OCR Further Statistics 2020 November — Question 3 9 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2020
SessionNovember
Marks9
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Mark schemeDownload PDF ↗
TopicWilcoxon tests
TypeWilcoxon rank-sum test (Mann-Whitney U test)
DifficultyChallenging +1.2 This is a straightforward application of the Mann-Whitney U test (Wilcoxon rank-sum test) from Further Statistics. Students must recognize the test from the context, calculate the test statistic from given rank sum, find critical value from tables, and state a standard assumption. While it requires knowledge of a specific further maths topic and multiple computational steps, it follows a standard procedure with no novel insight required—harder than typical A-level but routine for students who've studied this test.
Spec5.07b Sign test: and Wilcoxon signed-rank
Jo can use either of two different routes, A or B, for her journey to school. She believes that route A has shorter journey times. She measures how long her journey takes for 17 journeys by route A and 12 journeys by route B. She ranks the 29 journeys in increasing order of time taken, and she finds that the sum of the ranks of the journeys by route B is 219.
  1. Test at the 10\% significance level whether route A has shorter journey times than route B. [8]
  2. State an assumption about the 29 journeys which is necessary for the conclusion of the test to be valid. [1]
Question 3:
AnswerMarks Guidance
3(a) H : m = m , H : m < m where m and m are
0 A B 1 A B A B
the median journey times for A and B
AnswerMarks
W ~ N(180, 510)B1
B11.1
1.1OR: Median journey times equal, oe.
Allow if ms used but not defined
AnswerMarks
Both, can be implied, needs m = 12Allow “mean” or “average”
only if “population” stated
Allow √510 or 5102
Consider correct tail, either 219 or 141
(R = 219, m(m + n + 1) – R = 141)
m m
141.5−180
p = Φ   = 0.0441… BC
 510 
AnswerMarks
0.0441 < 0.1M1
M1
A1
AnswerMarks
A1ft1.1
1.1
1.1
AnswerMarks
1.1Find either P(≥ 219) (218.5) or
P(≤ 141) (141.5)
Needs some evidence. E.g.: 0.0421,
0.0401, 0.470 (no/wrong cc, √): M1
Explicit comparison. FT on wrong
AnswerMarks
p-value provided method correctUse of 0.9559 is M0 here.
For CV method see below
0.9559 > 0.9: A1A1 (M1A1)
0.9559 > 0.1: A1A0 M0A0
AnswerMarks Guidance
OR:CV 180 – z×√510 M1
141 (141.5) usedM1 unless 219 (218.5) used, in
z = 1.282 (CV = 151.05, 151.058..)A1 Stated or implied
141.5 < 151.05(85) or 218.5 > 208.95A1 CV and cc correct e.g. 141 < 150.55
Reject H .
0
AnswerMarks
Significant evidence that route B takes longerM1ft
A1ft
AnswerMarks
[8]1.1
2.2bCorrect first conclusion
Contextualised, not too definiteNeeds like-with-like, e.g.
0.9559 with 0.9
SC Sum of A’s ranks = 435 – 219 = 216 used: B1B0 M0M1A0A1 M1A1 max 5/8
AnswerMarks
Exxα: H : Journey times are the same, H : journey times for B are higher: B0
0 1
β: H : No evidence that median journey times are different, etc: B0
0
AnswerMarks
(b)Must be a random sample (of all journeys)
Or distributions must be same shape (necessary
AnswerMarks Guidance
assumption for Wilcoxon rank-sum test!)B1
[1]3.5b Or equivalent.
Allow “(journeys) independent”Not “representative”.
QuestionAnswer Marks 
Question 3:
3 | (a) | H : m = m , H : m < m where m and m are
0 A B 1 A B A B
the median journey times for A and B
W ~ N(180, 510) | B1
B1 | 1.1
1.1 | OR: Median journey times equal, oe.
Allow if ms used but not defined
Both, can be implied, needs m = 12 | Allow “mean” or “average”
only if “population” stated
Allow √510 or 5102
Consider correct tail, either 219 or 141
(R = 219, m(m + n + 1) – R = 141)
m m
141.5−180
p = Φ   = 0.0441… BC
 510 
0.0441 < 0.1 | M1
M1
A1
A1ft | 1.1
1.1
1.1
1.1 | Find either P(≥ 219) (218.5) or
P(≤ 141) (141.5)
Needs some evidence. E.g.: 0.0421,
0.0401, 0.470 (no/wrong cc, √): M1
Explicit comparison. FT on wrong
p-value provided method correct | Use of 0.9559 is M0 here.
For CV method see below
0.9559 > 0.9: A1A1 (M1A1)
0.9559 > 0.1: A1A0 M0A0
OR: | CV 180 – z×√510 | M1 | Allow √ errors | 180 + 1.282√510 etc is M0
141 (141.5) used | M1 | unless 219 (218.5) used, in
z = 1.282 (CV = 151.05, 151.058..) | A1 | Stated or implied | which case give M2(A1A1)
141.5 < 151.05(85) or 218.5 > 208.95 | A1 | CV and cc correct e.g. 141 < 150.55 | E.g. 219 > 209.45
Reject H .
0
Significant evidence that route B takes longer | M1ft
A1ft
[8] | 1.1
2.2b | Correct first conclusion
Contextualised, not too definite | Needs like-with-like, e.g.
0.9559 with 0.9
SC Sum of A’s ranks = 435 – 219 = 216 used: B1B0 M0M1A0A1 M1A1 max 5/8
Exx | α: H : Journey times are the same, H : journey times for B are higher: B0
0 1
β: H : No evidence that median journey times are different, etc: B0
0
(b) | Must be a random sample (of all journeys)
Or distributions must be same shape (necessary
assumption for Wilcoxon rank-sum test!) | B1
[1] | 3.5b | Or equivalent.
Allow “(journeys) independent” | Not “representative”.
Question | Answer | Marks | AO | Guidance
Jo can use either of two different routes, A or B, for her journey to school. She believes that route A has shorter journey times. She measures how long her journey takes for 17 journeys by route A and 12 journeys by route B. She ranks the 29 journeys in increasing order of time taken, and she finds that the sum of the ranks of the journeys by route B is 219.

\begin{enumerate}[label=(\alph*)]
\item Test at the 10\% significance level whether route A has shorter journey times than route B. [8]
\item State an assumption about the 29 journeys which is necessary for the conclusion of the test to be valid. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2020 Q3 [9]}}
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