A2 October 2020 Q6

EdexcelCurrent spec12 marksIncludes hypothesis testingConfidence Intervals t-Tests & Further Tests

6. A new employee, Kim, joins an existing employee, Jiang, to work in the quality control department of a company producing steel rods.
Each day a random sample of rods is taken, their lengths measured and a 95% confidence interval for the mean length of the rods, in metres, is calculated. It is assumed that the lengths of the rods produced are normally distributed.

Kim took a random sample of 25 rods and used the \(t\) distribution to obtain a 95% confidence interval of (1.193, 1.367) for the mean length of the rods.
Jiang commented that this interval was a little wider than usual and explained that they usually assume that the standard deviation does not change and can be taken as 0.175 metres.

(a) Test, at the 10% level of significance, whether or not Kim’s sample suggests that the standard deviation is different from 0.175 metres. State your hypotheses clearly. (9)

Using Kim’s sample and the normal distribution with a standard deviation of 0.175 metres,

(b) find a 95% confidence interval for the mean length of the rods. (3)

Mark scheme (a)

Scheme	Marks	AO
From CI \(\bar{x} = \dfrac{1.193 + 1.367}{2} = 1.28\) or width = 1.367 – 1.193 = 0.174	B1	1.1b
\(1.367 = \text{“}1.28\text{”} \pm 2.064 \times \dfrac{s}{\sqrt{25}}\) or \(\text{“}0.174\text{”} = 2 \times 2.064 \times \dfrac{s}{\sqrt{25}}\)	M1;A1	3.4 1.1b
\(\Rightarrow s = 0.210755\ldots\)	A1	1.1b
\(\mathrm{H}_0 : \sigma = 0.175 \qquad \mathrm{H}_1 : \sigma \neq 0.175\)	B1	2.5
\(\chi^2_{24} = \dfrac{24s^2}{\sigma^2} =,\ 34.8092\ldots\) awrt 34.8	M1, A1	3.3 1.1b
\(\chi^2_{24}(10\%)\) 2-tail CR is \(\chi^2_{24} \lt \underline{13.848}\) or \(\chi^2_{24} \gt \underline{36.415}\)	B1	2.1
34.8 is not significant so insufficient evidence that \(\sigma \neq 0.175\)	A1	2.2b
	(9)

Notes

1^st B1 for finding mean from CI or calculation of width of CI

1^st M1 for using the given \(t\) model to form an equation in \(s\). (Allow \(t\) for 2.064 where \(2 \lt t \lt 3\))

1^st A1 for correct use of \(t_{24} = 2.064\)

2^nd A1 for \(s = 0.21\) or better

2^nd B1 for correct hypotheses in terms of \(\sigma\).

2^nd M1 for selecting the appropriate model for this test

3^rd A1 for test statistic awrt 34.8

3^rd B1 for at least one correct critical value

4^th A1 for a correct conclusion confirming that assuming st. dev = 0.175 is OK

Mark scheme (b)

Scheme	Marks	AO
\(\text{“}1.28\text{”} \pm z \times \dfrac{0.175}{\sqrt{25}}\)	M1	3.3
\(z = 1.96\)	B1	1.1b
\(= (1.211\ldots,\ 1.349\ldots)\) = awrt (1.21, 1.35)	A1	1.1b
	(3)
	(12 marks)

Notes

M1 for use of correct formula with \(1.6 \lt z \lt 2\) (ft \(\bar{x}\) if found in (a))

B1 for \(z = 1.96\) or better used

A1 for an interval awrt (1.21, 1.35)