S4 June 2013 Q5

EdexcelOld spec17 marksIncludes hypothesis testingPoisson Distribution Quality of Tests

5. Water is tested at various stages during a purification process by an environmental scientist. A certain organism occurs randomly in the water at a rate of \(\lambda\) every 10 ml. The scientist selects a random sample of 20 ml of water to check whether there is evidence that \(\lambda\) is greater than 1. The criterion the scientist uses for rejecting the hypothesis that \(\lambda = 1\) is that there are 4 or more organisms in the sample of 20 ml.

(a) Find the size of the test. (2)

(b) When \(\lambda = 2.5\) find P(Type II error). (2)

A statistician suggests using an alternative test. The statistician’s test involves taking a random sample of 10 ml and rejecting the hypothesis that \(\lambda = 1\) if 2 or more organisms are present but accepting the hypothesis if no organisms are in the sample. If only 1 organism is found then a second random sample of 10 ml is taken and the hypothesis is rejected if 2 or more organisms are present, otherwise the hypothesis is accepted.

(c) Show that the power of the statistician’s test is given by \[1 - \mathrm{e}^{-\lambda} - \lambda(1 + \lambda)\mathrm{e}^{-2\lambda}\] (4)

Table 1 below gives some values, to 2 decimal places, of the power function of the statistician’s test.

\(\lambda\)	1.5	2	2.5	3	3.5	4
Power	0.59	0.75	0.86	\(r\)	0.96	0.97

Table 1

(d) Find the value of \(r\). (1)

Figure 1 shows a graph of the power function for the scientist’s test.

Figure 1: graph of power against lambda for the scientist’s test, lambda from 1.5 to 4 — Figure 1

(e) On the same axes draw the graph of the power function for the statistician’s test. (2)

Given that it takes 20 minutes to collect and test a 20 ml sample and 15 minutes to collect and test a 10 ml sample

(f) show that the expected time of the statistician’s test is slower than the scientist’s test for \(\lambda\mathrm{e}^{-\lambda} \gt \dfrac{1}{3}\) (4)

(g) By considering the times when \(\lambda = 1\) and \(\lambda = 2\) together with the power curves in part (e) suggest, giving a reason, which test you would use. (2)

Mark scheme (a)

Scheme	Marks
\(Y\) = no. of organisms in 20 ml. \(Y \sim \mathrm{Po}(2\lambda)\) Size \(= \mathrm{P}(Y \geqslant 4 \mid Y \sim \mathrm{Po}(2)),\ = 1 - \mathrm{P}(Y \leqslant 3) = 1 - 0.8571 = \underline{\mathbf{0.1429}}\)	M1, A1
	(2)

Notes

M1 for correct expression for size using Po(2)

Mark scheme (b)

Scheme	Marks
\(\mathrm{P}(\text{Type II error}) = 1 - \mathrm{P}(Y \geqslant 4 \mid Y \sim \mathrm{Po}(5)),\ = \mathrm{P}(Y \leqslant 3) = \underline{\mathbf{0.2650}}\)	M1, A1
	(2)

Notes

M1 for correct expression using Po(5)

Mark scheme (c)

Scheme	Marks
\(X\) = no. of organisms in 10 ml. \(X \sim \mathrm{Po}(\lambda)\) Power \(= \mathrm{P}(X \geqslant 2) + \mathrm{P}(X = 1) \times \mathrm{P}(X \geqslant 2)\)	M1
\(= \mathrm{P}(X \geqslant 2)\,[1 + \mathrm{P}(X = 1)] = \left[1 - \mathrm{e}^{-\lambda}(1 + \lambda)\right] \times \left[1 + \lambda\mathrm{e}^{-\lambda}\right]\)	M1A1
\(= 1 - \mathrm{e}^{-\lambda} - \lambda\mathrm{e}^{-\lambda} + \lambda\mathrm{e}^{-\lambda} - \lambda(1 + \lambda)\mathrm{e}^{-2\lambda} = 1 - \mathrm{e}^{-\lambda} - \lambda(1 + \lambda)\mathrm{e}^{-2\lambda}\)	A1cso
	(4)

Notes

1^st M1 for a correct expression in terms of probabilities
Alternate answer \(1 - [\mathrm{P}(X = 0) + \mathrm{P}(X = 1) \times \mathrm{P}(X \leqslant 1)]\)
2^nd M1 for an attempt at a correct equation in \(\lambda\)
1^st A1 for a correct expression in \(\lambda\)

Scheme	Marks
\(r = 0.92\)	B1
	(1)

Scheme	Marks
See Graph paper	B1B1
	(2)

Mark scheme (f)

Scheme	Marks
Expected time for statistician’s test: \(30 \times \mathrm{P}(X = 1) + 15 \times [1 - \mathrm{P}(X = 1)]\)	M1
\(= 30\lambda\mathrm{e}^{-\lambda} + 15\left(1 - \lambda\mathrm{e}^{-\lambda}\right) = 15\left(1 + \lambda\mathrm{e}^{-\lambda}\right)\)	A1
slower if: \(15\left(1 + \lambda\mathrm{e}^{-\lambda}\right) \gt 20,\ \Rightarrow \lambda\mathrm{e}^{-\lambda} \gt \dfrac{1}{3}\)	M1,A1cso
	(4)

Notes

1^st M1 for an attempt to calculate expected time
Alternate method \(15 + 15 \times \mathrm{P}(X = 1)\)
1^st A1 for a correct expression in terms of \(\lambda\)
2^nd M1 for attempt at correct inequality

Mark scheme (g)

Scheme	Marks
\(\lambda\mathrm{e}^{-\lambda}\) with \(\lambda = 1\) is 0.36…, with \(\lambda = 2\) is 0.27…so second(statisticians) test is slower if \(\lambda = 1\) but faster for \(\lambda = 2\). Second test is more powerful for all \(\lambda\)	B1
Choose second test - more powerful and faster for \(\lambda \geqslant 2\)	B1
	(2)
	(17 marks)

Notes

1^st B1 for a comment about power & timings
2^nd B1 for selecting second test