S4 June 2016 Q5

EdexcelOld spec14 marksIncludes hypothesis testingt-Tests & Further Tests

5. Fire brigades in cities \(X\) and \(Y\) are in similar locations. The response times, in minutes, during a particular month, for randomly selected calls are summarised in the table below.

	Sample size	Sample mean	Standard deviation \(s\)
\(X\)	9	14.8	6.76
\(Y\)	6	7.2	5.42

You may assume that the response times are from independent normal distributions.

Stating your hypotheses and showing your working clearly

(a) test, at the 10% level of significance, whether or not the variances of the populations from which the response times are drawn are the same, (5)

(b) test, at the 5% level of significance, whether or not the mean response time for the fire brigade in city \(X\) is more than 5 minutes longer than the mean response time for the fire brigade in city \(Y\). (8)

Mark scheme (a)

Scheme	Marks
\(\mathrm{H}_0 : \sigma^2_X = \sigma^2_Y \quad \mathrm{H}_1 : \sigma^2_X \ne \sigma^2_Y\)	B1
\(F_{8,5} = \dfrac{6.76^2}{5.42^2} = 1.556\)	M1A1
\(F_{8,5}\) is 4.82	B1
There is evidence that the variances are the same.	A1
	(5)

Notes

B1 both hypotheses

M1 Allow use of 6.76 and 5.42 instead of 6.76² and 5.42²

A1 awrt 1.56

Mark scheme (b)

Scheme	Marks
\(\mathrm{H}_0 : \mu_X = \mu_Y + 5 \quad \mathrm{H}_1 : \mu_X \gt \mu_Y + 5\)	B1
\({s_p}^2 = \dfrac{8 \times 6.76^2 + 5 \times 5.42^2}{13},= 39.42\ldots\) or \(s_p = 6.278\ldots\)	M1 A1
\(\left(t_{13} =\right)(\pm)\dfrac{14.8 - 7.2 - 5}{s_p\sqrt{\frac{1}{9} + \frac{1}{6}}} = (\pm)0.78578\ldots\) awrt 0.786	M1 M1dA1
Critical value \(t_{13}(5\%) = 1.771\)	B1
There is no evidence to Reject \(\mathrm{H}_0\) There is evidence that the fire brigade in \(X\) does not take more than 5 minutes longer than those in \(Y\).	A1cso
	(8)

Notes

B1 both hypotheses

M1 allow use of 6.76 and 5.42 instead of 6.76² and 5.42²

A1 awrt 39.4 or 6.28

B1 allow p value 0.650 instead of critical value

M1 use of correct formula with their \(S_p\) – condone missing 5
M1 use of correct formula with their \(S_p\)

(Corrected from the printed mark scheme: the critical value is printed as \(t_{13}(2.5\%) = 1.771\); for this one-tailed 5% test it is \(t_{13}(5\%) = 1.771\).)

Scheme	Marks
Test in part (b) requires the variances to be equal. The test in part (a) showed that the variances could be assumed to be equal.	B1
	(1)
	(14 marks)