A2 October 2020 Q5

EdexcelCurrent spec13 marksIncludes hypothesis testingChi-Squared Tests

5. A factory produces pins.

An engineer selects 40 independent random samples of 6 pins produced at the factory and records the number of defective pins in each sample.

Number of defective pins	0	1	2	3	4	5	6
Observed frequency	19	11	7	2	0	1	0

(a) Show that the proportion of defective pins in the 40 samples is 0.15 (2)

The engineer suggests that the number of defective pins in a sample of 6 can be modelled using a binomial distribution. Using the information from the sample above, a test is to be carried out at the 10% significance level, to see whether the data are consistent with the engineer’s suggested model.

The value of the test statistic for this test is 2.689

(b) Justifying the degrees of freedom used, carry out the test, at the 10% significance level, to see whether the data are consistent with the engineer’s suggested model.
State your hypotheses clearly. (8)

The engineer later discovers that the previously recorded information was incorrect.
The data should have been as follows.

Number of defective pins	0	1	2	3	4	5	6
Observed frequency	19	11	6	3	1	0	0

(c) Describe the effect this would have on the value of the test statistic that should be used for the hypothesis test.
Give reasons for your answer. (3)

Mark scheme (a)

Scheme	Marks	AO
\(p = \dfrac{(0) + 11 + 14 + 6 + (0) + 5 + (0)}{6 \times 40}\)	M1	2.1
\(p = \underline{\mathbf{0.15}}\)*	A1*cso	1.1b
	(2)

Notes

M1: Correct expression for \(p\) (may be seen in stages). Allow \(\tfrac{36}{240}\) but not \(\tfrac{6}{40}\) on its own

A1*cso: \(p = 0.15\) stated and no incorrect working seen

Mark scheme (b)

Scheme

Marks

\(X \sim \mathrm{B}(6, 0.15)\)

\(x\)	2	3	4	5	6
\(40 \times \mathrm{P}(X = x)\)	7.04…	1.65…	0.219…	0.015…	0.00…

3.4

Require \(40 \times \mathrm{P}(X \geqslant k) \gt 5\)
Exp. frequency for \(X \geqslant 2 = 8.94\ldots\) / \(X \geqslant 3 = 1.89\ldots\)

1.1b

Combine last 5 cells / only 3 cells in total

2.2a

2 is subtracted (as there are 2 restrictions) and the proportion used from data (and 1 equal totals)

2.4

\(3 - 2 = 1\) degree of freedom

1.1b

\(\mathrm{H}_0\): Binomial distribution is a suitable model
\(\mathrm{H}_1\): Binomial distribution is not a suitable model

3.4

Critical value \(\chi^2_{(1, 0.10)} = 2.705\) or 2.706

B1ft

1.1b

Test statistic is not in the critical region, insufficient evidence to reject \(\mathrm{H}_0\) (\(2.689 \lt 2.705/6\))
Data are consistent with binomial/engineer’s/suggested model.

B1ft

3.5a

(8)

Notes

M1: Attempting to find expected frequencies, at least 2 correct trunc. or rounded 1dp

M1: Recognising need to combine cells (Sight of awrt 8.94 implies M1M1)

A1: Combining cells for \(X \geqslant 2\) (to make 3 cells)

B1: Justifying why 2 is subtracted with \(p\) being calculated from data

A1: 1 degree of freedom

B1: Correct hypotheses (0.15 must not be included) Allow engineer’s model.

B1ft: Correct critical value (ft their df) May see \(\chi^2_{(2, 0.10)} = 4.605\) or \(\chi^2_{(3, 0.10)} = 6.251\)

B1ft: Correct inference (ft comparison of their CV with 2.689).
Condone \(p = 0.15\) included here. Do not allow contradictory statements to score here.
Hypotheses must be correct way round.

Mark scheme (c)

Scheme	Marks	AO
The total amount/proportion of defective pins remains the same.	M1	2.4
The cells for \(X \geqslant 2\) are still combined in the test.	M1	1.1b
So there is no change to the value of the test statistic.	A1	2.2a
	(3)
	(13 marks)

Notes

M1: Determining the number (\(N = 36\))/proportion (\(p = 0.15\)) of defective pins has not changed. e.g. \(11 + 12 + 9 + 4 = 36\). But not \(7 + 2 + 1 = 6 + 3 + 1\)

M1: Understanding the cells for \(X \geqslant 2\) are still combined in the test

A1: (dep on both M1s) Concluding that there is no change to the value of the test statistic.