A2 June 2023 Q3

EdexcelCurrent spec15 marksIncludes hypothesis testingChi-Squared Tests

3. In a class experiment, each day for 170 days, a child is chosen at random and spins a large cardboard coin 5 times and the number of heads is recorded.
The results are summarised in the following table.

Number of heads	0	1	2	3	4	5
Frequency	3	10	45	62	38	12

Marcus believes that a \(\mathrm{B}(5, 0.5)\) distribution can be used to model these data and he calculates expected frequencies, to 2 decimal places, as follows

Number of heads	0	1	2	3	4	5
Expected frequency	\(r\)	26.56	\(s\)	\(s\)	26.56	\(r\)

(a) Find the value of \(r\) and the value of \(s\) (3)

(b) Carry out a suitable test, at the 5% level of significance, to determine whether or not the \(\mathrm{B}(5, 0.5)\) distribution is a good model for these data.
You should state clearly your hypotheses, the test statistic and the critical value used. (6)

Nima believes that a better model for these data would be \(\mathrm{B}(5, p)\)

To test her model, Nima uses this value of \(p\), to calculate expected frequencies as follows

Number of heads	0	1	2	3	4	5
Expected frequency	2.07	14.65	41.44	58.63	41.47	11.74

The test statistic for Nima’s test is 1.62 (to 3 significant figures)

(d) State,

(i) giving your reasons, the degrees of freedom

(ii) the critical value

that Nima should use for a test at the 5% significance level. (3)

(e) With reference to Marcus’ and Nima’s test results, comment on

(i) the probability of the coin landing on heads,

(ii) the independence of the spins of the coin.

Give reasons for your answers. (2)

Mark scheme (a)

Scheme	Marks	AO
[\(X \sim \mathrm{B}(5, 0.5)\)] \(\mathrm{P}(X = 0) = \mathrm{P}(X = 5) = 0.03125\) or \(\mathrm{P}(X = 2)\) or \(\mathrm{P}(X = 3) = 0.3125\)	M1	1.1b
[multiply by 170 to get] \(r = \underline{\mathbf{5.31}}(25)\); \(\ s = \underline{\mathbf{53.1}}(25)\)	A1;A1	1.1b(x2)
	(3)

Notes

M1 for 1 correct probability which may be embedded (0.03125 or 0.3125 or \(0.5^5\) or \({}^5C_2\, 0.5^2 \times 0.5^3\))

1^st A1 for \(r\) = awrt 5.31 (condone \(\frac{85}{16}\))

2^nd A1 for \(s\) = awrt 53.1 (condone \(\frac{425}{8}\))

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0: \mathrm{B}(5, 0.5)\) is a suitable model \(\quad \mathrm{H}_1: \mathrm{B}(5, 0.5)\) is NOT a …	B1	2.5
\(\dfrac{(O_i - E_i)^2}{E_i}\): \(\dfrac{(3-\text{‘}5.31\text{’})^2}{\text{‘}5.31\text{’}} = 1.00\ldots\), \(\dfrac{(10-26.56)^2}{26.56} = 10.3\ldots\), \(\dfrac{(45-\text{‘}53.1\text{’})^2}{\text{‘}53.1\text{’}} = 1.23\ldots\), \(\dfrac{(62-\text{‘}53.1\text{’})^2}{\text{‘}53.1\text{’}} = 1.48\ldots\), \(\dfrac{(38-26.56)^2}{26.56} = 4.92\ldots\), \(\dfrac{(12-\text{‘}5.31\text{’})^2}{\text{‘}5.31\text{’}} = 8.41\ldots\) or \(\dfrac{O_i^2}{E_i}\): \(\dfrac{3^2}{\text{‘}5.31\text{’}} = 1.69\ldots\), \(\dfrac{10^2}{26.56} = 3.76\ldots\), \(\dfrac{45^2}{\text{‘}53.1\text{’}} = 38.1\ldots\), \(\dfrac{62^2}{\text{‘}53.1\text{’}} = 72.3\ldots\), \(\dfrac{38^2}{26.56} = 54.3\ldots\), \(\dfrac{12^2}{\text{‘}5.31\text{’}} = 27.1\ldots\)	M1	1.1b
\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i}\) or \(\displaystyle\sum\frac{O_i^2}{E_i} - 170 = 27.4\ldots\) awrt 27.4 or awrt 27.5	A1	1.1b
Degrees of freedom is \(6 - 1 = \underline{\mathbf{5}}\), and critical value is 11.07(0)	B1ft B1ft	1.1b(x2)
[Significant result] Marcus’ model/\(\mathrm{B}(5, 0.5)\) is not a good fit. (o.e.)	A1	2.2b
	(6)

Notes

1^st B1 for both hypotheses mentioning \(\mathrm{B}(5, 0.5)\) or Marcus’ distribution at least once

M1 for at least one correct (ft) term or expression of the test statistic (accept 2sf)

1^st A1 for awrt 27.4 or awrt 27.5 (correct value here scores M1A1)

2^nd B1 for 5 or ft if ‘their \(r\)’ < 5, then df (= 4 – 1) = 3

3^rd B1 for 11.07(0) (or better) for ft df = 4 \(\rightarrow\) 9.488 or df = 3 \(\rightarrow\) 7.815

A1 dep on 1^st M1 for a suitable conclusion in context rejecting \(\mathrm{B}(5, 0.5)\)/Marcus’ model
Must be compatible with their test statistic and their CV. Just ‘Bin is not a good fit’ is A0
A0 if inconsistent comments are seen e.g. “do not reject \(\mathrm{H}_0\), \(\mathrm{B}(5, 0.5)\) is not a good fit”

Mark scheme (c)

Scheme	Marks	AO
\(\hat{p} = \left[\dfrac{0 \times 3 + 1 \times 10 + \ldots + 5 \times 12}{170 \times 5}\right] = 0.58588\ldots\) awrt 0.586	B1	1.1b
	(1)

Notes

B1 for awrt 0.586 allow \(\frac{498}{850}\) o.e.

Mark scheme (d)

Scheme	Marks	AO
(i) Need to pool (first 2) cells (0 and 1 since \(E(0) \lt 5\)) and use of \(\hat{p}\)	M1	2.4
Degrees of freedom: 5 groups – 2 constraints = 3	A1	1.1b
(ii) Critical value is 7.815	B1ft	1.1b
	(3)

Notes

(i) M1 for both reasons, must mention pooling or show pooling or mention exp. value < 5 and use of estimated parameter

A1 for df = 3 (must have scored the M1 for this mark)

(ii) B1 for 7.815 (or better) allow this independent of the M1 only allow ft on df=4 \(\rightarrow\) 9.488

Mark scheme (e)

Scheme	Marks	AO
(i) Nima’s model is a good fit (since 1.62 < ‘7.815’)/Marcus’ is not and this suggests coin is biased/probability of head approx. 0.6	B1	2.4
(ii) Nima’s test suggests binomial is a good model and therefore independence of spins is a reasonable assumption	B1	2.2b
	(2)
	(15 marks)

Notes

(i) 1^st B1 for stating Nima’s (binomial) model is a good fit/do not reject \(\mathrm{H}_0\) for Nima’s model/Marcus’ model is not a good fit and suggest that coin is probably biased/ \(p \gt 0.5\) (\(p\) closer to ‘their (c)’)
Only comparing 1.62 with ‘27.4’ to reach \(p \gt 0.5\) is incorrect and scores B0

(ii) 2^nd B1 for mention of Nima’s test suggests Binomial distribution is suitable and that spins are independent (ignore reference to Marcus’ test for 2^nd B1)