S4 June 2014 Q5

EdexcelOld spec11 marksIncludes hypothesis testingQuality of Tests

5. A statistician believes a coin is biased and the probability, \(p\), of getting a head when the coin is tossed is less than 0.5

The statistician decides to test this by tossing the coin 10 times and recording the number, \(X\), of heads. He sets up the hypotheses \(\mathrm{H}_0 : p = 0.5\) and \(\mathrm{H}_1 : p \lt 0.5\) and rejects the null hypothesis if \(x \lt 3\)

(a) Find the size of the test. (1)

(b) Show that the power function of this test is \[(1 - p)^8\,(36p^2 + 8p + 1)\] (3)

Table 1 gives values, to 2 decimal places, of the power function for the statistician’s test.

\(p\)	0.1	0.15	0.2	0.25	0.3	0.35	0.4	0.45
Power	0.93	0.82	\(r\)	0.53	0.38	0.26	\(s\)	0.10

Table 1

(d) Draw the graph of the power function for the statistician’s test. (2)

(e) Find the range of values of \(p\) for which the probability of accepting the coin as unbiased, when in fact it is biased, is less than or equal to 0.4 (3)

Scheme	Marks
\(X \sim \mathrm{B}(10, 0.5)\)
Size \(= \mathrm{P}(\text{reject } \mathrm{H}_0 \mid p = 0.5)\)
\(= \mathrm{P}(X \lt 3 \mid p = 0.5)\)
\(= 0.0547\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
Power \(= \mathrm{P}(X = 2) + \mathrm{P}(X = 1) + \mathrm{P}(X = 0)\)	M1
\(= 45p^2(1 - p)^8 + 10p(1 - p)^9 + (1 - p)^{10}\)	A1
\(= (1 - p)^8\left(45p^2 + 10p(1 - p) + (1 - p)^2\right)\)
\(= (1 - p)^8(36p^2 + 8p + 1)\)	A1cso
	(3)

Notes

M1 for a correct expression/selection of probabilities

A1 for a fully correct expression

Scheme	Marks
\(r = 0.68\)	B1
\(s = 0.17\)	B1
	(2)

Scheme	Marks
	B1 points B1 curve
	(2)

Mark scheme (e)

Scheme	Marks
\(\mathrm{P}(\text{Type II error}) \leqslant 0.4\)	M1
\(1 - \text{power} \leqslant 0.4\)
Power \(\geqslant 0.6\)	A1
\(p \lt 0.23\)	A1
	(3)
	(11 marks)

Notes

M1 may be implied by Power \(\geqslant 0.6\) or correct value or by correct answer

A1 may be implied by correct answer

A1 allow number between 0.22 and 0.23 inclusive and either < or \(\leqslant\)