S3 June 2014 (R) Q5

EdexcelOld spec13 marksIncludes hypothesis testingt-Tests & Further Tests

5. A student believes that there is a difference in the mean lengths of English and French films. He goes to the university video library and randomly selects a sample of 120 English films and a sample of 70 French films. He notes the length, \(x\) minutes, of each of the films in his samples. His data are summarised in the table below.

	\(\Sigma x\)	\(\Sigma x^2\)	\(s^2\)	\(n\)
English films	10650	956 909	98.5	120
French films	6510	615 849	151	70

(a) Verify that the unbiased estimate of the variance, \(s^2\), of the lengths of English films is 98.5 minutes\(^2\) (2)

(b) Stating your hypotheses clearly, test, at the 1% level of significance, whether or not the mean lengths of English and French films are different. (7)

(d) The university video library contained 724 English films and 473 French films. Explain how the student could have taken a stratified sample of 190 of these films. (3)

Mark scheme (a)

Scheme	Marks
\({S_{\mathrm{E}}}^2 = \frac{1}{n - 1}\left(\sum x^2 - \frac{(\sum x)^2}{n}\right) = \frac{1}{119}\left(956909 - \frac{10650^2}{120}\right)\)	M1
\(= \frac{11721.5}{119} = 98.5\)	A1
	(2)

Alternative

Scheme	Marks
\({S_{\mathrm{E}}}^2 = \frac{n}{n - 1}\left(\frac{\sum x^2}{n} - \bar{x}^2\right) = \frac{120}{119}\left(\frac{956909}{120} - 88.75^2\right) = 98.5\)

Mark scheme (b)

Scheme	Marks
\(\mathrm{H_0}: \mu_{\mathrm{F}} = \mu_{\mathrm{E}}\) \(\mathrm{H_1}: \mu_{\mathrm{F}} \neq \mu_{\mathrm{E}}\)	B1
\(\bar{x}_E = \frac{10650}{120} = 88.75\) and \(\bar{x}_F = \frac{6510}{70} = 93\)	M1
Test statistic, \(z = \dfrac{93 - 88.75 - 0}{\sqrt{\frac{151}{70} + \frac{98.5}{120}}} = 2.4627\ldots\)	M1A1
Critical values, \(z = (\pm)2.5758\)	B1ft
Test stat is not in critical region Insufficient evidence to reject \(\mathrm{H_0}\) at 1% level	M1
No significant evidence of a difference in mean lengths of English and French films	A1ft
	(7)

Notes

1st B1 needs both \(\mathrm{H_0}\) and \(\mathrm{H_1}\), can be in words

2nd B1ft on their \(\mathrm{H_1}\)

1st M1 for attempt @ both means (\(\bar{x}_E\) may be in (a))

2nd M1 for attempt at correct test statistic, ft their values

3rd M1 for attempt to compare their test stat and critical values

A1 ft on their test and critical values but must include comment in context

Scheme	Marks
By CLT we can assume that the mean of a large sample has a Normal distribution	B1
	(1)

Mark scheme (d)

Scheme	Marks
On a list, label English films 1 – 724 and French films 1-473 (oe)	B1
Use random number table/generator to select \(\frac{724}{724 + 473} \times 190 = 115\) English films and \(\frac{473}{1197} \times 190 = 75\) French films	M1A1
	(3)
	(13 marks)

Notes

M1 requires use of random numbers and attempt to find correct sample sizes

A1 both 115 and 75 found.