\mathrm{H_0}\)’

No association between gender and happiness.

(10 marks)

Notes

(Corrected from the printed mark scheme: the degrees of freedom line is printed as \(\nu = (3 - 2)(2 - 1) = 2\).)

1st M1 for some use of \(\dfrac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}\). May be implied by at least 1 correct \(E_i\)

1st A1 awrt 13.5, 41.8, 30.7, 8.5, 26.2 and 19.3 Allow M1A0 for \(E_i\) rounded to integers

1st B1 for both hypotheses. Must mention “happiness” and “gender” at least once.
Use of “relationship” or “correlation” or “connection” is B0

2nd dM1 for at least 2 correct terms (in 3rd or 4th columns) or correct expressions with their \(E_i\)
Dependent on 1st M1. Accept 2sf accuracy for the M mark.

2nd A1 for all correct terms (2sf or better). May be implied by a correct ans
Allow truncation e.g. 44.2…

3rd A1 awrt 4.91 . Condone 4.915

2nd B1 for correct degrees of freedom (may be implied by a cv of 5.991)

3rd B1ft for cv that follows from their degrees of freedom

3rd M1 for a correct statement linking their test statistic and their cv
Contradictory statements score M0 e.g. “significant, do not reject \(\mathrm{H_0}\)”
Condone “reject \(\mathrm{H_1}\)”

4th A1 for a correct comment in context - must mention “gender” and “happiness”
Condone “relationship” or “connection” here but not “correlation”.
e.g. “There is no evidence of a relationship between gender and happiness”
No follow through. If e.g hypotheses are the wrong way around A0 here.

SC Use of calculator with no working may get M0A0B1M1A0A1B1B1M1A1

S3 June 2013 (R) Q4

EdexcelOld spec12 marksIncludes hypothesis testingChi-Squared Tests

4. John thinks that a person’s eye colour is related to their hair colour. He takes a random sample of 600 people and records their eye and hair colours. The results are shown in Table 1.

		Hair colour
		Black	Brown	Red	Blonde	Total
Eye colour	Brown	45	125	15	58	243
	Blue	34	90	10	58	192
	Hazel	20	38	16	26	100
	Green	6	29	7	23	65
	Total	105	282	48	165	600

Table 1

John carries out a \(\chi^2\) test in order to test whether eye colour and hair colour are related. He calculates the expected frequencies shown in Table 2.

		Hair colour
		Black	Brown	Red	Blonde
Eye colour	Brown	42.5	114.2	19.4	66.8
	Blue	33.6	90.2	15.4	52.8
	Hazel	17.5	47	8	27.5
	Green	11.4	30.6	5.2	17.9

Table 2

(a) Show how the value 47 in Table 2 has been calculated. (1)

(b) Write down the number of degrees of freedom John should use in this \(\chi^2\) test. (1)

Given that the value of the \(\chi^2\) statistic is 20.6, to 3 significant figures,

(c) find the smallest value of \(\alpha\) for which the null hypothesis will be rejected at the \(\alpha\)% level of significance. (1)

(d) Use the data from Table 1 to test at the 5% level of significance whether or not the proportions of people in the population with black, brown, red and blonde hair are in the ratio 2:6:1:3
State your hypotheses clearly. (9)

Mark scheme (d)

Scheme

Marks

\(\mathrm{H_0}\): hair colour occurs in the ratio 2:6:1:3
\(\mathrm{H_1}\): hair colour does not occur in the ratio 2:6:1:3

	black	brown	red	blonde
observed	105	282	48	165
expected	100	300	50	150
\(\dfrac{(O_i - E_i)^2}{E_i}\)	0.25	1.08	0.08	1.5
\(\dfrac{O_i^{\,2}}{E_i}\)	110.25	265.08	46.08	181.5

B1 expected
M1
A1

\(\displaystyle\sum \frac{(O_i - E_i)^2}{E_i} = 2.91\) or \(\displaystyle\sum \frac{O_i^{\,2}}{E_i} - 600 = 602.91 - 600 = 2.91\) (awrt 2.91)

\(\nu = 3\)

cv is 7.815

[2.91 < 7.815] so insufficient evidence to reject \(\mathrm{H_0}\) or not significant

dM1

There is evidence to suggest that hair colour does occur in the given ratio.

(9)

(12 marks)

Notes

1st B1 for both hypotheses. Must mention hair colour and ratio
e.g. “hair colour in the given ratio” Allow use of ditto

2nd B1 for all 4 correct expected frequencies

1st M1 for at least 2 correct calculations from 3rd or 4th row

1st A1 for all correct calculations to at least 3sf if row 4
If awrt 2.91 is seen with no incorrect working award B1M1A1A1

2nd dM1 Dep on 1st M1 for a correct statement linking their test statistic and their cv (cv > 3.5)

3rd A1 for a correct comment in context - must mention “hair colour” and “ratios” or “model”
e.g. “There is evidence of to support the given model” No follow through
If hypotheses are the wrong way round score A0.

S3 June 2013 Q4

EdexcelOld spec14 marksIncludes hypothesis testingChi-Squared Tests

4. Customers at a post office are timed to see how long they wait until being served at the counter. A random sample of 50 customers is chosen and their waiting times, \(x\) minutes, are summarised in Table 1.

Waiting time in minutes (\(x\))	Frequency
0–3	8
3–5	12
5–6	13
6–8	9
8–12	8

Table 1

(a) Show that an estimate of \(\bar{x} = 5.49\) and an estimate of \(s_x^2 = 6.88\) (3)

The post office manager believes that the customers’ waiting times can be modelled by a normal distribution.
Assuming the data is normally distributed, she calculates the expected frequencies for these data and some of these frequencies are shown in Table 2.

Waiting Time	\(x \lt 3\)	3–5	5–6	6–8	\(x \gt 8\)
Expected Frequency	8.56	12.73	7.56	\(a\)	\(b\)

Table 2

(b) Find the value of \(a\) and the value of \(b\). (3)

Mark scheme (a)

Scheme	Marks
\(\bar{x} = \dfrac{8 \times 1.5 + 12 \times 4 + 13 \times 5.5 + 9 \times 7 + 8 \times 10}{50} = \dfrac{274.5}{50} = 5.49\) (*)	B1cso
\(s^2 = \dfrac{8 \times 1.5^2 + 12 \times 4^2 + 13 \times 5.5^2 + 9 \times 7^2 + 8 \times 10^2}{49} - \dfrac{50}{49}5.49^2, = 6.88\) (*)	M1,A1cso
	(3)

Notes

B1cso for denominator of 50 and at least 3 products on num or 274.5 on num

M1 for a correct expression with at least 3 correct products on num or \(\dfrac{1844.25}{49} - \dfrac{1507.005}{49}\) or \(\dfrac{337.245}{49}\) or \(\left(\dfrac{7377}{200} - 5.49^2\right) \times \dfrac{50}{49}\) etc Allow 3sf accuracy

A1cso for 6.88 with M1 scored and no incorrect working seen

Mark scheme (b)

