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

Scheme	Marks
Area \(\propto \pi R^2\) so \(r = 81, -49 = \)32	M1, A1
\(s = 100 - \text{“32”} - 49\) or \(100 - 81 = \)19	B1ft
	(3)

Scheme	Marks
Not significant, Henry’s model is suitable	M1, A1
	(2)

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.

Scheme	Marks
\(\dfrac{39 \times 65}{100}\)	B1
	(1)

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)