S4 June 2011 Q5

EdexcelOld spec14 marksIncludes hypothesis testingConfidence Intervals t-Tests & Further Tests

5. The weights of the contents of breakfast cereal boxes are normally distributed.
A manufacturer changes the style of the boxes but claims that the weight of the contents remains the same.
A random sample of 6 old style boxes had contents with the following weights (in grams).

512 503 514 506 509 515

The weights, \(y\) grams, of the contents of an independent random sample of 5 new style boxes gave

\[\bar{y} = 504.8 \ \text{ and } \ s_y = 3.420\]

(a) Use a two-tail test to show, at the 10% level of significance, that the variances of the weights of the contents of the old and new style boxes can be assumed to be equal.
State your hypotheses clearly. (5)

(b) Showing your working clearly, find a 90% confidence interval for \(\mu_x - \mu_y\), where \(\mu_x\) and \(\mu_y\) are the mean weights of the contents of old and new style boxes respectively. (7)

Mark scheme (a)

Scheme	Marks
\(s_x^{\,2} = \dfrac{1559691 - 6 \times \left(\dfrac{3059}{6}\right)^2}{5} = 22.1666\ldots\)	M1
\(\mathrm{H}_0 : \sigma_x^{\,2} = \sigma_y^{\,2} \qquad \mathrm{H} : \sigma_x^{\,2} \neq \sigma_y^{\,2}\)	B1
\(\dfrac{s_x^{\,2}}{s_y^{\,2}} = 1.895\ldots\)	M1
\(F_{5,4} = 6.26\)	B1
\(\dfrac{s_x^{\,2}}{s_y^{\,2}} = 1.895\ldots\) awrt 1.90 and comment : not significant - variances of weights of the two boxes can be assumed equal.	A1
	(5)

Notes

1^st M1 for use of the correct formula for \(s_x^{\,2}\) with reasonable attempt at \(\sum x^2\) and \(\sum x\)
2^nd M1 for use of the correct test statistic. Allow use of 3.42 instead of 3.42². Top must be their variance.

Mark scheme (b)

Scheme	Marks
\(\bar{x} = 509.833\ldots \ \Rightarrow \ \bar{x} - \bar{y} = 5.03333\)	M1
\(s_p^{\,2} = \dfrac{5s_x^{\,2} + 4s_y^{\,2}}{9} = 17.513\ldots\) awrt 17.5	M1A1
5% two tail \(t\) value is \(t_9 = 1.833\)	B1
90% confidence interval is \(5.03\ldots \pm 1.833 \times \sqrt{17.513\ldots} \times \sqrt{\tfrac{1}{6} + \tfrac{1}{5}}\)	M1
\((0.388\ldots,\ 9.6782\ldots)\) awrt (0.388, 9.68)	A1, A1
	(7)

Notes

1^st M1 for attempting \(\bar{x} - \bar{y}\) can follow through their \(\bar{x}\)
2^nd M1 for attempt to find pooled estimate of variance
3^rd M1 for use of correct formula for CI allow any \(t\) value and ft their \(\bar{x}\) and \(s_p\)

Scheme	Marks
Zero is not in CI, there is evidence to reject the manufacturer’s claim Or the weight of the contents of the boxes has changed.	B1ft, B1ft
	(2)
	(14 marks)