Scheme	Marks
\(a = \quad 50 \times \mathrm{P}(6 \lt X \lt 8) = 50 \times \mathrm{P}(0.194.. \lt Z \lt 0.956..)\)	M1
\(a = 12.81\) (tables) or 12.68 (calc)	A1
\(b = 50 - (28.85 + a) \quad = 8.34\) (tables) or 8.47 (calc)	A1ft
	(3)

Notes

M1 a full method for \(a\) or \(b\) using the normal dist. Correct use of (6), 8, 5.49 and \(\sqrt{6.88}\) seen

1st A1 for \(a\) in range 12.68 ~ 12.81 or \(b\) in range 8.34~ 8.47 or awrt these values

2nd A1ft for 50 – 28.85 – their \(a\) (or \(b\)) (but requires M1). Allow awrt 3sf. Must add up to 50

Mark scheme (c)

Scheme

Marks

\(\mathrm{H}_0\): Normal distribution is a good fit \(\mathrm{H}_1\): Normal distribution is not a good fit

Class	\(O\)	\(E\)	\(\dfrac{O^2}{E}\)	\(\dfrac{(O - E)^2}{E}\)
0-3	8	8.56	7.4766…	0.0366…
3-5	12	12.73	11.31186….	0.0418…
5-6	13	7.56	22.354497…	3.9144…
6-8	9	12.68 or (12.81)	(6.32) ~ 6.38801..	1.0680...~ (1.13)
8-12	8	(8.34) or 8.47	7.556080…~ (7.67)	(0.013) ~ 0.0260..

M1

A1

\(\displaystyle\sum\frac{O^2}{E} - N = 5.087\ldots \sim 5.1400\ldots\) awrt (5.09 ~ 5.14)

\(\nu = 5 - 3 = 2\) (for 5 – 3 or 2 can be implied by 5.991 seen)

\(\chi^2_2(0.05) = 5.991\)

\(5.09 \lt 5.991\) so insufficient evidence to reject \(\mathrm{H}_0\)

Normal distribution is a good fit.

(8)

(14 marks)

Notes

1st B1 for both hypotheses. B0 if they include 5.49 or 6.88. Condone \(X\)~\(\mathrm{N}(\mu, \sigma^2)\) etc

1st M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 3 correct expressions or values.

1st A1 for at least 4 correct calcs - 3^rd or 4^th column. (2 dp or better and allow e.g. 7.47)
Allow any value in the ranges for the last two rows.

2nd A1 for a test statistic that is awrt 5.09 ~ 5.14. Award M1A1A1 if this is obtained.

2nd M1 for a correct statement based on their test statistic ( > 1) and their cv (> 3.8)
Contradictory statements score M0 e.g. “significant” do not reject \(\mathrm{H}_0\).

3rd A1 for a correct comment suggesting that normal model is suitable or manager’s belief is correct. No f t . Condone mention of 5.49 or 6.88 here. Hypotheses wrong way round scores A0

S3 June 2013 Q1

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

1. A doctor takes a random sample of 100 patients and measures their intake of saturated fats in their food and the level of cholesterol in their blood. The results are summarised in the table below.

Cholesterol level Intake of saturated fats	High	Low
High	12	8
Low	26	54

Using a 5% level of significance, test whether or not there is an association between cholesterol level and intake of saturated fats. State your hypotheses and show your working clearly. (10)

Mark scheme

Scheme

Marks

Cholesterol Level	High	Low
High	7.6	12.4	20
Low	30.4	49.6	80
	38	62	100

M1A1 (2)

\(\mathrm{H}_0\) : Cholesterol level is independent of intake of saturated fats(no association)
\(\mathrm{H}_1\) : Cholesterol level is not independent of intake of saturated fats (association)

B1 (1)

\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)	\(\dfrac{O^2}{E}\)
12	7.6	2.547… or \(\frac{242}{95}\)	18.947… or \(\frac{360}{19}\)
8	12.4	1.56129… or \(\frac{242}{155}\)	5.161… or \(\frac{160}{31}\)
26	30.4	0.6368… or \(\frac{121}{190}\)	22.236… or \(\frac{845}{38}\)
54	49.6	0.3903… or \(\frac{121}{310}\)	58.790… or \(\frac{3645}{62}\)

dM1
A1

\(\displaystyle\sum\frac{(O - E)^2}{E} = 5.1358234..\) or \(\dfrac{12^2}{7.6} + \dfrac{8^2}{12.4} + \dfrac{26^2}{30.4} + \dfrac{54^2}{49.6} - 100 = 5.14\) (awrt 5.14)

A1 (3)

\(\nu = (2 - 1)(2 - 1) = 1\)

\(\chi^2_1(0.05) = 3.841\)

B1 (2)

\(5.14 \gt 3.841\) so sufficient evidence to reject \(\mathrm{H}_0\) [Condone “accept \(\mathrm{H}_1\)”]

Association between cholesterol level and saturated fat intake

A1 (2)

(10 marks)

Notes

Minimum working use part marks: \(E_i\) (2), Hyp (1), 5.14 (3), 3.841 (2), Conclusion (2)

1st M1 for some use of \(\dfrac{\text{Row Total} \times \text{Col.Total}}{\text{Grand Total}}\). May be implied by correct \(E_i\)

1st A1 for all expected frequencies correct. Allow M1A0 for \(E_i\) rounded to integers

1st B1 for both hypotheses. Must mention “cholesterol” and “fats” at least once
Use of “relationship” or “correlation” or “connection” is B0

2nd dM1 for at least 2 correct terms (as in 3^rd or 4^th column) or correct expressions with their \(E_i\)
Dependent on 1^st M1 Accept 2sf accuracy for the M mark

2nd A1 for all correct terms. May be implied by a correct ans.(2 dp or better)
Allow truncation eg 2.54… 3rd A1 for awrt 5.14

2nd B1 for correct degrees of freedom (may be implied by a cv of 3.841)

3rd M1 for a correct statement linking their test statistic and their cv(cv could be 2.705 or > 3.5)
Contradictory statements score M0 e.g. “significant, do not reject \(\mathrm{H}_0\)”

4th A1 for a correct comment in context - must mention “cholestorol” and “fats” condone “relationship” or “connection” here but not “correlation”.
e.g. “There is evidence of a relationship between cholesterol level and fat intake”
No follow through. If e.g hypotheses are the wrong way round A0 here.

(corrected from the printed mark scheme: the first term of the alternative calculation is printed as \(\frac{1.2^2}{7.6}\); the observed frequency is 12, so it is \(\frac{12^2}{7.6}\))

S3 June 2012 Q6

EdexcelOld spec14 marksIncludes hypothesis testingChi-Squared Tests

6. A total of 100 random samples of 6 items are selected from a production line in a factory and the number of defective items in each sample is recorded. The results are summarised in the table below.

Number of defective items	0	1	2	3	4	5	6
Number of samples	6	16	20	23	17	10	8

(a) Show that the mean number of defective items per sample is 2.91 (2)

A factory manager suggests that the data can be modelled by a binomial distribution with \(n = 6\). He uses the mean from the sample above and calculates expected frequencies as shown in the table below.

