S3 June 2007 Q5

EdexcelOld spec14 marksIncludes hypothesis testingt-Tests & Further Tests

5. In a trial of diet \(A\) a random sample of 80 participants were asked to record their weight loss, \(x\) kg, after their first week of using the diet. The results are summarised by

\[\sum x = 361.6 \quad \text{and} \quad \sum x^2 = 1753.95\]

(a) Find unbiased estimates for the mean and variance of weight lost after the first week of using diet \(A\). (5)

The designers of diet \(A\) believe it can achieve a greater mean weight loss after the first week than a standard diet \(B\). A random sample of 60 people used diet \(B\). After the first week they had achieved a mean weight loss of 4.06 kg, with an unbiased estimate of variance of weight loss of 2.50 kg².

(b) Test, at the 5% level of significance, whether or not the mean weight loss after the first week using diet \(A\) is greater than that using diet \(B\). State your hypotheses clearly. (7)

(d) State an assumption you have made in carrying out the test in part (b). (1)

Mark scheme (a)

Scheme	Marks
\(\hat{\mu} = \bar{x} = \dfrac{361.6}{80},\ = \underline{4.52}\)	M1, A1
\(\hat{\sigma}^2 = s^2 = \dfrac{1753.95 - 80 \times \bar{x}^2}{79} = (1.51288\ldots)\)	M1A1ft
AWRT 1.51	A1
	(5)

Notes

2^nd M1 for a correct attempt at \(s\) or \(s^2\), A1ft for correct expression for \(s^2\), ft their mean.

N.B. \(\sigma^2_n = 1.49\ldots\) so \(\dfrac{80}{79} \times 1.49\ldots\) is M1A1ft

Mark scheme (b)

Scheme	Marks
\(\mathrm{H}_0 : \mu_A = \mu_B \qquad \mathrm{H}_1 : \mu_A \gt \mu_B\)	B1 B1
Denominator	M1
\(z = \dfrac{4.52 - 4.06}{\sqrt{\dfrac{1.51\ldots}{80} + \dfrac{2.50}{60}}} = \left(\dfrac{0.46}{\sqrt{0.0605\ldots}}\right)\) \(z\)	dM1
\(= (\pm)\ 1.8689\ldots\) AWRT (\(\pm\)) 1.87	A1
One tail c.v. is \(z = 1.6449\) (AWRT 1.645 or probability AWRT 0.0307 or 0.0308)	B1
(significant) there is evidence that diet \(A\) is better than diet \(B\) or evidence that (mean) weight lost in first week using diet \(A\) is more than with \(B\)	A1ft
	(7)

Notes

1^st B1 can be given for \(\mu_1 = \mu_2\), but 2^nd B1 must specify which is \(A\) or \(B\).

1^st M1 for the denominator, follow through their 1.51.
Must have square root can condone \(2.50^2\) but \(\sqrt{\dfrac{1.51^2}{80} + \dfrac{2.50^2}{60}}\) is M0.
Allow \(\sqrt{\dfrac{1.51}{79} + \dfrac{2.50}{59}}\) leading to AWRT 1.85 to score M1M1A0 in (b) and can score in (d).

2^nd dM1 for attempting the correct test statistic, dependent on denominator mark
1^st A1 for AWRT \(\pm\) 1.87, may be implied by a correct probability.
2^nd A1ft ft their test statistic vs their cv only if \(\mathrm{H}_1\) is correct and both Ms are scored

Mark scheme (c)

Scheme	Marks
CLT enables you to assume that \(\bar{A}\) and \(\bar{B}\) are normally distributed	B1
	(1)

Notes

B1 for stating either \(\bar{A}\) or \(\bar{B}\) (but not \(A\) or \(B\)) are normally distributed

Mark scheme (d)

Scheme	Marks
Assumed \(\sigma_A^{\,2} = s_A^{\,2}\) and \(\sigma_B^{\,2} = s_B^{\,2}\) (either)	B1
	(1)
	(14 marks)

Notes

B1 for either, can be stated in words in terms of variances or standard deviations.