A2 October 2021 Q2

EdexcelCurrent spec9 marksIncludes hypothesis testingt-Tests & Further Tests

2. A company produces two colours of candles, blue and white. The standard deviation of the burning times of the blue candles is 2.6 minutes and the standard deviation of the burning times of the white candles is 2.4 minutes.

Nissim claims that the mean burning time of blue candles is more than 5 minutes greater than the mean burning time of white candles.

A random sample of 90 blue candles is found to have a mean burning time of 39.5 minutes. A random sample of 80 white candles is found to have a mean burning time of 33.7 minutes.

(a) Stating your hypotheses clearly, use a suitable test to assess Nissim’s belief. Use a 1% level of significance. (6)

(b) Explain how the hypothesis test in part (a) would be carried out differently if the variances of the burning times of candles were unknown. (1)

The burning times for the candles may not follow a normal distribution.

(c) Describe the effect this would have on the calculations in the hypothesis test in part (a).
Give a reason for your answer. (2)

Mark scheme (a)

Scheme	Marks	AO
\(\mathrm{H}_0 : \mu_{b(lue)} = \mu_{w(hite)} + 5 \qquad \mathrm{H}_1 : \mu_{b(lue)} \gt \mu_{w(hite)} + 5\)	B1	2.1
\(s.e. = \sqrt{\dfrac{2.6^2}{90} + \dfrac{2.4^2}{80}}\)	M1	1.1b
\(z = \dfrac{39.5 - 33.7 - 5}{\sqrt{\dfrac{2.6^2}{90} + \dfrac{2.4^2}{80}}} = 2.085773\ldots\) awrt 2.09	M1 A1	3.1b 1.1b
CV = 2.3263 [ or \(p\)-value = 0.01849…]	B1	1.1b
Not significant, insufficient evidence to support Nissim’s claim.	A1	2.2b
	(6)

Notes

B1: Both hypotheses (oe) correct with correct notation (if using \(\mu_x\) and \(\mu_y\) these must be defined).

M1: Calculation of standard error

M1: Standardising using normal distribution test statistic for difference of two means with known variance

A1: awrt 2.09

B1: Correct critical value 2.3263 or better

A1: Drawing a correct inference in context

Scheme	Marks	AO
Use the \(t\)-test or (Since sample sizes are large,) use \(s^2\) as an approximation to \(\sigma^2\)	B1	2.4
	(1)

Mark scheme (c)

Scheme	Marks	AO
Since sample size is large, by the Central Limit Theorem, sample means will be (approximately) normally distributed	B1	2.4
…so no effect as the calculations in part (a) can still be carried out.	dB1	3.2b
	(2)
	(9 marks)

Notes

B1: Understanding that the assumptions required for the hypothesis test, by CLT sample means follow a normal distribution

dB1: (dep on previous B1) Correct evaluation