AS June 2018 Q1

EdexcelAS paperCurrent spec10 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

1. A researcher is investigating the distribution of orchids in a field. He believes that the Poisson distribution with a mean of 1.75 may be a good model for the number of orchids in each square metre. He randomly selects 150 non-overlapping areas, each of one square metre, and counts the number of orchids present in each square.

The results are recorded in the table below.

Number of orchids in each square metre	0	1	2	3	4	5	6
Number of squares	30	42	35	26	11	6	0

He calculates the expected frequencies as follows

Number of orchids in each square metre	0	1	2	3	4	5	More than 5
Number of squares	26.07	45.62	39.91	23.28	10.19	3.57	\(r\)

(a) Find the value of \(r\) giving your answer to 2 decimal places. (1)

The researcher will test, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution with mean 1.75

(b) State clearly the hypotheses required to test whether or not this Poisson distribution is a suitable model for these data. (1)

The test statistic for this test is 2.0 and the number of degrees of freedom to be used is 4

(d) Stating your critical value clearly, determine whether or not these data support the researcher’s belief. (2)

The researcher works in another field where the number of orchids in each square metre is known to have a Poisson distribution with mean 1.5

He randomly selects 200 non-overlapping areas, each of one square metre, in this second field, and counts the number of orchids present in each square.

(e) Using a Poisson approximation, show that the probability that he finds at least one square with exactly 6 orchids in it is 0.506 to 3 decimal places. (4)

Scheme	Marks	AO
1.36 or 1.37	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(\mathrm{H}_0\): \(\mathrm{Po}(1.75)\) is a suitable model \(\mathrm{H}_1\): \(\mathrm{Po}(1.75)\) is not a suitable model	B1	3.4
	(1)

Notes

B1: For both hypotheses correct. Must have \(\mathrm{Po}(1.75)\) or Poisson with mean 1.75 and be attached to \(\mathrm{H}_0\) and \(\mathrm{H}_1\) the right way round.

Mark scheme (c)

Scheme	Marks	AO
Cells are combined for expected frequencies < 5 so combine the last 3 cells	B1	2.4
subtract 1 since totals agree	B1	2.4
	(2)

Notes

B1: Explaining why there are 5 classes. Must mention combine the 3 cells when frequencies < 5 or to combine the 3 cells to make frequency > 5

B1: Explaining why 1 is subtracted. Must say/show 1 is subtracted and Totals agree or Total frequency must be 150 or only need 4 pieces of data to find the other or \(\lambda\) is known or 1.75 is given.

NB B0 for “only 1 constraint” on its own.

Mark scheme (d)

Scheme	Marks	AO
\(\chi^2_4 = 9.488\)	B1	1.1b
therefore, the researcher’s belief is supported or evidence that Po(1.75) is a good model for the number of orchids in each square metre	B1ft	3.5a
	(2)

Notes

B1: awrt 9.49

B1ft: ft their critical value only. For drawing the correct conclusion – condone missing 1.75.
If hypotheses are the wrong way round or there are no hypotheses in (b) award B0

Mark scheme (e)

Scheme	Marks	AO
\(\mathrm{P}(\text{exactly 6 orchids}) =\) awrt 0.00353	B1	1.1b
\(X \sim \mathrm{B}(200, \text{“}0.00353\text{”})\) mean \(= 200 \times \text{“}0.00353\text{”} =\) awrt 0.706	M1	3.3
\(Y \sim \mathrm{Po}(\text{“}0.706\text{”})\) \(1 - \mathrm{P}(Y = 0) = 1 - \mathrm{e}^{-\text{“}0.706\text{”}}\)	M1	3.4
\(= 0.506\)*	A1*	2.1
	(4)
	(10 marks)

Notes

B1: awrt 0.00353. May be implied by awrt 0.706 for mean.

M1: Selecting the model \(\mathrm{B}(200, \text{“their P(exactly 6 orchids)”})\) and using \(np\) (\(0 \lt p \lt 1\)) to find the mean. May be implied by awrt 0.706

M1: Using the model Po(their \(np\)) and using or writing \(1 - \mathrm{P}(Y = 0)\) or \(1 - \mathrm{P}(Y \leqslant 0)\) or \(1 - \mathrm{e}^{-\text{“}0.706\text{”}}\)

A1*: only award if the previous 3 marks have been awarded. and 0.506 stated.