S3 June 2009 Q6

EdexcelOld spec10 marksIncludes hypothesis testingt-Tests & Further Tests

6. The lengths of a random sample of 120 limpets taken from the upper shore of a beach had a mean of 4.97 cm and a standard deviation of 0.42 cm. The lengths of a second random sample of 150 limpets taken from the lower shore of the same beach had a mean of 5.05 cm and a standard deviation of 0.67 cm.

(a) Test, using a 5% level of significance, whether or not the mean length of limpets from the upper shore is less than the mean length of limpets from the lower shore. State your hypotheses clearly. (8)

(b) State two assumptions you made in carrying out the test in part (a). (2)

Mark scheme (a)

Scheme	Marks
\(\mu_{\mathrm{U}}\) ~ mean length of upper shore limpets, \(\mu_{\mathrm{L}}\) ~ mean length of lower shore limpets \(\mathrm{H}_0 : \mu_{\mathrm{u}} = \mu_{\mathrm{L}}\) \(\mathrm{H}_1 : \mu_{\mathrm{u}} \lt \mu_{\mathrm{L}}\) both	B1
s.e. \(= \sqrt{\dfrac{0.42^2}{120} + \dfrac{0.67^2}{150}}\)	M1
\(= 0.0668\)	A1
\(z = \dfrac{5.05 - 4.97}{0.0668} = (\pm)1.1975\) awrt \(\pm\) 1.20	dM1 A1
Critical region is \(z \geqslant 1.6449\), or probability = awrt (0.115 or 0.116) \(z = \pm 1.6449\)	B1
(\(1.1975 \lt 1.6449\)) therefore not in critical region / accept \(\mathrm{H}_0\)/not significant (or \(\mathrm{P}(Z \geqslant 1.1975) = 0.1151,\ 0.1151 \gt 0.05\) or z not in critical region )	M1
There is no evidence that the limpets on the upper shore are shorter than the limpets on the lower shore.	A1
	(8)

Notes

1st B1 If \(\mu_1, \mu_2\) used then it must be clear which refers to upper shore. Accept sensible choice of letters such as \(u\) and \(l\).

1st M1 Condone minor slips e.g. \(\dfrac{0.67^2}{120}\) or \(\dfrac{0.67}{150} + \dfrac{0.42^2}{120}\) etc i.e. swapped \(n\) or one sd and one variance but M0 for \(\sqrt{\dfrac{0.67}{150} + \dfrac{0.42}{120}}\)

1st A1 can be scored for a fully correct expression. May be implied by awrt 1.20

2nd dM1 is dependent upon the 1^st M1 but can ft their se value if this mark is scored.

2nd A1 for awrt (\(\pm\)) 1.20

3rd M1 for a correct statement based on their \(z\) value and their cv. No cv is M0A0
If using probability they must compare their \(p\) (<0.5) with 0.05 (o.e) so can allow 0.884< 0.95 to score this 3^rd M1 mark.
May be implied by their contextual statement and M1A0 is possible.

3rd A1 for a correct comment to accept null hypothesis that mentions length of limpets on the two shores.

Mark scheme (b)

Scheme	Marks
Assume the populations or variables are independent	B1
Standard deviation of sample = standard deviation of population [Mention of Central Limit Theorem does NOT score the mark]	B1
	(2)
	(10 marks)

Notes

1st B1 for one correct statement. Accept ”samples are independent”

2nd B1 for both statements