Number of defective items	0	1	2	3	4	5	6
Expected frequency	1.87	10.54	24.82	\(a\)	22.01	8.29	\(b\)

(b) Calculate the value of \(a\) and the value of \(b\) giving your answers to 2 decimal places. (4)

(c) Test, at the 5% level, whether or not the binomial distribution is a suitable model for the number of defective items in samples of 6 items.
State your hypotheses clearly. (8)

Mark scheme (b)

Scheme	Marks
\(p = \dfrac{2.91}{6} = 0.485\)	B1
\(a = 100 \times {}^6\mathrm{C}_3 \times 0.485^3 \times 0.515^3 = 31.17\)	M1A1
\(b = 100 \times 0.485^6 = 1.3(0)\)	A1
	(4)

Notes

0.485 can be implied by at least 1 correct answer.

Accept awrt 2dp for final answers

Clear use of Binomial and x100 required for method.

Mark scheme (c)

Scheme

Marks

\(\mathrm{H}_0\) : Binomial is a good fit
\(\mathrm{H}_1\) : Binomial is a not a good fit

Number of defective items	0 or 1	2	3	4	5 or 6
\(O\)	22	20	23	17	18
\(E\)	12.41	24.82	31.17	22.01	9.59

\(\displaystyle\sum\frac{(O - E)^2}{E} = \frac{(22 - 12.41)^2}{12.41} + \frac{(20 - 24.82)^2}{24.82} + \ldots + \frac{(18 - 9.59)^2}{9.59} = 18.998\ldots\) awrt 19.0

M1A1

\(\nu = 5 - 2 = 3\) degrees of freedom

\(\chi^2_3(5\%) = 7.815\)

B1ft

\(18.998\ldots \gt 7.815\) so reject \(\mathrm{H}_0\)

Binomial is a not a good fit (and is not a good model for the number of defective items in samples of size 6)

(8)

(14 marks)

Notes

Parameters in hyps award B0

1st M1 for combining either 0 and 1 or 5 and 6 or both. Require at least 1 value in a combined correct.

2nd M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 2 correct expressions or values.

2nd A1 for a correct comment suggesting that Binomial model is not suitable. No ft
Condone parameters here.

S3 June 2012 Q4

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

4. Two breeds of chicken are surveyed to measure their egg yield. The results are shown in the table below.

Egg yield Breed	Low	Medium	High
Leghorn	22	52	26
Cornish	14	32	4

Showing each stage of your working clearly, test, at the 5% significance level, whether or not there is an association between egg yield and breed of chicken. State your hypotheses clearly. (10)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\) : Egg yield and breed of chicken are independent (not associated)
\(\mathrm{H}_1\) : Egg yield and breed of chicken are dependent (associated)

Egg Yield Breed	Low	Medium	High	Total
Leghorn	\(\dfrac{100 \times 36}{150} = 24\)	\(\dfrac{100 \times 84}{150} = 56\)	\(\dfrac{100 \times 30}{150} = 20\)	100
Cornish	\(\dfrac{50 \times 36}{150} = 12\)	\(\dfrac{50 \times 84}{150} = 28\)	\(\dfrac{50 \times 30}{150} = 10\)	50
Total	36	84	30	150

M1A1

\(O\)	\(E\)	\(\displaystyle\sum\frac{(O - E)^2}{E}\)	\(\displaystyle\sum\frac{O^2}{E}\)
22	24	0.166667	20.2
52	56	0.285714	48.3
26	20	1.8	33.8
14	12	0.333333	16.3
32	28	0.571429	36.6
4	10	3.6	1.6

M1A1

\(\displaystyle\sum\frac{(O - E)^2}{E} = 6.757\ldots\) or \(\displaystyle\sum\frac{O^2}{E} - 150 = 6.757\ldots\)

\(\nu = 2, \chi^2_2(5\%) = 5.991\)

B1B1ft

(\(6.757 \gt 5.991\) so sufficient evidence to) reject \(\mathrm{H}_0\)

Egg yield and breed of chicken are dependent (associated)

(10 marks)

Notes

B1 for both hypotheses. Must mention “yield” and “breed” in both but condone ditto marks.
Use of “relationship” or “correlation” or “connection” is B0

1st M1 for some use of \(\dfrac{\text{Row Total} \times \text{Col.Total}}{\text{Grand Total}}\). May be implied by a correct \(E_i\)

1st A1 for all expected frequencies correct

2nd M1 for at least two correct terms or correct expressions with their \(E_i\)

2nd A1 for all correct terms. May be implied by a correct answer (2 sf or better)

3rd M1 for a correct statement linking their test statistic and their cv . Must be \(\chi^2\) not normal.

4th A1 for a correct comment in context - must mention “egg yield” and “breed of chicken” - condone “relationship” or “connection” here but not “correlation”. No follow through e.g. “There is no evidence of a relationship between egg yield and breed of chicken” is A0 whatever their test stat or cv.

(corrected from the printed mark scheme: the alternative calculation is printed as \(\sum\frac{O^2}{E} - 100\); the grand total is 150, so it is \(\sum\frac{O^2}{E} - 150 = 156.757\ldots - 150 = 6.757\ldots\))

S3 June 2011 Q5

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

5. The number of hurricanes per year in a particular region was recorded over 80 years. The results are summarised in Table 1 below.

No of hurricanes, \(h\)	0	1	2	3	4	5	6	7
Frequency	0	2	5	17	20	12	12	12

Table 1

(a) Write down two assumptions that will support modelling the number of hurricanes per year by a Poisson distribution. (2)

(b) Show that the mean number of hurricanes per year from Table 1 is 4.4875 (2)

(c) Use the answer in part (b) to calculate the expected frequencies \(r\) and \(s\) given in Table 2 below to 2 decimal places. (3)

\(h\)	0	1	2	3	4	5	6	7 or more
Expected frequency	0.90	4.04	\(r\)	13.55	\(s\)	13.65	10.21	13.39

Table 2

(d) Test, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution. State your hypotheses clearly. (6)

Mark scheme (c)

Scheme

Marks

No of hurricanes, \(h\)	0	1	2	3	4	5	6	7+
\(80\mathrm{P}(X = h)\)	0.9	4..038	\(r\)=9.06…	13.55	\(s\)=15.205	13.647…	10.206…	13.388…

M1A1A1

(3)

Notes

M for 80xPoisson probability with 4.4875 and either 2 or 4.

1st A1 for awrt 9.06 and 2^nd A1 for awrt 15.20 or 15.21

Mark scheme (d)

Scheme

Marks

No of hurricanes, \(h\)	0	1	2	3	4	5	6	7+
Combine to give expected frequencies >5	13.9991			13.55	15.205…	13.647…	10.206…	13.388…
Observed	7			17	20	12	12	12
\(\dfrac{(O - E)^2}{E}\)	3.499…			0.876…	1.511…	0.198…	0.315…	0.143…
\(\dfrac{O_i^2}{E_i}\)	3.500...			21.322...	26.306...	10.551...	14.108...	10.755..

\(\mathrm{H}_0\): Poisson distribution is a good fit o.e.
\(\mathrm{H}_1\): Poisson distribution is not a good fit o.e.

