S3 June 2011 Q7

EdexcelOld spec16 marksIncludes hypothesis testingConfidence Intervals t-Tests & Further Tests

7. Roastie’s Coffee is sold in packets with a stated weight of 250 g. A supermarket manager claims that the mean weight of the packets is less than the stated weight. She weighs a random sample of 90 packets from their stock and finds that their weights have a mean of 248 g and a standard deviation of 5.4 g.

(a) Using a 5% level of significance, test whether or not the manager’s claim is justified. State your hypotheses clearly. (5)

(b) Find the 98% confidence interval for the mean weight of a packet of coffee in the supermarket’s stock. (4)

(c) State, with a reason, the action you would recommend the manager to take over the weight of a packet of Roastie’s Coffee. (2)

Roastie’s Coffee company increase the mean weight of their packets to \(\mu\) g and reduce the standard deviation to 3 g. The manager takes a sample of size \(n\) from these new packets. She uses the sample mean \(\overline{X}\) as an estimator of \(\mu\).

(d) Find the minimum value of \(n\) such that \(\mathrm{P}\left(\left|\overline{X} - \mu\right| \lt 1\right) \geqslant 0.98\) (5)

Mark scheme (a)

Scheme	Marks
\(\mathrm{H}_0 : \mu = 250, \ \mathrm{H}_1 : \mu \lt 250\),	B1
\(z = \dfrac{248 - 250}{\frac{5.4}{\sqrt{90}}}\)	M1
\(= -3.513\ldots\) awrt -3.51	A1
Critical value -1.6449	B1
\(-3.513.. \lt -1.6449\) so sufficient evidence to reject \(\mathrm{H}_0\) Manager’s claim is justified.	A1
	(5)

Notes

1st B1 for \(\mathrm{H}_0\) and for \(\mathrm{H}_1\) (must be <250) They must use \(\mu\) not \(x\), \(p\), \(\lambda\) or \(\bar{x}\) etc

1st M1 for attempt at standardising using 248, 250 and sd. Can accept \(\pm\).
Critical region: 250-0.936=249.064 for M1A1 (and compare with 248.)

3rd B1 for \(\pm\)1.6449 seen (or probability of 0.0002 or better)

2nd A1 for a correct contextualised comment. Must mention “Manager” and “claim” or “weight” and “stated weight”. No follow through.

Mark scheme (b)

Scheme	Marks
98% CI for \(\mu\) is \(248 \pm 2.3263 \times \dfrac{5.4}{\sqrt{90}}\)	M1B1
= awrt (247,249) dependent upon \(z\) value awrt 2.33	A1A1
	(4)

Notes

2.3263 or better for B mark. Any \(z\) value replacing 2.3263 award M.

Scheme	Marks
Hypothesis test is significant or CI does not contain stated weight.	B1
(Manager should ask the company to investigate if their) stated weight is too high o.e.	B1
	(2)

Mark scheme (d)

Scheme	Marks
\(P\left(\left\|\bar{x} - \mu\right\| \lt 1\right) = 0.98\) \(\dfrac{1}{\frac{3}{\sqrt{n}}} = 2.3263\)	M1 A1
\(n = (3 \times 2.3263)^2 = 48.7\ldots\)	dM1A1
Sample size 49 required.	A1
	(5)
	(16 marks)

Notes

1st M for LHS = \(z\) value >1

1st A for RHS awrt 2.33

2nd A1 for answers in the range 48.7-48.9

3rd A1 don’t condone \(\geqslant\)