OCR Further Statistics 2022 June — Question 8 7 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWilcoxon tests
TypeCritical region or test statistic properties
DifficultyChallenging +1.8 This question requires knowledge of the normal approximation to the Wilcoxon signed-rank test, including the formulas for mean and variance of W+ under H0, and working backwards from critical values to find both sample size and significance level. It demands multi-step algebraic manipulation and understanding of hypothesis testing theory beyond routine application, typical of Further Maths statistics but still within established techniques.
Spec5.07b Sign test: and Wilcoxon signed-rank

8 The critical region for an \(r\) \% two-tailed Wilcoxon signed-rank test, based on a large sample of size \(n\), is \(\left\{ W _ { + } \leqslant 113 \right\} \cup \left\{ W _ { + } \geqslant 415 \right\}\).
  1. Show that \(n = 32\).
  2. Using a suitable approximation, determine the value of \(r\).

Question 8:
AnswerMarks Guidance
8(a)  = 264
¼n(n + 1) = 264
AnswerMarks
 n = 32 or –33, but n > 0 so 32 only AGB1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
2.2aAllow even if no CC used
Use formula for mean
AnswerMarks
Solve n2 + n – 1056 = 0 to obtain 32Or: 113 + 415 = ½n(n + 1): M2
Can be implied by both 32 and –33
Just verification: SC B1
AnswerMarks
(b)Variance = 124 n ( n + 1 ) ( 2 n + 1 ) = 2860
(113.5 – 264)/2860 (= 0.002445)
 r = 20.002445  100%
AnswerMarks
= 0.489 (%)B1
M1*
depM1
A1
AnswerMarks
[4]3.3
3.4
3.1b
AnswerMarks
2.2aVariance 2860 stated or implied
Standardise, their parameters (or use 414.5)
Double their p-value (p < 0.5), oe
AnswerMarks
0.49 or better (0.48 is probably from no cc)No or wrong cc: (0.503% or 0.518%):
B1M1M1A0
Allow 0.50% only if correct CC seen
Question 8:
8 | (a) |  = 264
¼n(n + 1) = 264
 n = 32 or –33, but n > 0 so 32 only AG | B1
M1
A1
[3] | 3.1a
1.1
2.2a | Allow even if no CC used
Use formula for mean
Solve n2 + n – 1056 = 0 to obtain 32 | Or: 113 + 415 = ½n(n + 1): M2
Can be implied by both 32 and –33
Just verification: SC B1
(b) | Variance = 124 n ( n + 1 ) ( 2 n + 1 ) = 2860
(113.5 – 264)/2860 (= 0.002445)
 r = 20.002445  100%
= 0.489 (%) | B1
M1*
depM1
A1
[4] | 3.3
3.4
3.1b
2.2a | Variance 2860 stated or implied
Standardise, their parameters (or use 414.5)
Double their p-value (p < 0.5), oe
0.49 or better (0.48 is probably from no cc) | No or wrong cc: (0.503% or 0.518%):
B1M1M1A0
Allow 0.50% only if correct CC seen
8 The critical region for an $r$ \% two-tailed Wilcoxon signed-rank test, based on a large sample of size $n$, is $\left\{ W _ { + } \leqslant 113 \right\} \cup \left\{ W _ { + } \geqslant 415 \right\}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $n = 32$.
\item Using a suitable approximation, determine the value of $r$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2022 Q8 [7]}}