\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i} = 6.545..\) or \(\dfrac{O_i^2}{E_i} = 86.545 - 80 = 6.545..\) (awrt 6.55 or 6.54)

\(\nu = 6 - 2 = 4\)

cv is 9.488 (ft their \(\nu\) i.e. \(\chi_\nu^{\ 2}(0.05)\))

B1ft

\(6.545 \lt 9.488\) so insufficient evidence to reject \(\mathrm{H}_0\)
(Hurricanes) can be modelled by a Poisson distribution

(6)

(13 marks)

Notes

1st M1 for some pooling and attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 3 correct expressions or values.

1st B1 no value for parameter permitted

2nd A1 for a correct comment suggesting that Poisson model is suitable. No ft

S3 June 2011 Q3

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

3. A factory manufactures batches of an electronic component. Each component is manufactured in one of three shifts. A component may have one of two types of defect, \(D_1\) or \(D_2\), at the end of the manufacturing process. A production manager believes that the type of defect is dependent upon the shift that manufactured the component. He examines 200 randomly selected defective components and classifies them by defect type and shift. The results are shown in the table below.

Defect type Shift	\(D_1\)	\(D_2\)
First shift	45	18
Second shift	55	20
Third shift	50	12

Stating your hypotheses, test, at the 10% level of significance, whether or not there is evidence to support the manager’s belief. Show your working clearly. (10)

Mark scheme

Scheme

Marks

Defect Type Shift	D₁	D₂
First Shift	47.25	15.75	63
Second Shift	56.25	18.75	75
Third Shift	46.5	15.5	62
	150	50	200

M1A1

\(\mathrm{H}_0\) : Type of defect is independent of Shift (no association)
\(\mathrm{H}_1\) : Type of defect is not independent of Shift (association)

\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)	\(\dfrac{O_i^2}{E_i}\)
45	47.25	0.1071…	42.857...
18	15.75	0.3214…	20.571..
55	56.25	0.02777…	53.777...
20	18.75	0.0833…	21.333...
50	46.5	0.2634…	53.763...
12	15.5	0.7903…	9.290...

M1A1

\(\dfrac{(O - E)^2}{E} = 1.5934..\) or \(\dfrac{O_i^2}{E_i} - 200 = 201.5934 - 200 = 1.5934..\) awrt 1.59

\(\nu = (3 - 1)(2 - 1) = 2\)

\(\chi^2_2(0.10) = 4.605\)

B1ft

\(1.59 \lt 4.605\) so insufficient evidence to reject \(\mathrm{H}_0\)

Insufficient evidence to support manager’s belief /claim.

(10 marks)

Notes

1st M1 for some use of \(\dfrac{\text{Row Total} \times \text{Col.Total}}{\text{Grand Total}}\) May be implied by correct \(E_i\)

1st A1 for all expected frequencies correct

B1 for both hypotheses. Must mention “defect” and “shift” at least once
Use of “relationship” or “correlation” or “connection” is B0

2nd M1 for at least two correct terms (as in 3^rd or 4^th column) or correct expressions with their \(E_i\)

2nd A1 for all correct terms. May be implied by a correct answer.(2 dp or better-allow eg 0.10…)

3rd M1 for a correct statement linking their test statistic and their cv . Must be \(\chi^2\) not normal.

4th A1 for a correct comment in context - must mention “manager’s belief” or “shift” and “defect type” - condone “relationship” or “connection” here but not “correlation”. No follow through e.g. “There is evidence of a relationship between shift and type of defect”

S3 June 2010 Q6

EdexcelOld spec12 marksIncludes hypothesis testingChi-Squared Tests

6. A total of 228 items are collected from an archaeological site. The distance from the centre of the site is recorded for each item. The results are summarised in the table below.

Distance from the centre of the site (m)	0–1	1–2	2–4	4–6	6–9	9–12
Number of items	22	15	44	37	52	58

Test, at the 5% level of significance, whether or not the data can be modelled by a continuous uniform distribution. State your hypotheses clearly. (12)

Mark scheme

Scheme

Marks

Distance from centre of site (m)	0-1	1-2	2-4	4-6	6-9	9-12
\(b - a\)	1	1	2	2	3	3
No of artefacts	22	15	44	37	52	58

\(\mathrm{P}(a \leqslant X \lt b)\)

\(\frac{1}{12}\)

\(\frac{1}{6}\)

\(\frac{1}{4}\)

\(228 \times \mathrm{P}(a \leqslant X \lt b)\)

Class	\(O_i\)	\(E_i\)	\(\dfrac{(O_i - E_i)^2}{E_i}\)	\(\dfrac{O_i^2}{E_i}\)
0-1	22	19	\(\frac{9}{19} = 0.4736\ldots\)	25.57…
1-2	15	19	\(\frac{16}{19} = 0.8421\ldots\)	11.84…
2-4	44	38	\(\frac{36}{38} = 0.9473\ldots\)	50.94…
4-6	37	38	\(\frac{1}{38} = 0.0263\ldots\)	36.02…
6-9	52	57	\(\frac{25}{57} = 0.4385\ldots\)	47.43…
9-12	58	57	\(\frac{1}{57} = 0.0175\ldots\)	59.01…

M1

A1

\(\mathrm{H}_0\): continuous uniform distribution is a good fit
\(\mathrm{H}_1\): continuous uniform distribution is not a good fit

\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i} = \frac{313}{114} = 2.75\) or \(\displaystyle\sum\frac{O_i^2}{E_i} - 228 = 230.745\ldots - 228 = \ldots\) (awrt 2.75)

dM1A1

\(\nu = 6 - 1 = 5\)

\(\chi^2_5(0.05) = 11.070\) (ft their \(\nu\) i.e. \(\chi_\nu^2(0.05)\))

B1ft

\(2.75 \lt 11.070\), insufficient evidence to reject \(\mathrm{H}_0\)

Continuous uniform distribution is a suitable model

(12 marks)

Notes

1st M1 for calculation of at least 3 widths and attempting proportions/probs. or for 1:2:3 ratio seen

1st A1 for correct probabilities

2nd A1 for all correct expected frequencies

2nd M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 3 correct expressions or values.
Follow through their \(E_i\) provided they are not all = 38

3rd A1 for a correct set of calcs - 3^rd or 4^th column. (2 dp or better and allow e.g. 0.94…)

3rd dM1 dependent on 2^nd M1 for attempting a correct sum or calculation (must see at least 3 terms and +)

The first three Ms and As can be implied by a test statistic of awrt 2.75

4th M1 for a correct statement based on their test statistic ( > 1) and their cv (> 3.8)
Contradictory statements score M0 e.g. “significant” do not reject \(\mathrm{H}_0\).

5th A1 for a correct comment suggesting that continuous uniform model is suitable. No ft

S3 June 2010 Q5

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

5. A random sample of 100 people were asked if their finances were worse, the same or better than this time last year. The sample was split according to their annual income and the results are shown in the table below.

Finances Annual income	Worse	Same	Better
Under £15 000	14	11	9
£15 000 and above	17	20	29

Test, at the 5% level of significance, whether or not the relative state of their finances is independent of their income range. State your hypotheses and show your working clearly. (10)

Mark scheme

Scheme

Marks

Finances Income	Worse	Same	Better
Under £15 000	10.54	10.54	12.92	34
£15 000 and above	20.46	20.46	25.08	66
	31	31	38	100

M1
A1

\(\mathrm{H}_0\) : State of finances and income are independent (not associated)
\(\mathrm{H}_1\) : State of finances and income are not independent (associated)

\(O_i\)	\(E_i\)	\(\dfrac{(O_i - E_i)^2}{E_i}\)	\(\dfrac{O_i^2}{E_i}\)
14	10.54	1.1358….	18.59..
11	10.54	0.0200….	11.48..
9	12.92	1.1893…	6.269..
17	20.46	0.5851…	14.12..
20	20.46	0.0103…	19.55..
29	25.08	0.6126…	33.53..

