S2 June 2008 Q3

EdexcelOld spec5 marksIncludes hypothesis testingPoisson Distribution

3. A test statistic has a Poisson distribution with parameter \(\lambda\).

Given that

\[\mathrm{H}_0 : \lambda = 9,\ \ \mathrm{H}_1 : \lambda \ne 9\]

(a) find the critical region for the test statistic such that the probability in each tail is as close as possible to 2.5%. (3)

(b) State the probability of incorrectly rejecting \(\mathrm{H}_0\) using this critical region. (2)

Mark scheme (a)

Scheme	Marks
\(X \sim \mathrm{Po}(9)\) may be implied by calculations in part a or b	M1
\(\mathrm{P}(X \leqslant 3) = 0.0212\) \(\mathrm{P}(X \geqslant 16) = 0.0220\)
CR \(X \leqslant 3;\ \cup\ X \geqslant 16\)	A1; A1
	(3)

Notes

M1 for using Po (9) – other values you might see which imply Po (9) are 0.0550, 0.0415, 0.9780, 0.9585, 0.9889, 0.0111, 0.0062 or may be assumed by at least one correct region.
A1 for \(X \leqslant 3\) or \(X \lt 4\) condone c1 or CR instead of \(X\)
A1 for \(X \geqslant 16\) or \(X \gt 15\)

They must identify the critical regions at the end and not just have them as part of their working. Do not accept \(\mathrm{P}(X \leqslant 3)\) etc gets A0

Mark scheme (b)

Scheme	Marks
P(rejecting Ho) \(= 0.0212 + 0.0220\)	M1
\(= 0.0432\) or 0.0433	A1 cao
	(2)
	(5 marks)

Notes

(b) if they use 0.0212 and 0.0220 they can gain these marks regardless of the critical regions in part a. If they have not got the correct numbers they must be adding the values for their critical regions. (both smaller than 0.05) You may need to look these up. The most common table values for lambda = 9 are in this table

\(x\)	2	3	4	5	14	15	16	17	18
	0.0062	0.0212	0.0550	0.1157	0.9585	0.9780	0.9889	0.9947	0.9976

A1 awrt 0.0432 or 0.0433

Special case
If you see 0.0432 / 0.0433 and then they go and do something else with it eg 1 – 0.0432 award M1 A0