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)

Scheme	Marks
\(\dfrac{282 \times 100}{600}\) (Do not accept 282 – 114.2 – 90.2 – 30.6 (o.e.))	B1
	(1)

Scheme	Marks
9	B1
	(1)

Scheme	Marks
2.5 or better (Do not accept 0.025)	B1
	(1)

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.