M1
A1

\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i} = 3.553\ldots\) or \(\displaystyle\sum\frac{O_i^2}{E_i} - 100 = 103.553\ldots - 100 = 3.553\ldots\) (awrt 3.55)

\(\nu = (3 - 1)(2 - 1) = 2\)

cv is 5.991

\(3.553 \lt 5.991\) so insufficient evidence to reject \(\mathrm{H}_0\) or not significant

There is no evidence of association between state of finances and income.

(10 marks)

Notes

1st M1 for some use of \(\dfrac{\text{Row Total} \times \text{Col.Total}}{\text{Grand Total}}\). May be implied by correct \(E_i\)

1st A1 for all expected frequencies correct

B1 for both hypotheses. Must mention “state” or “finances” and “income” at least once
Use of “relationship” or “correlation” or “connection” is B0

2nd M1 for at least two correct terms (as in 3^rd or 4^th column) or correct expressions with their \(E_i\)

2nd A1 for all correct terms. May be implied by a correct answer.(2 dp or better-allow eg 1.13…)

3rd M1 for a correct statement linking their test statistic and their cv. Must be \(\chi^2\) not normal.

4th A1 for a correct comment in context - must mention “state” or “finances” and “income” condone “relationship” or “connection” here but not “correlation”. No follow through.
e.g. “There is no evidence of a relationship between finances and income”

S3 June 2009 Q5

EdexcelOld spec12 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

5. The number of goals scored by a football team is recorded for 100 games. The results are summarised in Table 1 below.

Number of goals	Frequency
0	40
1	33
2	14
3	8
4	5

Table 1

(a) Calculate the mean number of goals scored per game. (2)

The manager claimed that the number of goals scored per match follows a Poisson distribution. He used the answer in part (a) to calculate the expected frequencies given in Table 2.

Number of goals	Expected Frequency
0	34.994
1	\(r\)
2	\(s\)
3	6.752
\(\geqslant 4\)	2.221

Table 2

(b) Find the value of \(r\) and the value of \(s\) giving your answers to 3 decimal places. (3)

Mark scheme (a)

Scheme	Marks
\(\lambda = \dfrac{0 \times 40 + 1 \times 33 + 2 \times 14 + 3 \times 8 + 4 \times 5}{100} = 1.05\)	M1 A1
	(2)

Notes

M1 for an attempt to find the mean- at least 2 terms on numerator seen
Correct answer only will score both marks

Mark scheme (b)

Scheme	Marks
Using Expected frequency \(= 100 \times \mathrm{P}(X = x) = 100 \times \dfrac{\mathrm{e}^{-1.05}1.05^x}{x!}\) gives	M1
\(r = 36.743\) awrt 36.743 or 36.744	A1
\(s = 19.290\) 19.29 or awrt 19.290	A1
	(3)

Notes

M1 for use of correct formula (ft their mean). 1^st A1 for \(r\), 2^nd A1 for \(s\) (19.29 OK)

Mark scheme (c)

Scheme

Marks

\(\mathrm{H}_0\) : Poisson distribution is a suitable model
\(\mathrm{H}_1\) : Poisson distribution is not a suitable model

Number of goals	Frequency	Expected frequency
0	40	34.994
1	33	36.743
2	14	19.290
3	8	6.752	8.972443
\(\geqslant 4\)	5	2.221	8.972443

\(\upsilon = 4 - 1 - 1 = 2\)

B1ft

CR : \(\chi^2_2(0.05) \gt 5.991\)

\(\displaystyle\sum\frac{(\mathrm{O} - \mathrm{E})^2}{\mathrm{E}} = \frac{(40 - 34.9937)^2}{34.9937} + \ldots + \frac{(13 - 8.972443)^2}{8.972443}\)
[\(= 0.7161\ldots + 0.3813\ldots + 1.4508\ldots + 1.80789\ldots\)]

\(= 4.356.\) (ans in range 4.2 – 4.4)

Not in critical region
Number of goals scored can follow a Poisson distribution / managers claim is justified

A1 ft

(7)

(12 marks)

Notes

1st B1 Must have both hypotheses and mention Poisson at least once
inclusion of their value for mean in hypotheses is B0 but condone in conclusion

1st M1 for an attempt to pool \(\geqslant 4\)

2nd B1ft for \(n - 1 - 1 = 2\) i.e realising that they must subtract 2 from their \(n\)

3rd B1 for 5.991 only

2nd M1 for an attempt at the test statistic, at least 2 correct expressions/values (to 3sf)

1st A1 for answers in the range 4.2~4.4

2nd A1 for correct comment in context based on their test statistic and their cv that mentions goals or manager. Dependent on 2^nd M1
Condone mention of Po(1.05) in conclusion
Score A0 for inconsistencies e.g. “significant” followed by “manager’s claim is justified”

S3 June 2008 Q6

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests

6. Ten cuttings were taken from each of 100 randomly selected garden plants. The numbers of cuttings that did not grow were recorded.

The results are as follows

No. of cuttings which did not grow	0	1	2	3	4	5	6	7	8, 9 or 10
Frequency	11	21	30	20	12	3	2	1	0

(a) Show that the probability of a randomly selected cutting, from this sample, not growing is 0.223 (2)

A gardener believes that a binomial distribution might provide a good model for the number of cuttings, out of 10, that do not grow.

He uses a binomial distribution, with the probability 0.2 of a cutting not growing. The calculated expected frequencies are as follows

No. of cuttings which did not grow	0	1	2	3	4	5 or more
Expected frequency	\(r\)	26.84	\(s\)	20.13	8.81	\(t\)

(b) Find the values of \(r\), \(s\) and \(t\). (4)

(c) State clearly the hypotheses required to test whether or not this binomial distribution is a suitable model for these data. (2)

The test statistic for the test is 4.17 and the number of degrees of freedom used is 4.

