A2 June 2022 Q5

EdexcelCurrent spec8 marksIncludes hypothesis testingConfidence Intervals t-Tests & Further Tests

5. The concentration of an air pollutant is measured in micrograms/m³

Samples of air were taken at two different sites and the concentration of this particular air pollutant was recorded.

For Site \(A\) the summary statistics are shown below.

	number of samples	\(s_A^2\)
Site \(A\)	13	6.39

For Site \(B\) there were 9 samples of air taken.

A test of the hypothesis \(\mathrm{H}_0 : \sigma_A^2 = \sigma_B^2\) against the hypothesis \(\mathrm{H}_1 : \sigma_A^2 \neq \sigma_B^2\) is carried out using a 2% level of significance.

(a) State a necessary assumption required to carry out the test. (1)

Given that the assumption in part (a) holds,

(b) find the set of values of \(s_B^2\) that would lead to the null hypothesis being rejected, (4)

(c) find a 99% confidence interval for the variance of the concentration of the air pollutant at Site \(\boldsymbol{A}\). (3)

Mark scheme (a)

Scheme	Marks	AO
The concentration of air pollutant (for each site) follows a normal distribution	B1	2.4
	(1)

Notes

B1: Correct modelling assumption. Allow air pollutant samples are independent.
Samples of air pollutant are normally distributed is B0

Mark scheme (b)

Scheme	Marks	AO
\(F_{12,8\,(0.01)} = 5.67 \qquad \text{or} \qquad F_{8,12\,(0.01)} = 4.5(0)\)	B1	3.3
\(\dfrac{6.39}{s_B^2} \gt \text{‘}5.67\text{’} \qquad \dfrac{s_B^2}{6.39} \gt \text{‘}4.50\text{’}\)	M1M1	1.1b 2.1
\((0 \lt)\, s_B^2 \lt 1.12698\ldots,\ \ s_B^2 \gt 28.755\)	A1	1.1b
	(4)

Notes

B1: Setting up either \(F\) statistic

M1: Either correct inequality (condone equation)

M1: Understanding both correct inequalities are needed (condone equation)

A1: Correct range of values with awrt 1.13 and awrt 28.8
\(s_B^2 \lt 1.12698\ldots\) and \(s_B^2 \gt 28.755\) is A0

Mark scheme (c)

Scheme	Marks	AO
\(\chi^2_{12,\,0.005} = 28.3 \quad \text{or} \quad \chi^2_{12,\,0.995} = 3.074\)	M1	3.3
\(\dfrac{12 \times 6.39}{\text{“}28.3\text{”}} \lt \sigma_A^2 \lt \dfrac{12 \times 6.39}{\text{“}3.074\text{”}}\)	M1	1.1b
\(2.7095\ldots \lt \sigma_A^2 \lt 24.94469\ldots\)	A1	1.1b
	(3)
	(8 marks)

Notes

M1: For realising a \(\chi^2\) distribution must be used as a model with at least one value correct. Allow \(\chi^2_{12,\,0.01} = 26.217\) or \(\chi^2_{12,\,0.99} = 3.571\) for this mark.

M1: Setting up 99% CI

A1: Correct CI with awrt 2.71 and awrt 24.9