A2 June 2024 Q5

EdexcelCurrent spec10 marksIncludes hypothesis testingQuality of Tests

5. Some of the components produced by a factory are defective. The management requires that no more than 3% of the components produced are defective.
Niluki monitors the production process and takes a random sample of \(n\) components.

(a) Write down the hypotheses Niluki should use in a test to assess whether or not the proportion of defective components is greater than 0.03 (1)

Niluki defines the random variable \(D_n\) to represent the number of defective components in a sample of size \(n\). She considers two tests A and B

In test A, Niluki uses \(n = 100\) and if \(D_{100} \geqslant 5\) she rejects \(\mathrm{H}_0\)

(b) Find the size of test A (2)

In test B, Niluki uses \(n = 80\) and

if \(D_{80} \geqslant 5\) she rejects \(\mathrm{H}_0\)
if \(D_{80} \leqslant 3\) she does not reject \(\mathrm{H}_0\)
if \(D_{80} = 4\) she takes a second random sample of size 80 and if \(D_{80} \geqslant 1\) in this second sample then she rejects \(\mathrm{H}_0\) otherwise she does not reject \(\mathrm{H}_0\)

Given that the actual proportion of defective components is 0.06

(d)

(i) find the power of test A

(ii) find the expected number of components sampled using test B (3)

Given also that, when the actual proportion of defective components is 0.06, the power of test B is 0.713

(e) suggest, giving your reasons, which test Niluki should use. (1)

Scheme	Marks	AO
\(\mathrm{H}_0: p = 0.03 \quad \mathrm{H}_1: p \gt 0.03\)	B1	2.5
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(D_{100} \sim \mathrm{B}(100, 0.03)\) [\(\mathrm{P}(D_{100} \geqslant 5) = 1 - \mathrm{P}(D_{100} \leqslant 4)\)]	M1	3.3
\(= 0.18214\ldots\) awrt 0.182	A1	1.1b
	(2)

Notes

M1 for sight or use of the correct model. Allow for 1 – 0.81785… or 1 – 0.91916..

A1 for awrt 0.182

SC Use of Poisson approximation
Allow M1 for \(\mathrm{Po}(3)\) and A1 for answer of 0.1847 or better

Mark scheme (c)

Scheme	Marks	AO
\(\mathrm{P}(D_{80} \geqslant 5) + \mathrm{P}(D_{80} = 4) \times \mathrm{P}(D_{80} \geqslant 1)\)	M1	2.1
\(= 0.09279\ldots + 0.12654\ldots \times 0.91255\ldots\)	A1	1.1b
\(= 0.20826\ldots\) awrt 0.208	A1	1.1b
	(3)

Notes

M1 for a correct expression for required probability. May be implied by 1^st A1

1^st A1 for a correct numerical expression – values to 2 s.f. or better

2^nd A1 for awrt 0.208

SC Use of Poisson approximation
Allow M1 and 1^st A1 for awrt 0.21 but 2^nd A0 (\(\mathrm{Po}(2.4)\) gives 0.2099…)

Mark scheme (d)

Scheme	Marks	AO
(i) Test A: [\(X \sim \mathrm{B}(100, 0.06)\) \(\mathrm{P}(X \geqslant 5) =\)] 0.72322… awrt 0.723	B1	1.2
(ii) Expected number \(= 80 + 80 \times \mathrm{P}(Y = 4)\) with \(Y \sim \mathrm{B}(80, 0.06)\)	M1	3.4
\(= 94.87\ldots =\) 95	A1	1.1b
	(3)

Notes

(i) B1 for power of test A = awrt 0.723

(ii) M1 Correct expression with \(Y \sim \mathrm{B}(80, 0.06)\)
or for use of \(\mathrm{B}(80, 0.06)\) to obtain a probability of 0.814 or 0.186
Implied by sight of 95 or better

A1 for 95 (accept awrt 94.9)
NB: May see \(0.814 \times 80 + 0.186 \times 160 =\) awrt 94.9 or 95, which is M1A1

SC Use of Poisson approximation
(d)(i) Allow B1 for 0.7149 or better from \(\mathrm{Po}(6)\)
(d)(ii) Allow M1 for expression (answer should be 94.56…) A0 for answer.

Mark scheme (e)

Scheme	Marks	AO
Tests of similar size and power but Test B involves sampling fewer components so would advise to use test B.	B1	2.4
	(1)
	(10 marks)

Notes

B1 for a choice backed up by a suitable reason:
If concluding B, they need to mention similar power and a smaller sample size
If concluding A, they need to mention smaller size and greater power
NB: We do not ft the candidate’s incorrect values for size or power