(d) Explain fully why there are 4 degrees of freedom. (2)

(e) Stating clearly the critical value used, carry out the test using a 5% level of significance. (3)

Mark scheme (a)

Scheme	Marks
\(p = \dfrac{0 \times 11 + 1 \times 21 + \ldots}{10 \times (11 + 21 + \ldots) \text{ or } 10 \times 100},\ = \dfrac{223}{1000} = 0.223\) (*) (Accept \(\dfrac{223}{1000}\))	M1, A1cso
	(2)

Notes

M1 Must show clearly how to get either 223 or 1000. As printed or better.
A1cso for showing how to get both 223 and 1000 and reaching \(p = 0.223\)

Mark scheme (b)

Scheme	Marks
\(r = (0.8)^{10} \times 100 = 10.7374\) awrt 10.74	M1A1
\(s = \dbinom{10}{2}(0.8)^8 \times (0.2)^2 \times 100 = 30.198\ldots\) awrt 30.2	A1
\(t = 100 - [r + s + 26.84 + 20.13 + 8.81] =\) awrt 3.28	A1cao
	(4)

Notes

M1 for any correct method (a correct expression) seen for \(r\) or \(s\).
1^st A1 for correct value for \(r\) awrt 10.74
2^nd A1 for \(s\) = awrt 30.2
3^rd A1 for \(t = 3.28\) only

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0\) : Binomial ([\(n = 10\)], \(p = 0.2\)) is a suitable model for these data	B1
\(\mathrm{H}_1\) : Binomial ([\(n = 10\)], \(p = 0.2\)) is NOT a suitable model for these data	B1
	(2)

Notes

B1 for each. The value of \(p\) must be mentioned at least once. Accept B(10, 0.2)
If hypotheses are correct but with no value of \(p\) then score B0B1
Minimum is \(X \sim \mathrm{B}(10, 0.2)\). If just B(10, 0.2) and not B(10, 0.2) award B1B0

Mark scheme (d)

Scheme	Marks
Since \(t \lt 5\), the last two groups are combined	M1
and \(\nu = 4 = 5 - 1\)	A1
	(2)

Notes

M1 for combining groups (must be stated or implied by a new table with combined cell seen)
A1 for the calculation 4 = 5 - 1

Mark scheme (e)

Scheme	Marks
Critical value \(\chi^2_4(5\%) = 9.488\)	B1
Not significant or do not reject null hypothesis	M1
The binomial distribution with \(p = 0.2\) is a suitable model for the number of cuttings that do not grow	A1
	(3)
	(13 marks)

Notes

M1 for a correct statement based on 4.17 and their cv (context not required) (may be implied)
Use of 4.17 as a critical value scores B0M0A0
A1 for a correct interpretation in context and \(p = 0.2\) and cuttings mentioned.

S3 June 2008 Q2

EdexcelOld spec11 marksIncludes hypothesis testingChi-Squared Tests

2. Students in a mixed sixth form college are classified as taking courses in either Arts, Science or Humanities. A random sample of students from the college gave the following results

		Course
		Arts	Science	Humanities
Gender	Boy	30	50	35
Gender	Girl	40	20	42

Showing your working clearly, test, at the 1% level of significance, whether or not there is an association between gender and the type of course taken. State your hypotheses clearly. (11)

Mark scheme

Scheme

Marks

\(\dfrac{115 \times 70}{217} = 37.0967\ldots\) or \(\dfrac{1150}{31}\) (etc) \(\dfrac{1265}{31}, \dfrac{1020}{31}, \dfrac{1122}{31}\)

Expected (Obs)	A	S	H
Boy	37.1 (30)	37.1 (50)	40.8 (35)
Girl	32.9 (40)	32.9 (20)	36.2 (42)

A1A1

\(\mathrm{H}_0\) : There is no association between course and gender
\(\mathrm{H}_1\) : There is some association between course and gender (both)

\(\displaystyle\sum \frac{(O - E)^2}{E} = \frac{(37.1 - 30)^2}{37.1} + \frac{(32.9 - 40)^2}{32.9} + \ldots + \frac{(36.2 - 42)^2}{36.2}\)

M1A1ft

\(= 1.358 + 4.485 + 0.824 + 1.532 + 5.058 + 0.929 = 14.189\ldots\) awrt 14.2

\(\nu = (3 - 1)(2 - 1) = 2\), \(\chi^2_2(1\%)\) critical value is 9.210 (condone 9.21)

B1, B1ft

Significant result or reject null hypothesis

There is evidence of an association between course taken and gender
[Correct answers only score full marks]

A1ft

(11)

(11 marks)

Alternative

Scheme	Marks
\(\displaystyle\sum \frac{O^2}{E} - N = \frac{30^2}{37.1} + \frac{40^2}{32.9} + \ldots + \frac{42^2}{36.2} - 217\)	M1A1ft

Notes

1^st M1 for some use of the \(\dfrac{\text{row total} \times \text{col total}}{\text{grand total}}\) formula

1^st A1 for one correct row or one correct column of expected frequencies to nearest integer
2^nd A1 for all expected frequencies correct to awrt 1 dp (Allow exact fractions)

1^st B1 for hypotheses. Independence is OK. Must mention courses and gender at least once.
Use of \(\rho\) or “correlation” is B0 but allow ISW.

2^nd M1 for an attempt to calculate test statistic. At least one correct expression, ft expected freq.
3^rd A1 follow through expected frequencies for at least 3 expressions

3^rd M1 for a correct statement relating their test statistic and their cv (may be implied by comment)
5^th A1 for a contextualised comment relating their test statistic and their cv. Ignore their \(\mathrm{H}_0\) or \(\mathrm{H}_1\) or assume that they were correct. Must mention courses and gender

S3 June 2007 Q4

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests

4. A quality control manager regularly samples 20 items from a production line and records the number of defective items \(x\). The results of 100 such samples are given in Table 1 below.

\(x\)	0	1	2	3	4	5	6	7 or more
Frequency	17	31	19	14	9	7	3	0

Table 1

(a) Estimate the proportion of defective items from the production line. (2)

The manager claimed that the number of defective items in a sample of 20 can be modelled by a binomial distribution. He used the answer in part (a) to calculate the expected frequencies given in Table 2.

\(x\)	0	1	2	3	4	5	6	7 or more
Expected frequency	12.2	27.0	\(r\)	19.0	\(s\)	3.2	0.9	0.2

Table 2

(b) Find the value of \(r\) and the value of \(s\) giving your answers to 1 decimal place. (3)

(d) Explain what the analysis in part (c) tells the manager about the occurrence of defective items from this production line. (1)

Mark scheme (a)

Scheme	Marks
\(\dfrac{0 \times 17 + 1 \times 31 + \ldots}{17 + 31 + \ldots} = \left(\dfrac{200}{100} = 2\right)\), \(\hat{p} = \dfrac{2}{20} = \underline{0.1}\) (Accept \(\dfrac{2}{20}\) or 2 per 20)	M1, A1
	(2)

Notes

M1 for attempt to find mean or \(\hat{p}\) (as printed or better). The 0.1 must be seen in part (a).

Mark scheme (b)

Scheme	Marks
e.g. \(r = 100 \times \dbinom{20}{2}(0.1)^2(0.9)^{18}\)	M1
\(r = 28.5\), \(s\) = AWRT 9	A1, A1
	(3)

Notes

M1 for correct expression for \(r\) or \(s\) using the binomial distribution. Follow through their \(\hat{p}\).

Mark scheme (c)

Scheme

Marks

\(x\)	0	1	2	3	\(\geqslant 4\)
\(O_i\)	17	31	19	14	19
\(E_i\)	12.2	27.0	28.5	19.0	13.3
\(\dfrac{(O - E)^2}{E}\)	1.89	0.59	3.17	1.32	2.44

Pooling

\(\displaystyle\sum \frac{(O - E)^2}{E} =\) AWRT 9.4

M1A1c.a.o.

