S3 June 2011 Q5

EdexcelOld spec13 marksIncludes hypothesis testingChi-Squared Tests Poisson Distribution

5. The number of hurricanes per year in a particular region was recorded over 80 years. The results are summarised in Table 1 below.

No of hurricanes, \(h\)	0	1	2	3	4	5	6	7
Frequency	0	2	5	17	20	12	12	12

Table 1

(a) Write down two assumptions that will support modelling the number of hurricanes per year by a Poisson distribution. (2)

(b) Show that the mean number of hurricanes per year from Table 1 is 4.4875 (2)

(c) Use the answer in part (b) to calculate the expected frequencies \(r\) and \(s\) given in Table 2 below to 2 decimal places. (3)

\(h\)	0	1	2	3	4	5	6	7 or more
Expected frequency	0.90	4.04	\(r\)	13.55	\(s\)	13.65	10.21	13.39

Table 2

(d) Test, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution. State your hypotheses clearly. (6)

Scheme	Marks
Hurricanes: occur singly / are independent or occur at random /are a rare event / at a constant rate	B1B1
	(2)

Scheme	Marks
From data \(\dfrac{1 \times 2 + 2 \times 5 + 3 \times 17 + .. + 7 \times 12}{80} = 4.4875\)	M1A1
	(2)

Mark scheme (c)

Scheme

Marks

No of hurricanes, \(h\)	0	1	2	3	4	5	6	7+
\(80\mathrm{P}(X = h)\)	0.9	4..038	\(r\)=9.06…	13.55	\(s\)=15.205	13.647…	10.206…	13.388…

M1A1A1

(3)

M for 80xPoisson probability with 4.4875 and either 2 or 4.

1st A1 for awrt 9.06 and 2^nd A1 for awrt 15.20 or 15.21

Mark scheme (d)

Scheme

Marks

No of hurricanes, \(h\)	0	1	2	3	4	5	6	7+
Combine to give expected frequencies >5	13.9991			13.55	15.205…	13.647…	10.206…	13.388…
Observed	7			17	20	12	12	12
\(\dfrac{(O - E)^2}{E}\)	3.499…			0.876…	1.511…	0.198…	0.315…	0.143…
\(\dfrac{O_i^2}{E_i}\)	3.500...			21.322...	26.306...	10.551...	14.108...	10.755..

\(\mathrm{H}_0\): Poisson distribution is a good fit o.e.
\(\mathrm{H}_1\): Poisson distribution is not a good fit o.e.

\(\displaystyle\sum\frac{(O_i - E_i)^2}{E_i} = 6.545..\) or \(\dfrac{O_i^2}{E_i} = 86.545 - 80 = 6.545..\) (awrt 6.55 or 6.54)

\(\nu = 6 - 2 = 4\)

cv is 9.488 (ft their \(\nu\) i.e. \(\chi_\nu^{\ 2}(0.05)\))

B1ft

\(6.545 \lt 9.488\) so insufficient evidence to reject \(\mathrm{H}_0\)
(Hurricanes) can be modelled by a Poisson distribution

(6)

(13 marks)

1st M1 for some pooling and attempting \(\dfrac{(O - E)^2}{E}\) or \(\dfrac{O^2}{E}\), at least 3 correct expressions or values.

1st B1 no value for parameter permitted

2nd A1 for a correct comment suggesting that Poisson model is suitable. No ft