October 2021 Paper 2 Q11

OCR ACurrent spec12 marksIncludes hypothesis testingIncludes samplingData Processing Normal Distribution

11 Zac is planning to write a report on the music preferences of the students at his college. There is a large number of students at the college.

(a) State one reason why Zac might wish to obtain information from a sample of students, rather than from all the students. [1]

(b) Amaya suggests that Zac should use a sample that is stratified by school year.
Give one advantage of this method as compared with random sampling, in this context. [1]

Zac decides to take a random sample of 60 students from his college. He asks each student how many hours per week, on average, they spend listening to music during term. From his results he calculates the following statistics.

Mean	Standard deviation	Median	Lower quartile	Upper quartile
21.0	4.20	20.5	18.0	22.9

(c) Sundip tells Zac that, during term, she spends on average 30 hours per week listening to music.
Discuss briefly whether this value should be considered an outlier. [3]

(d) Layla claims that, during term, each student spends on average 20 hours per week listening to music. Zac believes that the true figure is higher than 20 hours. He uses his results to carry out a hypothesis test at the 5% significance level.
Assume that the time spent listening to music is normally distributed with standard deviation 4.20 hours.
Carry out the test. [7]

Scheme	Marks
Population large oe	B1
	[1]

Mark scheme (b)

Scheme	Marks
eg: Includes students from all years (or ages) Numbers in years in correct proportions Different years might like different music	B1
	[1]

Notes

or: Different years may have different numbers of students

NOT “It’s more representative” or “Takes all students into account”
“You get a range of people” “It avoids bias”

Mark scheme (c)

Scheme	Marks
\(21 + 2 \times 4.2 = 29.4\)	B1
\(22.9 + 1.5(22.9 - 18.0) = 30.25\)	B1
Unclear whether 30 is an outlier	B1
	[3]

Notes

B1: Allow “30 is more than 2 sds away from the mean”

B1: Allow “30 is less than 1.5×IQR from UQ”

B1: or eg "It depends which definition you use."
Any comment implying uncertainty
Ignore comments about mean ± 3 sds or mean ± 1 sd

Mark scheme (d)

Scheme	Marks
\(\mathrm{H}_0: \mu = 20\)	B1
\(\mathrm{H}_1: \mu \gt 20\) where \(\mu\) = pop mean time spent oe	B1
\(\overline{X} \sim \mathrm{N}\left(20, \dfrac{4.2^2}{60}\right)\) and \(\overline{X} = 21\)	M1
\(\mathrm{P}(\overline{X} \gt 21) = 0.0326\)	A1
Compare 0.05	A1
Reject \(\mathrm{H}_0\) Condon Accept \(\mathrm{H}_1\)	M1
There is evidence that (mean) time spent is \(\gt 20\) hours or eg there is evidence to support Zac’s belief	A1f
	[7]
NB Use of \(4.2^2/60\) as sd gives \(p = 0.000335\)

Notes

NB Allow 2 sf throughout

B1 B1: Allow other letters, not \(X\) unless defined. Not \(\overline{X}\)
B1B0 for 1 error eg 2-tail or:

2-tail: B1B0
undefined \(\mu\): B1B0
not in terms of parameter: B1B0
\(\mu\) = sample mean implied: B1B0
Not include value 20: B0B0
eg \(\mathrm{H}_0 = 20\) etc: B0B0

M1: Correct distribution and value of \(\overline{X}\).
stated or implied eg by 0.0326 or 0.967 or 20.9 or 1.84 or 0.000335 even if within incorrect statement eg \(\mathrm{P}(X = 21) = 0.0326\)
Condone \(\dfrac{4.2^2}{\sqrt{60}}\) or \(\dfrac{4.2^2}{60^2}\) or \(\dfrac{4.2}{60}\)

A1: BC Allow 2 sf, ie 0.033

A1: Dep 0.0326 or 1.84 or 0.9674 or \(\mathrm{P}(X \gt 21)\) or \(\mathrm{P}(X \geqslant 21)\) soi
Must compare like with like,
eg NOT prob cf \(z\)-value or large prob cf small prob or CV cf wrong end of acceptance region

M1: Dependent on clearly valid comparison of like with like.
Dep 0.0326 or 0.9674 or 20.9 or 1.84 or \(\mathrm{P}(X \gt 21)\) or \(\mathrm{P}(X \geqslant 21)\) soi (corrected from the printed mark scheme: it printed “0.0485 or 0.951”, values that do not arise in this question)
May be implied by conclusion,
eg “There is evidence that mean time is \(\gt 20\) hours” M1A1

A1f: In context, not definite;
eg “Mean time is \(\gt 20\) hours": A0
But “There is evidence to reject \(\mathrm{H}_0\) and that mean time is \(\gt 20\) h” M1A1
Allow opposite conclusion, ft their values, if above conditions met and loses 2^nd A1. But potentially can score all the other 6 marks

Alternative methods for M1A1A1

Scheme	Marks
or \(\dfrac{a - 20}{4.2 \div \sqrt{60}} = 1.645\) \((a = 20.9)\)	M1
CV = 20.9	A1
\(21 \gt 20.9\) or 21 not in acceptance region	A1

Scheme	Marks
or \(\dfrac{21 - 20}{4.2 \div \sqrt{60}}\) \((= 1.84)\)	M1
\(z_{calc} = 1.84\)	A1
Compare 1.645	A1

Scheme	Marks
\(\overline{X} \sim \mathrm{N}\left(20, \dfrac{4.2^2}{60}\right)\) and \(\overline{X} = \dfrac{1260}{60}\) or 21	M1
\(\mathrm{P}(\overline{X} \lt 21) = 0.9674\)	A1
Compare 0.95	A1

For each M1: Condone \(\dfrac{4.2^2}{\sqrt{60}}\) or \(\dfrac{4.2^2}{60^2}\) or \(\dfrac{4.2}{60}\)
\(\mathrm{P}(\overline{X} \lt 21) = 0.9674\): BC