\(\nu = 5 - 2 = 3, \quad \chi^2_3(5\%) = 7.815\)

B1ft, B1ft

\(\mathrm{H}_0\) : Binomial distribution is a good/suitable model/fit [Condone: B(20, 0.1) is…]
\(\mathrm{H}_1\) : Binomial distribution is not a suitable model both

(Significant result) Binomial distribution is not a suitable model

A1cao

(7)

Notes

1^st M1 for some pooling (accept \(x \geqslant 5\), obs.freq. …14, 9, 10 and exp.freq. 19.0, \(s\), 4.3)

2^nd M1 for calculation of test statistic (N.B. \(x \geqslant 5\) gives 14.5). One correct term seen.

1^st B1ft for number of classes – 2 (N.B. \(x \geqslant 5\) will have 6 – 2 = 4)

2^nd B1ft for the appropriate tables value, ft their degrees of freedom. (NB \(\chi^2_4(5\%) = 9.488\))

3^rd B1 (for hypotheses) allow just “\(X \sim \mathrm{B}(20, 0.1)\)” for null etc.

2^nd A1 for correctly rejecting Binomial model. No ft and depends on 2^nd M1.

Mark scheme (d)

Scheme	Marks
defective items do not occur independently or not with constant probability	B1ft
	(1)
	(13 marks)

Notes

B1ft for independence or constant probability – must mention defective items or defectives
Follow through their conclusion in (c). So if they do not reject they may say “defectives occur with probability 0.1”. Stating the value implies constant probability.

S3 June 2007 Q2

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

2. The Director of Studies at a large college believed that students’ grades in Mathematics were independent of their grades in English. She examined the results of a random group of candidates who had studied both subjects and she recorded the number of candidates in each of the 6 categories shown.

	Maths grade A or B	Maths grade C or D	Maths grade E or U
English grade A or B	25	25	10
English grade C to U	15	30	15

(a) Stating your hypotheses clearly, test the Director’s belief using a 10% level of significance. You must show each step of your working. (9)

The Head of English suggested that the Director was losing accuracy by combining the English grades C to U in one row. He suggested that the Director should split the English grades into two rows, grades C or D and grades E or U as for Mathematics.

(b) State why this might lead to problems in performing the test. (1)

Mark scheme (a)

Scheme

Marks

\(\mathrm{H}_0\) : Maths grades are independent of English grades or No association …
\(\mathrm{H}_1\) : Maths and English grades are dependent or There is an association …

Expected Frequencies e.g. \(\dfrac{60 \times 40}{120} = 20\)

20	27.5	12.5
20	27.5	12.5

M1 A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 2 \times \left(\frac{5^2}{20} + \frac{2.5^2}{27.5} + \frac{2.5^2}{12.5}\right),\ = 3.9545\ldots\) AWRT 3.95 or 3.955

M1, A1

\(\nu = (3 - 1)(2 - 1) = 2\); \(\chi^2_2(10\%)\) c.v. \(= 4.605\)

B1; B1

\(3.95 \lt 4.605\) or not significant or do not reject \(\mathrm{H}_0\) (allow reject \(\mathrm{H}_1\))

Insufficient evidence of an association between English and maths grades
or there is support for the Director’s belief
or Student’s grades in maths and English are independent

(9)

Notes

1^st B1 for both hypotheses in terms of independence or association and in context.
Must mention Maths and English in at least one of the hypotheses.
“relationship” or “correlation” or “connection” or “link” is B0

1^st M1 for some correct calculation seen
1^st A1 for all expected frequencies correct. Accept answers without formula seen.

2^nd M1 for some evidence seen of attempt to calculate test statistic.
At least one correct term seen. Follow through their expected frequencies.
2^nd A1 for AWRT 3.95. Answers only please escalate!

3^rd M1 for correct comparison or statement – may be implied by correct conclusion.
3^rd A1 for conclusion in context using “association” or “independence” in connection with grades.
Don’t insist on seeing English or maths mentioned here.
Use ISW for comments if a false statement and correct statement are seen.

Mark scheme (b)

Scheme	Marks
May have some expected frequencies <5 (and hence need to pool rows/cols)	B1
	(1)
	(10 marks)

Notes

B1 If they just say expected frequencies are “small” they must go onto mention need to pool.

S3 June 2006 Q8

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests

8. Five coins were tossed 100 times and the number of heads recorded. The results are shown in the table below.

Number of heads	0	1	2	3	4	5
Frequency	6	18	29	34	10	3

(a) Suggest a suitable distribution to model the number of heads when five unbiased coins are tossed. (2)

(b) Test, at the 10% level of significance, whether or not the five coins are unbiased. State your hypotheses clearly. (11)

Mark scheme (b)

Scheme

Marks

\(\mathrm{H}_0 : \mathrm{B}(5, 0.5)\) is a suitable model (good fit)
\(\mathrm{H}_1 : \mathrm{B}(5, 0.5)\) is not a suitable model (not a good fit)
ft for \(\hat{p} = 0.466\).

B1ft

No. of heads	0	1	2	3	4	5
Expected	3.125	15.625	31.25	31.25	15.625	3.125
Actual	6	18	29	34	10	3

\(100\,\mathrm{P}(X = x)\) for Bin, 1 correct = A1, All correct = A1, 3sf or better

M1 A1 A1

	\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)
0 or 1	24	18.75	1.47
2	29	31.25	0.162
3	34	31.25	0.242
4 or 5	13	18.75	\(1.76\dot{3}\)

grouped \(O\) and \(E\). All correct 2sf or better.

M1 A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 3.637\dot{3}\) \(\sum\) required, awrt 3.64

M1 A1

\(\nu = 4 - 1 = 3\), \(\chi^2_3(0.10) = 6.251\)

B1 B1ft

Insufficient evidence to reject \(\mathrm{H}_0\)
\(\mathrm{B}(5, 0.5)\) is a suitable model.
No evidence that coins are biased

A1ft

(11)

(13 marks)

Notes

Ungrouped gives awrt 5.44, \(\nu = 5\), \(\chi^2_5 = 9.236\)

S3 June 2006 Q6

EdexcelOld spec11 marksIncludes hypothesis testingChi-Squared Tests

6. A research worker studying colour preference and the age of a random sample of 50 children obtained the results shown below.

Age in years	Red	Blue	Totals
4	12	6	18
8	10	7	17
12	6	9	15
Totals	28	22	50

Using a 5% significance level, carry out a test to decide whether or not there is an association between age and colour preference. State your hypotheses clearly. (11)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\) : No association between age and colour (independent)

\(\mathrm{H}_1\) : Association between age and colour (Not independent)

\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)
12	10.08	0.3657…
6	7.92	0.4654…
10	9.52	0.0242…
7	7.48	0.0308…
6	8.4	0.6857…
9	6.6	0.8727…

at least one \(\dfrac{\text{RT} \times \text{CT}}{\text{GT}}\), All
\(\dfrac{(O - E)^2}{E}\) 3sf or better

M1 A1
M1 A1

\(\displaystyle\sum \frac{(O - E)^2}{E} = 2.4446\ldots\) \(\sum\), awrt 2.44

M1 A1

\(\nu = (3 - 1)(2 - 1) = 2\), \(\chi^2_2 = 5.991\)

B1 B1ft

Insufficient evidence to reject \(\mathrm{H}_0\).
No association between age and colour

A1ft

(11)

(11 marks)

Notes

(cv from correct table for ft)

S3 January 2006 Q6

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

6. An area of grass was sampled by placing a 1 m × 1 m square randomly in 100 places. The numbers of daisies in each of the squares were counted. It was decided that the resulting data could be modelled by a Poisson distribution with mean 2. The expected frequencies were calculated using the model.

