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