A2 June 2024 Q2

EdexcelCurrent spec7 marksIncludes hypothesis testingPoisson Distribution Quality of Tests

2. The number of errors made by a secretary is modelled by a Poisson distribution with a mean of 2.4 per 100 words.

A 100-word piece of work completed by the secretary is selected at random.

(a) Find the probability that

(i) there are exactly 3 errors,

(ii) there are fewer than 2 errors. (2)

After a long holiday, a randomly selected piece of work containing 250 words completed by the secretary is examined to see if the rate of errors has changed.

(b) Stating your hypotheses clearly, and using a 5% level of significance, find the critical region for a suitable test. (4)

Scheme	Marks	AO
(i) 0.20901… awrt 0.209	B1	3.4
(ii) 0.30844.. awrt 0.308	B1	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0: \lambda = 2.4\) (or \(\mu = 6\)) \(\qquad \mathrm{H}_1: \lambda \neq 2.4\) (or \(\mu \neq 6\))	B1	2.5
[\(E\) = no. of errors] \(E \sim \mathrm{Po}(6)\)	M1	3.3
\(\mathrm{P}(E \leqslant 1) = 0.0174\) or \(\mathrm{P}(E \leqslant 2) = 0.0620\) and \(\mathrm{P}(E \leqslant 11) = 0.980\) or \(\mathrm{P}(E \geqslant 12) = 0.0201\)	M1	3.4
Critical region: \(E \leqslant 1\) or \(E \geqslant 12\)	A1	1.1b
	(4)

Notes

B1 for both hypotheses correct in terms of \(\lambda\) or \(\mu\) (allow \(\lambda = 6\) etc)

1^st M1 for selecting the correct model. Sight or use of \(\mathrm{Po}(6)\)

2^nd M1 for use of the correct model with two probs correct to 2.s.f. (accept \(\mathrm{P}(E \geqslant 12) = 0.02\))
Must see attempt at lower and upper limit. Probabilities may be seen in (c).

A1 for correct critical region (both parts). Allow \(E \leqslant 1\) and \(E \geqslant 12\) or \(E \leqslant 1, E \geqslant 12\) etc
Writing CR as probability statements is A0
NB: Completely correct CR implies M1M1A1

SC: 1-tailed test
B0 as hypotheses are incorrect
M1 for sight or use of \(\mathrm{Po}(6)\)
M1 (dep on \(\mathrm{H}_1\)) for sight of \(\mathrm{P}(E \leqslant 1) = 0.0174\) or \(\mathrm{P}(E \geqslant 11) = 0.0426\), in line with their \(\mathrm{H}_1\)
A1 for CR: \(E \leqslant 1\) or CR: \(E \geqslant 11\), in line with their hypotheses

Mark scheme (c)

Scheme	Marks	AO
[\(\mathrm{P}(\text{Type I error}) = 0.0174 + 0.0201 =\)] 0.0375 (Calc gives: 0.017351… + 0.0200919… = 0.037443…)	B1ft	1.2
	(1)
	(7 marks)

Notes

B1ft for 0.0375 or 0.0374 or summing their two appropriate probs (ft their CR)
NB: If candidate uses a 1-tailed test, this mark cannot be gained