S3 June 2012 Q6

EdexcelOld spec14 marksIncludes hypothesis testingChi-Squared Tests

6. A total of 100 random samples of 6 items are selected from a production line in a factory and the number of defective items in each sample is recorded. The results are summarised in the table below.

Number of defective items	0	1	2	3	4	5	6
Number of samples	6	16	20	23	17	10	8

(a) Show that the mean number of defective items per sample is 2.91 (2)

A factory manager suggests that the data can be modelled by a binomial distribution with \(n = 6\). He uses the mean from the sample above and calculates expected frequencies as shown in the table below.

Number of defective items	0	1	2	3	4	5	6
Expected frequency	1.87	10.54	24.82	\(a\)	22.01	8.29	\(b\)

(b) Calculate the value of \(a\) and the value of \(b\) giving your answers to 2 decimal places. (4)

(c) Test, at the 5% level, whether or not the binomial distribution is a suitable model for the number of defective items in samples of 6 items.
State your hypotheses clearly. (8)

Scheme	Marks
Mean \(= \dfrac{1 \times 16 + 2 \times 20 + \ldots + 6 \times 8}{100} = 2.91\) ag	M1A1
	(2)

Mark scheme (b)

Scheme	Marks
\(p = \dfrac{2.91}{6} = 0.485\)	B1
\(a = 100 \times {}^6\mathrm{C}_3 \times 0.485^3 \times 0.515^3 = 31.17\)	M1A1
\(b = 100 \times 0.485^6 = 1.3(0)\)	A1
	(4)

Notes

0.485 can be implied by at least 1 correct answer.

Accept awrt 2dp for final answers

Clear use of Binomial and x100 required for method.

Mark scheme (c)

Scheme

Marks

\(\mathrm{H}_0\) : Binomial is a good fit
\(\mathrm{H}_1\) : Binomial is a not a good fit

Number of defective items	0 or 1	2	3	4	5 or 6
\(O\)	22	20	23	17	18
\(E\)	12.41	24.82	31.17	22.01	9.59

\(\displaystyle\sum\frac{(O - E)^2}{E} = \frac{(22 - 12.41)^2}{12.41} + \frac{(20 - 24.82)^2}{24.82} + \ldots + \frac{(18 - 9.59)^2}{9.59} = 18.998\ldots\) awrt 19.0

M1A1

\(\nu = 5 - 2 = 3\) degrees of freedom

\(\chi^2_3(5\%) = 7.815\)

B1ft

\(18.998\ldots \gt 7.815\) so reject \(\mathrm{H}_0\)

Binomial is a not a good fit (and is not a good model for the number of defective items in samples of size 6)

(8)

(14 marks)

Notes

Parameters in hyps award B0

1st M1 for combining either 0 and 1 or 5 and 6 or both. Require at least 1 value in a combined correct.

2nd M1 for attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 2 correct expressions or values.

2nd A1 for a correct comment suggesting that Binomial model is not suitable. No ft
Condone parameters here.