S3 June 2013 Q6

EdexcelOld spec11 marksIncludes hypothesis testingt-Tests & Further Tests

6. Fruit-n-Veg4U Market Gardens grow tomatoes. They want to improve their yield of tomatoes by at least 1 kg per plant by buying a new variety. The variance of the yield of the old variety of plant is 0.5 kg² and the variance of the yield for the new variety of plant is 0.75 kg². A random sample of 60 plants of the old variety has a mean yield of 5.5 kg. A random sample of 70 of the new variety has a mean yield of 7 kg.

(a) Stating your hypotheses clearly test, at the 5% level of significance, whether or not there is evidence that the mean yield of the new variety is more than 1 kg greater than the mean yield of the old variety. (9)

(b) Explain the relevance of the Central Limit Theorem to the test in part (a). (2)

Mark scheme (a)

Scheme	Marks
\(\mathrm{H}_0 : \mu_{new} - \mu_{old} = 1\)	B1
\(\mathrm{H}_1 : \mu_{new} - \mu_{old} \gt 1\)	B1
\(z = \dfrac{7 - 5.5 - 1}{\sqrt{\dfrac{0.5}{60} + \dfrac{0.75}{70}}} = 3.62254\ldots\) (awrt 3.62)	M1 A1A1 A1
Critical value \(z = 1.6449\) (allow \(\pm\))	B1
[\(3.62 \gt 1.6449\)] so sufficient evidence to reject \(\mathrm{H}_0\)	dM1
Evidence that the mean yield of new variety is more than 1 kg greater than the old variety.	A1
	(9)

Notes

1st & 2nd B1 for hypotheses. Accept \(\mu_1, \mu_2\) or \(\mu_A, \mu_B\) etc if there is some indication of which is which e.g. \(A \sim \mathrm{N}(\mu_A, 0.5)\)

1st M1 for an attempt at se. Condone switching 0.5 and 0.75 \(\sqrt{\dfrac{0.5 \text{ or } 0.75}{60} + \dfrac{0.75 \text{ or } 0.5}{70}}\)

1st A1 for a correct expression for denominator of test statistic or 0.138… or \(\sqrt{0.0190\ldots}\)

2nd A1 for a correct numerator of test statistic (must have the \(-1\))

3rd A1 for awrt 3.62
[Allow – 3.62 from numerator of 5.5 – 7 − − 1 and compatible \(\mathrm{H}_1\)]

3rd B1 for \(\pm\) 1.6449 seen or probability of 0.0002 (tables) or 0.000145…(calc) [allow 0.0001]

2nd dM1 dep. on 1^st M1 for a correct statement based on their normal cv and their test statistic

2nd A1 for correct comment in context. Must mention “yield” and “varieties” or “old” and “new” and “1”
If second B mark is B0 award A0 here

ALT Pooled estimate: If they calculate \(s_p = \sqrt{0.41845\ldots} = 0.64688\ldots\) allow 1^st M1, 1^st A1 for expression (or awrt 0.114) and 2^nd A1 if numerator correct but A0 for test statistic (4.39)

Mark scheme (b)

Scheme	Marks
Mean yield is normally distributed	B1
Sample size is large. Must state or imply that in this case sample size is large	B1
	(2)
	(11 marks)

Notes

1st B1 for mention of mean (yield) and normal (distribution)

2nd B1 for mention of sample (size) being large in this case