OCR MEI Further Statistics Minor Specimen — Question 5 10 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of Spearman’s rank correlation coefficien
TypeEffect of data changes
DifficultyStandard +0.3 This is a standard Spearman's rank correlation question with routine calculations and a straightforward hypothesis test. Part (i) requires basic interpretation, parts (ii-iii) are textbook procedures (ranking, formula application, comparing to critical values), and part (iv) tests understanding of when Pearson vs Spearman is appropriate. The 10-mark allocation and multi-part structure are typical, but no novel insight or complex reasoning is required—slightly easier than average due to its procedural nature.
Spec5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank

Each contestant in a talent competition is given a score out of \(20\) by a judge. The organisers suspect that the judge's scores are associated with the age of the contestant. Table \(5.1\) and the scatter diagram in Fig. \(5.2\) show the scores and ages of a random sample of \(7\) contestants.
ContestantABCDEFG
Age6651392992214
Score1211151716189
Table 5.1 \includegraphics{figure_1} Fig. 5.2 Contestant G did not finish her performance, so it is decided to remove her data.
  1. Spearman's rank correlation coefficient between age and score, including all \(7\) contestants, is \(-0.25\). Explain why Spearman's rank correlation coefficient becomes more negative when the data for contestant G is removed. [1]
  2. Calculate Spearman's rank correlation coefficient for the \(6\) remaining contestants. [3]
  3. Using this value of Spearman's rank correlation coefficient, carry out a hypothesis test at the \(5\%\) level to investigate whether there is any association between age and score. [5]
  4. Briefly explain why it may be inappropriate to carry out a hypothesis test based on Pearson's product moment correlation coefficient using these data. [1]

Question 5:
AnswerMarks Guidance
5(i) Spearman’s correlation coefficient will become more
negative as there is a stronger tendency for score to go
AnswerMarks Guidance
down as age goes up.E1
[1]2.4 oe
straight line does not get
the mark.
AnswerMarks Guidance
5(ii) Contestant A B C D E F
Age rank 6 5 4 3 1 2
Score rank 2 1 3 5 4 6
r = –0.7714 (4 s.f.)
AnswerMarks
SM1
M1
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1N
For ranking Age
EFor ranking Score
AnswerMarks
BCcan be reversed as long as
consistent
can be reversed as long as
consistent
AnswerMarks Guidance
5(iii) H : there is no association between age and score.
0
H : there is an association between age and score.
1
For n = 6 5% critical value is 0.8857
E
Since 0.7714< 0.8857
P
Do not reject the null hypothesis
There is insufficient evidence to suggest at the 5% level
AnswerMarks
that there is association between age and score.B1
B1
I
C
B1
M1
A1FT
AnswerMarks
[5]M
3.3
2.5
3.4
1.1
AnswerMarks
2.2bB1 for H
0
B1 for H and population soi
1
NB H H not in terms of ρ
0 1
B1 for (cid:114) 0.8857
M1 for comparison with
critical value, provided
AnswerMarks
r< 1
s
A1 for conclusion in words FT
their r and sensible critical
s
AnswerMarks
valuehypotheses as shown in
the answer column should
be understood to imply
population
AnswerMarks Guidance
5(iv) S
Cannot tell if it is bivariate NormalE1
[1]3.5b E.g. score must be an integer
ContestantA B
Age rank6 5
Score rank2 1
Question 5:
5 | (i) | Spearman’s correlation coefficient will become more
negative as there is a stronger tendency for score to go
down as age goes up. | E1
[1] | 2.4 | oe | Saying it is closer to a
straight line does not get
the mark.
5 | (ii) | Contestant A B C D E F
Age rank 6 5 4 3 1 2
Score rank 2 1 3 5 4 6
r = –0.7714 (4 s.f.)
S | M1
M1
A1
[3] | 1.1
1.1
1.1 | N
For ranking Age
EFor ranking Score
BC | can be reversed as long as
consistent
can be reversed as long as
consistent
5 | (iii) | H : there is no association between age and score.
0
H : there is an association between age and score.
1
For n = 6 5% critical value is 0.8857
E
Since 0.7714< 0.8857
P
Do not reject the null hypothesis
There is insufficient evidence to suggest at the 5% level
that there is association between age and score. | B1
B1
I
C
B1
M1
A1FT
[5] | M
3.3
2.5
3.4
1.1
2.2b | B1 for H
0
B1 for H and population soi
1
NB H H not in terms of ρ
0 1
B1 for (cid:114) 0.8857
M1 for comparison with
critical value, provided
|r| < 1
s
A1 for conclusion in words FT
their r and sensible critical
s
value | hypotheses as shown in
the answer column should
be understood to imply
population
5 | (iv) | S
Cannot tell if it is bivariate Normal | E1
[1] | 3.5b | E.g. score must be an integer
Contestant | A | B | C | D | E | F
Age rank | 6 | 5 | 4 | 3 | 1 | 2
Score rank | 2 | 1 | 3 | 5 | 4 | 6
Each contestant in a talent competition is given a score out of $20$ by a judge. The organisers suspect that the judge's scores are associated with the age of the contestant. Table $5.1$ and the scatter diagram in Fig. $5.2$ show the scores and ages of a random sample of $7$ contestants.

\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
Contestant & A & B & C & D & E & F & G \\
\hline
Age & 66 & 51 & 39 & 29 & 9 & 22 & 14 \\
\hline
Score & 12 & 11 & 15 & 17 & 16 & 18 & 9 \\
\hline
\end{tabular}

\textbf{Table 5.1}

\includegraphics{figure_1}

\textbf{Fig. 5.2}

Contestant G did not finish her performance, so it is decided to remove her data.

\begin{enumerate}[label=(\roman*)]
\item Spearman's rank correlation coefficient between age and score, including all $7$ contestants, is $-0.25$. Explain why Spearman's rank correlation coefficient becomes more negative when the data for contestant G is removed. [1]

\item Calculate Spearman's rank correlation coefficient for the $6$ remaining contestants. [3]

\item Using this value of Spearman's rank correlation coefficient, carry out a hypothesis test at the $5\%$ level to investigate whether there is any association between age and score. [5]

\item Briefly explain why it may be inappropriate to carry out a hypothesis test based on Pearson's product moment correlation coefficient using these data. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor  Q5 [10]}}