AS June 2023 Q4

EdexcelAS paperCurrent spec12 marksIncludes hypothesis testingChi-Squared Tests

4. Table 1 below shows the number of car breakdowns in the Snoreap district in each of 60 months.

Number of car breakdowns	0	1	2	3	4	5
Frequency	12	11	19	14	3	1

Table 1

Anja believes that the number of car breakdowns per month in Snoreap can be modelled by a Poisson distribution. Table 2 below shows the results of some of her calculations.

Number of car breakdowns	0	1	2	3	4	\(\geqslant 5\)
Observed frequency \((O_i)\)	12	11	19	14	3	1
Expected frequency \((E_i)\)	9.92			9.64	4.34

Table 2

(a) State suitable hypotheses for a test to investigate Anja’s belief. (1)

(b) Explain why Anja has changed the label of the final column to \(\geqslant 5\) (1)

(d) Find the value of \(\dfrac{(O_i - E_i)^2}{E_i}\) when the number of car breakdowns is

(i) 1

(ii) 3 (2)

(e) Explain why Anja used 3 degrees of freedom for her test. (2)

The test statistic for Anja’s test is 6.54 to 2 decimal places.

(f) Stating the critical value and using a 5% level of significance, complete Anja’s test. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0\): No. of car breakdowns per month follows a Poisson distribution \(\mathrm{H}_1\): No. of car breakdowns per month does not follow a Poisson distribution	B1	2.5
	(1)

Notes

B1 correct hypotheses mentioning “breakdowns” and “Poisson”. \(\mathrm{Po}(1.8)\) o.e. is B0

Mark scheme (b)

Scheme	Marks	AO
A Poisson distribution will assign some probability to (all) values greater than 5 (o.e.)	B1	2.4
	(1)

Notes

B1 for a suitable reason mentioning the Poisson distribution taking values beyond 5
e.g. sample space for Poisson being \([0, \infty)\)

Mark scheme (c)

Scheme	Marks	AO
Need \(\lambda\): \(\left[\hat{\lambda} =\right] \dfrac{0 \times 12 + 1 \times 11 + 2 \times 19 + 3 \times 14 + 4 \times 3 + 5 \times 1}{(12 + 11 + 19 + 14 + 3 + 1)}\) or 1.8	M1	1.1b
[Under \(\mathrm{H}_0\) \(X \sim \mathrm{Po}(1.8)\)] \(E_1 = 60 \times \mathrm{P}(X = 1) = 17.85(227\ldots)\)	M1	3.4 1.1b
\(E_2 = 60 \times \mathrm{P}(X = 2) = 16.06(705\ldots)\)	A1	1.1b
\(E_{\geqslant 5} = 60 - \displaystyle\sum_0^4 E_i = 2.18(43996\ldots)\)	B1ft
	(4)

Notes

1^st M1 for an expression for the mean with least 3 correct products and correct denominator
M0 if they have found \(\lambda\) by working backwards from \(\mathrm{P}(X = 0) \Rightarrow 60\mathrm{e}^{-\lambda} = 9.92\)

2^nd M1 for a correct method for finding \(E_1\) or \(E_2\) (implied by one correct value)

1^st A1 for awrt 17.85 and awrt 16.07

B1ft for awrt 2.18, or using 60 – (sum of their \(E_i\)) or \(60 \times (1 - \mathrm{P}(X \leqslant 4))\)

Mark scheme (d)

Scheme	Marks	AO
\(\dfrac{(11 - \text{“}17.85\ldots\text{”})^2}{\text{“}17.85\ldots\text{”}} = 2.6287\ldots\) or \(\dfrac{(14 - 9.64)^2}{9.64} = 1.97195\ldots\)	M1	1.1b
awrt 2.63 and awrt 1.97	A1	1.1b
	(2)

Notes

M1 for either correct expression (ft their \(E_1\))

A1 for awrt 2.63 and awrt 1.97

Mark scheme (e)

Scheme	Marks	AO
Need to combine last two columns since \(E_i\) are \(\lt 5\)	B1	1.1a
Degrees of freedom therefore 5 – 2 since mean for Poisson estimated from \(O_i\)	B1	1.1a
	(2)

Notes

1^st B1 for explaining need to pool columns since \(E_i \lt 5\)
allow candidates describing combining last three columns as long as they say \(E_i \lt 5\)

2^nd B1 for mentioning mean/rate/parameter/\(\lambda\) estimated from \(O_i\) and 2 constraints

Mark scheme (f)

Scheme	Marks	AO
\(\chi_3^2(5\%) = 7.815\)	B1	1.1b
(Not significant) insufficient evidence to reject Anja’s belief	B1	2.2b
	(2)
	(12)

Notes

1^st B1 for correct cv of 7.815 (or better)

2^nd B1 for correct conclusion in context mentioning “breakdowns” and “Poisson”
Allow equivalent words for belief, such as “theory” etc.
Must be consistent with their cv
Allow \(\mathrm{Po}(1.8)\) in conclusion instead of Poisson