S4 June 2013 (R) Q4

EdexcelOld spec15 marksIncludes hypothesis testingt-Tests & Further Tests

4. A company carries out an investigation into the strengths of rods from two different suppliers, Ardo and Bards. Independent random samples of rods were taken from each supplier and the force, \(x\) kN, needed to break each rod was recorded. The company wrote the results on a piece of paper but unfortunately spilt ink on it so some of the results can not be seen.
The paper with the results on is shown below.

Ink-stained results. Ardo: 13.1, 13.6, 13.2, 13.8, 12.8, 13.5, 13.8. Bards: 15.3, 15.5, 14.1, 15.4, 14.2, 15.4, the rest hidden. Ardo n_A = 7, mean 13.4; Bards n_B = 9, mean 14.8; variances hidden. Pooled estimate of variance = 0.261

(a)

(i) Use the data from Ardo to calculate an unbiased estimate, \(s_A^2\), of the variance.

(ii) Hence find an unbiased estimate, \(s_B^2\), of the variance for the sample of 9 values from Bards. (4)

(b) Stating your hypotheses clearly, test at the 10% level of significance whether or not there is a difference in variability of strength between the rods from Ardo and the rods from Bards.
(You may assume the two samples come from independent normal distributions.) (5)

(c) Use a 5% level of significance to test whether the mean strength of rods from Bards is more than 0.9 kN greater than the mean strength of rods from Ardo. (6)

Mark scheme (a)

Scheme	Marks
(i) Ardo \(s^2 = \dfrac{1}{6}\left(1257.78 - 7(13.4)^2\right)\)	M1
\(= 0.143\ldots\) awrt 0.143	A1
(ii) Bards \(0.261 = \dfrac{6 \times 0.143\ldots + 8 \times s^2}{7 + 9 - 2}\)	M1
\(s^2 = 0.349\ldots\)	A1
	(4)

Notes

(i) M1 for attempt to calculate \(s^2\)

(ii) M1 use of correct formula for \({s_p}^2\)
A1 awrt 0.349 / 0.3495

Mark scheme (b)

Scheme	Marks
\(\mathrm{H}_0 : \sigma_1^2 = \sigma_2^2,\ \mathrm{H}_1 : \sigma_1^2 \ne \sigma_2^2\)	B1
critical values \(\mathrm{F}_{8,6} = 4.15\) \(\left(\dfrac{1}{\mathrm{F}_{8,6}} = 0.241\right)\)	B1
\(\dfrac{s_2^2}{s_1^2} = \dfrac{0.349}{0.143} = \text{awrt } 2.44\) \(\left(\dfrac{s_1^2}{s_2^2} = \dfrac{0.143}{.349} = 0.41\right)\)	M1; A1
Since 2.44… (0.424) is not in the critical region we accept \(\mathrm{H}_0\) and conclude there is no evidence that the two variances are different	A1cso
	(5)

Notes

1st B1 allow \(\mathrm{H}_0 : \sigma_1 = \sigma_2,\ \mathrm{H}_1 : \sigma_1 \ne \sigma_2\)

M1 For use of a correct formula

Mark scheme (c)

Scheme	Marks
\(\mathrm{H}_0 : \mu_B - \mu_A = 0.9; \quad \mathrm{H}_1 : \mu_B - \mu_A \gt 0.9\) both	B1
CR: \(t_{14}(0.05) \gt 1.761\) 1.761	B1
\(t = \pm\dfrac{14.8 - 13.4 - 0.9}{\sqrt{0.261\left(\frac{1}{7} + \frac{1}{9}\right)}} = \pm 1.94\ldots\)	M1 A1
awrt \(\pm 1.94\)	A1
Since 1.94… is in the critical region we reject \(\mathrm{H}_0\) and conclude that the mean strength of rods from Bards is more than 0.9 kN than that from *Ardo*.	A1 ft
	(6)
	(15 marks)

Notes

B1 must use \(\mu\). If not use \(A\) and \(B\) it must be clear which is which

M1 for attempt at correct test statistic – matching their hypotheses

1st A1 correct test statistic for their hypotheses