The following table shows the observed and expected frequencies.

Number of daisies	Observed frequency	Expected frequency
0	8	13.53
1	32	27.07
2	27	\(r\)
3	18	\(s\)
4	10	9.02
5	3	3.61
6	1	1.20
7	0	0.34
\(\geqslant 8\)	1	\(t\)

(a) Find values for \(r\), \(s\) and \(t\). (4)

(b) Using a 5% significance level, test whether or not this Poisson model is suitable. State your hypotheses clearly. (7)

An alternative test might have been to estimate the population mean by using the data given.

Mark scheme (b)

Scheme	Marks
\(\mathrm{H}_0\) : A Poisson model Po(2) is a suitable model. \(\mathrm{H}_1\) : A Poisson model Po(2) is not a suitable model. both	B1
Amalgamate data	M1
\(\displaystyle\sum \frac{(O - E)^2}{E} = 3.28\) (awrt)	M1 A1
\(\nu = 6 - 1 = 5\)	B1
\(\chi^2_5(5\%) = 11.070\) (follow through their degrees of freedom)	B1ft
\(3.28 \lt 11.070\) There is insufficient evidence to reject \(\mathrm{H}_0\), Po(2) is a suitable model.	A1ft
	(7)

Notes

(corrected from the printed mark scheme: the comparison is printed as “3.25 < 11.070”; the test statistic found above is 3.28)

S3 January 2006 Q4

EdexcelOld spec9 marksIncludes hypothesis testingChi-Squared Tests

4. People over the age of 65 are offered an annual flu injection. A health official took a random sample from a list of patients who were over 65. She recorded their gender and whether or not the offer of an annual flu injection was accepted or rejected. The results are summarised below.

Gender	Accepted	Rejected
Male	170	110
Female	280	140

Using a 5% significance level, test whether or not there is an association between gender and acceptance or rejection of an annual flu injection. State your hypotheses clearly. (9)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\) : No association between gender and acceptance
\(\mathrm{H}_1\) : gender and acceptance are associated

	Accept	Not accept	Total
Males	170 (180)	110 (100)	280
Females	280 (270)	140 (150)	420
Totals	450	250	700

Expected Values

M1 A1

\(O\)	\(E\)	\(\dfrac{(O - E)^2}{E}\)
170	180	0.5556
110	100	1.0000
280	270	0.3704
140	150	0.6667

\(\displaystyle\sum \frac{(O - E)^2}{E} = 2.59\) (Yates’ 2.34) (Condone use of Yates’)

M1 A1

\(\nu = 1\); (5%) = 3.841

B1; B1

\(3.841 \gt 2.59\). There is insufficient evidence to reject \(\mathrm{H}_0\)
There is no association between a persons gender and their acceptance (of the offer of a flu jab.)

M1
A1ft

(9)

(9 marks)

S3 June 2005 Q5

EdexcelOld spec12 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

5. The number of times per day a computer fails and has to be restarted is recorded for 200 days. The results are summarised in the table.

Number of restarts	Frequency
0	99
1	65
2	22
3	12
4	2

Test whether or not a Poisson model is suitable to represent the number of restarts per day. Use a 5% level of significance and state your hypothesis clearly. (12)

Mark scheme

Scheme

Marks

\(\mathrm{H}_0\): Poisson distribution is a suitable model
\(\mathrm{H}_1\): Poisson distribution is not a suitable model both

\(\hat{\lambda} = \dfrac{(0 \times 99) + (1 \times 65) + \cdots + (4 \times 2)}{200} = \dfrac{153}{200} = \underline{0.765}\)

M1 A1

Using \(\mathrm{P}(X = x) = \dfrac{0.765^x \mathrm{e}^{-0.765}}{x!}\) where \(X\) represents the number of restarts gives \(200 \times \mathrm{P}(X = x)\)

\(X\)	Observed Frequency		Expected Frequency
0	99		93.06678…
1	65		71.19604…
2	22		27.23250…
3	12	} 14	6.94428…	} 8.50468
\(\geqslant 4\)	2	} 14	1.56040…	} 8.50468

0, 1, 2: A1, A1 (−1 e.e.); 3, \(\geqslant 4\) combined: A1

A1, A1
(−1 e.e.)
A1

\(\nu = 4 - 1 - 1 = 2\); CR: \(\chi^2_2 \gt 5.991\) from Poisson
\(\nu = 4 - 1 = 3\); CR: \(\chi^2 \gt 7.815\) from Poisson(0.765)

B1; B1ft

\(\displaystyle\sum \frac{(O - E)^2}{E} = 5.47368\ldots\) Use of \(\sum (O - E)^2/E\)

5.40 – 5.50

5.47 is not in the critical region.
Number of computer failures per day can be modelled by a Poisson distribution

A1ft

(12)

(12 marks)

S3 June 2005 Q3

EdexcelOld spec11 marksIncludes hypothesis testingChi-Squared Tests

3. A researcher carried out a survey of three treatments for a fruit tree disease. The contingency table below shows the results of a survey of a random sample of 60 diseased trees.

	No action	Remove diseased branches	Spray with chemicals
Tree died within 1 year	10	5	6
Tree survived for 1–4 years	5	9	7
Tree survived beyond 4 years	5	6	7

Test, at the 5% level of significance, whether or not there is any association between the treatment of the trees and their survival. State your hypotheses and conclusion clearly. (11)

Mark scheme

Scheme

Marks

	No action	Remove diseased branches	Spray with chemicals	Totals
Tree died within 1 year	10 (7)	5 (7)	6 (7)	21
Survived 1–4 years	5 (7)	9 (7)	7 (7)	21
Survived > 4 years	5 (6)	6 (6)	7 (6)	18
Totals	20	20	20	60

\(\dfrac{\text{RT} \times \text{CT}}{\text{GT}}\)

\(6 \times 7\)

\(3 \times 6\)

\(\mathrm{H}_0\): Treatment & Survival are independent (not associated)
\(\mathrm{H}_1\): Treatment & Survival are not independent (associated)

B1 both

\(\alpha = 0.05\)
\(\nu = (3 - 1) \times (3 - 1) = 4\)

CR: \(\chi^2 \gt 9.488\)

B1ft

\(\displaystyle\sum \frac{(O - E)^2}{E} = \frac{9}{7} + \frac{4}{7} + \frac{1}{7} + \frac{4}{7} + \frac{4}{7} + 0 + \frac{1}{6} + 0 + \frac{1}{6}\) Using \(\sum \frac{(O - E)^2}{E}\) Any 2 values

M1 A1

\(= 3.47619\ldots\) AWRT 3.48

Since 3.47619… is NOT in the critical region (i.e. \(\lt 9.488\)) there is insufficient evidence to reject \(\mathrm{H}_0\). Comparison

There is no evidence of association between treatment and length of survival. Conclusion

A1ft

(11)

(11 marks)