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.

Scheme	Marks	AO
\(\mathrm{B}(3, 0.5)\)	B1	3.3
	(1)

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.

\(x\)	0	1	2	3
\(\mathrm{P}(X = x)\)	0.125	0.375	0.375	0.125