S4 June 2009 Q2

EdexcelOld spec12 marksIncludes hypothesis testingt-Tests & Further Tests

2. An emission-control device is tested to see if it reduces CO₂ emissions from cars. The emissions from 6 randomly selected cars are measured with and without the device. The results are as follows.

Car	\(A\)	\(B\)	\(C\)	\(D\)	\(E\)	\(F\)
Emissions without device	151.4	164.3	168.5	148.2	139.4	151.2
Emissions with device	148.9	162.7	166.9	150.1	140.0	146.7

(a) State an assumption that needs to be made in order to carry out a \(t\)-test in this case. (1)

(b) State why a paired \(t\)-test is suitable for use with these data. (1)

(c) Using a 5% level of significance, test whether or not there is evidence that the device reduces CO₂ emissions from cars. (8)

(d) Explain, in context, what a type II error would be in this case. (2)

Scheme	Marks
The differences are normally distributed	B1
	(1)

Scheme	Marks
The data is collected in pairs or small sample size and variance unknown or samples not independent	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(d\): 2.5, 1.6, 1.6, −1.9, −0.6, 4.5 at least 2 correct	M1
\((\Sigma d = 7.7,\ \Sigma d^2 = 35.59)\quad \bar{d} = \pm 1.2833,\ \mathrm{sd} = 2.2675.\ \ (\mathrm{Var} = 5.141)\)	A1, A1
\(\mathrm{H}_0 : \mu_d = 0,\ \mathrm{H}_1 : \mu_d \gt 0\) (\(\mathrm{H}_1 : \mu_d \lt 0\) if \(d\) −2.5, −1.6, −1.6 etc) both depend on their \(d\)’s	B1
\(t = \dfrac{\pm 1.2833\sqrt{6}}{2.2675} = \pm 1.386\ldots\) formula and substitution, 1.38 – 1.39	M1, A1
Critical value \(t_5(5\%) = 2.015\) (1 tail)	B1
Not significant. Insufficient evidence to support that the device reduces CO₂ emissions.	A1 ft
	(8)

Notes

awrt \(\pm 1.28\), 2.27

Mark scheme (d)

Scheme	Marks
The idea that the device reduces CO₂ emissions has been rejected when in fact it does reduce emissions. OR Concluding that the device does not reduce emissions when in fact it does (if not in context can get B1 only)	B1 B1
	(2)
	(12 marks)