S2 June 2008 Q6

EdexcelOld spec13 marksIncludes hypothesis testingPoisson Distribution

6. A call centre agent handles telephone calls at a rate of 18 per hour.

(a) Give two reasons to support the use of a Poisson distribution as a suitable model for the number of calls per hour handled by the agent. (2)

(b) Find the probability that in any randomly selected 15 minute interval the agent handles

(i) exactly 5 calls,

(ii) more than 8 calls. (5)

The agent received some training to increase the number of calls handled per hour. During a randomly selected 30 minute interval after the training the agent handles 14 calls.

(c) Test, at the 5% level of significance, whether or not there is evidence to support the suggestion that the rate at which the agent handles calls has increased. State your hypotheses clearly. (6)

Mark scheme (a)

Scheme	Marks
Calls occur singly Calls occur at a constant rate Calls occur independently or randomly.	B1 B1
	(2)

Notes

any two of the 3 only need calls once

B1 B1 They must use calls at least once. Independently and randomly are the same reason.
Award the first B1 if they only gain 1 mark.

Special case if they don’t put in the word calls but write two correct statements award B0B1

Mark scheme (b)

Scheme	Marks
(i) \(X \sim \mathrm{Po}(4.5)\) used or seen in (i) or (ii)	M1
\(\mathrm{P}(X = 5) = \mathrm{P}(X \leqslant 5) - \mathrm{P}(X \leqslant 4)\) \(= 0.7029 - 0.5321\)	M1
\(= 0.1708\)	A1
	(3)
(ii) \(\mathrm{P}(X \gt 8) = 1 - \mathrm{P}(X \leqslant 8)\) \(= 1 - 0.9597\)	M1
\(= 0.0403\)	A1
	(2)

Notes

(b) correct answers only score full marks

(i) M1 Po (4.5) may be implied by them using it in their calculations in (i) or (ii)
M1 for \(\mathrm{P}(X \leqslant 5) - \mathrm{P}(X \leqslant 4)\) or \(\dfrac{\mathrm{e}^{-\lambda}\lambda^5}{5!}\)
A1 only awrt 0.171

(ii) M1 for \(1 - \mathrm{P}(X \leqslant 8)\)
A1 only awrt 0.0403

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0 : \lambda = 9\ (\lambda = 18)\) \(\mathrm{H}_1 : \lambda \gt 9\ (\lambda \gt 18)\) may use \(\lambda\) or \(\mu\)	B1
\(X \sim \mathrm{Po}(9)\) may be implied	B1
\(\mathrm{P}(X \geqslant 14) = 1 - \mathrm{P}(X \leqslant 13)\) \(\left[\mathrm{P}(X \geqslant 14) = 1 - 0.9261 = 0.0739\right]\)	M1
\(= 1 - 0.9261\) \(\mathrm{P}(X \geqslant 15) = 1 - 0.9585 = 0.0415\) \(= 0.0739\) CR \(X \geqslant 15\)	A1
\(0.0739 \gt 0.05\) \(14 \leqslant 15\) Accept \(\mathrm{H}_0\). or it is not significant or a correct statement in context from their values	M1
There is insufficient evidence to say that the number of calls per hour handled by the agent has increased.	A1
	(6)
	(13 marks)

Notes

att \(\mathrm{P}(X \geqslant 14)\) | \(\mathrm{P}(X \geqslant 15)\) awrt 0.0739

B1 both. Must be one tail test. They may use \(\lambda\) or \(\mu\) and either 9 or 18 and match \(\mathrm{H}_0\) and \(\mathrm{H}_1\)

M1 Po (9) may be implied by them using it in their calculations.
M1 attempt to find \(\mathrm{P}(X \geqslant 14)\) eg \(1 - \mathrm{P}(X \leqslant 13)\) or \(1 - \mathrm{P}(X \lt 14)\)
A1 correct probability or CR

To get the next2 marks the null hypothesis must state or imply that \((\lambda) = 9\) or 18

M1 for a correct statement based on their probability or critical region or a correct contextualised statement that implies that.

A1. This depends on their M1 being awarded for accepting \(\mathrm{H}_0\). Conclusion in context. Must have calls per hour has not increased. Or the rate of calls has not increased.
Any statement that has the word calls in and implies the rate not increasing
e.g. no evidence that the rate of calls handled has increased
Saying the number of calls has not increased gains A0 as it does not imply rate
NB this is an A mark on EPEN

They may also attempt to find \(\mathrm{P}(X \lt 14) = 0.9261\) and compare with 0.95