S4 June 2010 Q1

EdexcelOld spec13 marksIncludes hypothesis testingConfidence Intervals t-Tests & Further Tests

1. A teacher wishes to test whether playing background music enables students to complete a task more quickly. The same task was completed by 15 students, divided at random into two groups. The first group had background music playing during the task and the second group had no background music playing.
The times taken, in minutes, to complete the task are summarised below.

	Sample size \(n\)	Standard deviation \(s\)	Mean \(\bar{x}\)
With background music	8	4.1	15.9
Without background music	7	5.2	17.9

You may assume that the times taken to complete the task by the students are two independent random samples from normal distributions.

(a) Stating your hypotheses clearly, test, at the 10% level of significance, whether or not the variances of the times taken to complete the task with and without background music are equal. (5)

(b) Find a 99% confidence interval for the difference in the mean times taken to complete the task with and without background music. (7)

Experiments like this are often performed using the same people in each group.

Mark scheme (a)

Scheme	Marks
\(\mathrm{H}_0 : \sigma_1^2 = \sigma_2^2,\ \mathrm{H}_1 : \sigma_1^2 \neq \sigma_2^2\)	B1
critical values \(\mathrm{F}_{6,7} = 3.87\ \left(\dfrac{1}{\mathrm{F}_{6,7}} = 0.258\right)\)	B1
\(\dfrac{s_2^2}{s_1^2} = \dfrac{5.2^2}{4.1^2};\ = 1.61\quad \left(\dfrac{s_1^2}{s_2^2} = \dfrac{4.1^2}{5.2^2} = 0.622\right)\)	M1; A1
Since 1.61 (0.622) is not in the critical region we accept \(\mathrm{H}_0\) and conclude there is no evidence that the two variances are different	A1ft
	(5)

Notes

B1 Allow \(\sigma_1 = \sigma_2\) and \(\sigma_1 \neq \sigma_2\)
B1 must match their F
M1 for \(\dfrac{s_2^2}{s_1^2}\) or other way up
A1 awrt 1.61(0.622)

Mark scheme (b)

Scheme	Marks
\(\mathrm{Sp}^2 = \dfrac{7 \times 4.1^2 + 6 \times 5.2^2}{7 + 6} = 21.53\ldots\)	M1A1
\(t_{13} = 3.012\)	B1
99% CI \(= (17.9 - 15.9) \pm 3.012 \times \sqrt{21.53} \times \sqrt{\dfrac{1}{8} + \dfrac{1}{7}}\)	M1A1ft
\(= \pm(9.23,\ -5.233)\), [ or accept: [0, 9.23] or [−9.23, 0] ] awrt 9.23, −5.23	A1A1
	(7)

Notes

M1 A1 \(\mathrm{Sp}^2\) may be seen in part a
B1 3.012 only
M1 for \((17.9 - 15.9) \pm t\text{ value} \times \sqrt{\mathrm{S_p}^2} \times \sqrt{\dfrac{1}{8} + \dfrac{1}{7}}\)
A1ft their \(\mathrm{Sp}^2\)
A1 awrt 9.23/−9.23 A1 awrt −5.23/5.23

Mark scheme (c)

Scheme	Marks
a person will be quicker at the task second time through/ times not independent/ familiar with the task/groups are not independent	B1
	(1)
	(13 marks)

Notes

B1 any correct sensible comment