Standard +0.3 This is a straightforward application of the Wilcoxon rank-sum test with clearly stated hypotheses, given rank sum, and standard significance level. Students need to calculate the test statistic, find critical values from tables, and make a conclusion. While it requires knowledge of a specific non-parametric test (slightly specialized for Further Maths), the execution is mechanical with no conceptual complications or multi-step problem-solving required.
8 A university course was taught by two different professors. Students could choose whether to attend the lectures given by Professor \(Q\) or the lectures given by Professor \(R\). At the end of the course all the students took the same examination.
The examination marks of a random sample of 30 students taught by Professor \(Q\) and a random sample of 24 students taught by Professor \(R\) were ranked. The sum of the ranks of the students taught by Professor \(Q\) was 726 .
Test at the 5\% significance level whether there is a difference in the ranks of the students taught by the two professors.
p > 0.025, 2p > 0.05, z < 1.96, or 1.96 used in CV
Do not reject H .
0
Insufficient evidence of a difference between the
Answer
Marks
ranks.
B1
M1ft
A1ft
Answer
Marks
[10]
1.1
1.1
Answer
Marks
2.2b
Explicit correct comparison
Correct first conclusion, needs
correct method and like-with-like
Answer
Marks
Contextualised, not too definite
Needs like-with-like
(e.g. p must be < 0.5)
ft on wrong ts, or 1-tail/2-tail
confusions, e.g. p compared
with 0.05 or not explicit, or z
≥ 1.645
SC 726 or 594 used: can get B1; M0A0; M1A1; M1A0 [0.1235, 0.1253 or 0.1272]; B1M1A1, total 7/10
For 726 and N(825, 3300) giving p = 0.0432: B1; M0A0; M0A0; M1A1; B1M1A1, total 6/10
Question 8:
8 | H : m = m , H : m m , where m and m are
0 Q R 1 Q R Q R
the medians of the rankings given to Q and R
Sum of ranks = ½ 54 55 = 1485
R = 1485 – 726 = 759 [or 561]
m
R ~ N(660,
m
… 3300)
P(R 759) = 0.0432 (3 s.f.)
m
[or z = 1.715] | B1
M1
A1
M1
A1
M1
A1 | 1.1
1.1
1.1
3.1b
3.3
3.4
1.1 | Allow m undefined. If verbal, must
mention medians, m or distribution.
Allow m = 0 as opposed to m = m
d Q R
Find sum of ranks
Correct value of R seen
m
normal, mean their ½ 24 55
Both parameters correct
Standardise, their R
m
Correct test statistic (0.0432)
0.0424 or 0.0416 (no/wrong cc):
M1A0 | Not anything that might be
unless symbol clearly
defined as median.
Not “there is no difference in
the ranks …”
Allow even if 726 used later
Allow SD/Var muddle
(Same for P(R ≤ 561)
m
Allow z [1.71, 1.715],
allow z = 1.72 only if cc
demonstrated correct
Alternatively: | Not 759 – or 726 – …; not wrong
CV 660 + 1.96√3300 [= 772.6] | M1 | tail for comparison, but allow | Or 561.5 > 547.4
758.5 < 772.6 | A1 | Needs correct cc | Wrong z-value: M1A1ft B0
p > 0.025, 2p > 0.05, z < 1.96, or 1.96 used in CV
Do not reject H .
0
Insufficient evidence of a difference between the
ranks. | B1
M1ft
A1ft
[10] | 1.1
1.1
2.2b | Explicit correct comparison
Correct first conclusion, needs
correct method and like-with-like
Contextualised, not too definite | Needs like-with-like
(e.g. p must be < 0.5)
ft on wrong ts, or 1-tail/2-tail
confusions, e.g. p compared
with 0.05 or not explicit, or z
≥ 1.645
SC 726 or 594 used: can get B1; M0A0; M1A1; M1A0 [0.1235, 0.1253 or 0.1272]; B1M1A1, total 7/10
For 726 and N(825, 3300) giving p = 0.0432: B1; M0A0; M0A0; M1A1; B1M1A1, total 6/10
8 A university course was taught by two different professors. Students could choose whether to attend the lectures given by Professor $Q$ or the lectures given by Professor $R$. At the end of the course all the students took the same examination.
The examination marks of a random sample of 30 students taught by Professor $Q$ and a random sample of 24 students taught by Professor $R$ were ranked. The sum of the ranks of the students taught by Professor $Q$ was 726 .
Test at the 5\% significance level whether there is a difference in the ranks of the students taught by the two professors.
\hfill \mbox{\textit{OCR Further Statistics 2019 Q8 [10]}}