S3 June 2014 (R) Q6

EdexcelOld spec17 marksIncludes hypothesis testingChi-Squared Tests

6. Bags of £1 coins are paid into a bank. Each bag contains 20 coins.

The bank manager believes that 5% of the £1 coins paid into the bank are fakes. He decides to use the distribution \(X \sim \mathrm{B}(20, 0.05)\) to model the random variable \(X\), the number of fake £1 coins in each bag.

(a) State the assumptions necessary for the binomial distribution to be an appropriate model in this case. (2)

The bank manager checks a random sample of 150 bags of £1 coins and records the number of fake coins found in each bag. His results are summarised in Table 1.

Number of fake coins in each bag	0	1	2	3	4 or more
Observed frequency	43	62	26	13	6
Expected frequency	53.8	56.6	\(r\)	8.9	\(s\)

Table 1

(b) Calculate the values of \(r\) and \(s\), giving your answers to 1 decimal place. (3)

(c) Carry out a hypothesis test, at the 5% significance level, to see if the data supports the bank manager’s statistical model. State your hypotheses clearly. (7)

The assistant manager thinks that a binomial distribution is a good model but suggests that the proportion of fake coins is higher than 5%. She calculates the actual proportion of fake coins in the sample and uses this value to carry out a new hypothesis test on the data. Her expected frequencies are shown in Table 2.

Number of fake coins in each bag	0	1	2	3	4 or more
Observed frequency	43	62	26	13	6
Expected frequency	44.5	55.7	33.2	12.5	4.1

Table 2

(d) Explain why there are 2 degrees of freedom in this case. (2)

(e) Given that she obtains a \(\chi^2\) test statistic of 2.67, test the assistant manager’s hypothesis that the binomial distribution is a good model for the number of fake coins in each bag. Use a 5% level of significance and state your hypotheses clearly. (3)

Scheme	Marks
Independence of each occurrence (of a fake coin)	B1
Constant probability for each occurrence (of a fake)	B1
	(2)

Mark scheme (b)

Scheme	Marks
\(r = 150 \times \mathrm{P}(X = 2) = 150 \times \dbinom{20}{2} \times 0.05^2 \times 0.95^{18}\)	M1
\(r = 28.3015\ldots\) awrt 28.3	A1
\(s = 150 - (53.8 + 56.6 + 28.3 + 8.9) = 2.4\)	A1ft
	(3)

Notes

M1A1 for one of \(r\) or \(s\) correct

A1ft for other value if using 150 - … and answer must be >0

Mark scheme (c)

Scheme

Marks

\(\mathrm{H_0}\): Bin(20, 0.05) is a suitable model
\(\mathrm{H_1}\): Bin(20, 0.05) is not a suitable model

Combining last two groups

	\(\geqslant 3\)
Observed frequency	19
Expected frequency	11.3

\(\nu = 4 - 1 = 3\)

Critical value, \(\chi^2(0.05) = 7.815\) (accept 9.488 if their \(\nu = 4\))

B1ft

Test statistic, \(\sum \frac{(O - E)^2}{E} = \frac{(43 - 53.8)^2}{53.8} + \frac{(62 - 56.6)^2}{56.6} + \cdots\)

\(= 2.168\ldots + 0.515\ldots + 0.186\ldots + 5.246\ldots\)
\(= 8.117\) (accept 10.16 if groups not combined)

A1ft

In critical region, sufficient evidence to reject \(\mathrm{H_0}\), accept \(\mathrm{H_1}\)
Significant evidence at 5% level to reject the manager’s model

A1ft

(7)

Notes

1st B1 can be in words but must include \(p = 0.05\)

3rd B1 ft on their \(\nu\)

Test statistic alternative method
Test stat \(= \sum \frac{O^2}{E} - 150 = \frac{43^2}{53.8} + \frac{62^2}{56.6} + \cdots - 150 = 8.117\ldots\)

1st A1 ft if their groups not combined

2nd A1 ft their test and critical values but must be comment in context e.g. mention of “manager’s model” or “fake coins”

Mark scheme (d)

Scheme	Marks
\(\nu = 4 - 2 = 2\)
4 classes due to pooling	B1
2 restrictions (equal total and mean/proportion)	B1
	(2)

Notes

1st B1 evidence that pooling is required

2nd B1 must have correct reasons for restrictions.

Mark scheme (e)

Scheme	Marks
\(\mathrm{H_0}\): Binomial distribution is a good model \(\mathrm{H_1}\): Binomial distribution is not a good model	B1
Critical value, \(\chi^2(0.05) = 5.991\)	B1
Test statistic is not in critical region, insufficient evidence to reject \(\mathrm{H_0}\) Accept the assistant manager’s model for the number of fake coins per bag.	B1
	(3)
	(17 marks)