| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Non-parametric tests |
| Type | Normality assessment for test choice |
| Difficulty | Standard +0.3 This is a straightforward application of choosing between parametric and non-parametric tests. Part (a) requires interpreting a given normal probability plot (routine skill), and part (b) involves executing a standard sign test or Wilcoxon test with clear hypotheses and critical values from tables. The question guides students through the decision-making process rather than requiring independent problem-solving, making it easier than average for Further Statistics. |
| Spec | 5.07b Sign test: and Wilcoxon signed-rank |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | The data do not appear to be close to a straight line |
| Answer | Marks |
|---|---|
| distributed | E1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 3.5b | Condone ‘Not close to line of best fit’ |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | Wilcoxon signed-rank test |
| Answer | Marks |
|---|---|
| suggest that median diameter is not 50 cm | B1 |
| Answer | Marks |
|---|---|
| [8] | 1.1a |
| Answer | Marks |
|---|---|
| 3.5a | Can be implied by correct procedure being carried out. |
| Answer | Marks | Guidance |
|---|---|---|
| Diameter | D − 50 | Rank |
| 82.6 | 32.6 | 7 |
| 79.2 | 29.2 | 6 |
| 77.8 | 27.8 | 5 |
| 38.4 | –11.6 | 1 |
| 88.1 | 38.1 | 8 |
| 32.2 | –17.8 | 2 |
| 26.5 | –23.5 | 3 |
| 23.4 | –26.6 | 4 |
| 94.3 | 44.3 | 9 |
| 104.2 | 54.2 | 10 |
Question 4:
4 | (a) | The data do not appear to be close to a straight line
Which suggests that the data may not be Normally
distributed | E1
E1
[2] | 1.1
3.5b | Condone ‘Not close to line of best fit’
Dep on first E mark
4 | (b) | Wilcoxon signed-rank test
H : population median is 50
0
H : population median is not 50
1
Diameter D − 50 Rank
82.6 32.6 7
79.2 29.2 6
77.8 27.8 5
38.4 –11.6 1
88.1 38.1 8
32.2 –17.8 2
26.5 –23.5 3
23.4 –26.6 4
94.3 44.3 9
104.2 54.2 10
W = 1 + 2 + 3 + 4 = 10
-
(W = 5 + 6 + 7 + 8 + 9 + 10 = 45)
+
Test statistic = W = 10
-
Critical value = 8
So do not reject H . There is insufficient evidence to
0
suggest that median diameter is not 50 cm | B1
B1
B1
M1
M1
A1
B1
A1
[8] | 1.1a
3.3
2.1
1.1
1.1
1.1
3.4
3.5a | Can be implied by correct procedure being carried out.
Allow if any attempt at ranking even if not using D – 50
No marks for a t test or Normal distribution test
Median used. Do not accept ‘average’ unless mention ‘mean
= median (due to symmetry)’
Both hypotheses correct including ‘population’ at least once
For attempt at ranking. Dep on finding D – 50.
Attempt to calculate either W or W . Allow with their ranks
+ -
Independent of all other marks
Do not FT incorrect test statistic or critical value
Diameter | D − 50 | Rank
82.6 | 32.6 | 7
79.2 | 29.2 | 6
77.8 | 27.8 | 5
38.4 | –11.6 | 1
88.1 | 38.1 | 8
32.2 | –17.8 | 2
26.5 | –23.5 | 3
23.4 | –26.6 | 4
94.3 | 44.3 | 9
104.2 | 54.2 | 104 A wood contains a large number of mature beech trees. The diameters in centimetres of a random sample of 10 of these trees are as follows.\\
$\begin{array} { l l l l l l l l l l } 82.6 & 79.2 & 77.8 & 38.4 & 88.1 & 32.2 & 26.5 & 23.4 & 94.3 & 104.2 \end{array}$
A tree surgeon wants to know if the average diameter of mature beech trees in this wood is 50 cm . The tree surgeon produces a Normal probability plot for these data.\\
\includegraphics[max width=\textwidth, alt={}, center]{4caa7409-cb32-41da-ad64-012a45753296-4_796_1230_589_230}
\begin{enumerate}[label=(\alph*)]
\item Explain why the tree surgeon should not carry out a test based on the $t$ distribution.
\item Carry out a suitable test at the $5 \%$ significance level to investigate whether the average diameter of mature beech trees in this wood is 50 cm .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2022 Q4 [10]}}