S3 June 2017 Q2

EdexcelOld spec10 marksIncludes hypothesis testingChi-Squared Tests

Figure 1: a circular dial marked from 0 to 345 degrees in steps of 15, with a pointer from the centre — Figure 1

The pointer shown in Figure 1 is spun so that it comes to rest between 0 and 360 degrees.

Linda claims that it is equally likely to come to rest at any point between 0 and 360 degrees. She spins the pointer 100 times and her results are summarised in the table below. She calculates expected frequencies for some of the possible outcomes and these are also given in the table below.

Angle (degrees)	0–45	45–90	90–180	180–315	315–360
Frequency	18	16	18	29	19
Expected frequency	12.5	\(a\)	\(b\)	\(c\)	12.5

(a) Find the values of the missing expected frequencies \(a\), \(b\) and \(c\). (2)

(b) Stating your hypotheses clearly and using a 5% level of significance, test whether or not Linda’s claim is supported by these data. (8)

Mark scheme (a)

Scheme	Marks
\(a = \dfrac{45}{360} \times 100 = 12.5\)	M1
\(b = \dfrac{90}{360} \times 100 = 25\)
\(c = \dfrac{135}{360} \times 100 = 37.5\)	A1
	(2)

Notes

M1 For one correct calculation

A1 12.5, 25, 37.5 only

Mark scheme (b)

Scheme

Marks

\(\mathrm{H}_0 :\) Continuous uniform distribution is a suitable model
\(\mathrm{H}_1 :\) Continuous uniform distribution is not a suitable model

\(O\)	\(E\)	\(\dfrac{(O-E)^2}{E}\)	\(\dfrac{O^2}{E}\)
18	12.5	2.42	25.92
16	12.5	0.98	20.48
18	25	1.96	12.96
29	37.5	1.927	22.43
19	12.5	3.38	28.88
100	100	10.67	110.67

M1A1

\(\displaystyle\sum \frac{(O-E)^2}{E} = 10.67\) or \(\displaystyle\sum \frac{O^2}{E} - 100 = 10.67\)

\(\nu = 5 - 1 = 4,\ \chi^2_4(5\%) = 9.488\)

B1,B1ft

(10.7>9.488) Reject \(\mathrm{H}_0\)

Linda’s claim is not supported.

A1ft

(8)

(10 marks)

Notes

B1 Both. Accept in terms of Linda’s claim. Accept U[0,360).

M1 for attempting \(\dfrac{(O-E)^2}{E}\) or \(\dfrac{O^2}{E}\) with at least 3 correct expressions or values.

A1 for all values correct to 2dp or as fractions. Can be implied by correct answer if all not listed.

A1 awrt 10.7

B1,B1ft 4 can be implied by awrt 9.488 seen or \(p =\) awrt 0.03

M1 Correct for their test stat and cv or thier \(p\) to 5% . Can be implied by correct conclusion from their test stat and cv.

A1ft A correct comment suggesting that continuous uniform model is not suitable or Linda’s claim is not correct. Linda’s claim must be described in full if Linda is not mentioned.
Follow through from their test stat and cv, but hypotheses must be correct.