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 \(p = 0.638\)