| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Single time period probability |
| Difficulty | Standard +0.3 This is a straightforward hypothesis test selection and execution problem. Students must recognize that a t-test is appropriate (normal distribution supported by K-S test, small sample, population SD unknown), then perform a standard two-tailed t-test with given summary statistics. The calculations are routine and the context is clear, making this slightly easier than average for Further Maths statistics. |
| Spec | 2.05d Sample mean as random variable2.05e Hypothesis test for normal mean: known variance |
| 116.9 | 114.9 | 110.9 | 113.9 | 114.9 |
| 117.9 | 112.9 | 99.9 | 114.9 | 103.9 |
| 123.9 | 105.7 | 108.9 | 102.9 | 112.7 |
| \(| l |\) | Statistics |
| n | 15 |
| Mean | 111.6733 |
| \(\sigma\) | 6.1877 |
| s | 6.4048 |
| \(\Sigma \mathrm { x }\) | 1675.1 |
| \(\Sigma \mathrm { x } ^ { 2 }\) | 187638.31 |
| Min | 99.9 |
| Q 1 | 105.7 |
| Median | 112.9 |
| Q 3 | 114.9 |
| Max | 123.9 |
| \(n\) | 15 | |||
| \(p > 0.15\) | |||
| Null hypothesis |
| |||
| Alternative hypothesis |
|
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (i) | Kolmogorov-Smirnov p-value [greater than |
| Answer | Marks |
|---|---|
| t-test | E1 |
| Answer | Marks |
|---|---|
| [3] | 3.5a |
| Answer | Marks | Guidance |
|---|---|---|
| 3.1b | M | SC2 For Wilcoxon test with |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (ii) | H : (cid:80)(cid:32)110.2 H : (cid:80)(cid:122)110.2 |
| Answer | Marks |
|---|---|
| that in the UK | B1 |
| Answer | Marks |
|---|---|
| [8] | 1.1a |
| Answer | Marks |
|---|---|
| 3.5a | Both hypotheses |
| Answer | Marks |
|---|---|
| Conclusion in context | Allow hypotheses based on median if |
| Answer | Marks | Guidance |
|---|---|---|
| Price | -110.2 | rank |
| 116.9 | 6.7 | 11 |
| 117.9 | 7.7 | 13 |
| 123.9 | 13.7 | 15 |
| 114.9 | 4.7 | 8 |
| 112.9 | 2.7 | 4 |
| 105.7 | -4.5 | 6 |
| 110.9 | 0.7 | 1 |
| 99.9 | -10.3 | 14 |
| 108.9 | -1.3 | 2 |
| 113.9 | 3.7 | 5 |
| 114.9 | 4.7 | 8 |
| 102.9 | -7.3 | 12 |
| 114.9 | 4.7 | 8 |
| 103.9 | -6.3 | 10 |
| 112.7 | 2.5 | 3 |
Question 7:
7 | (i) | Kolmogorov-Smirnov p-value [greater than
0.15] indicates that the data could be from a
Normal distribution
Sample small with unknown population
variance
t-test | E1
E1
E
B1
[3] | 3.5a
I
C
2.4
3.1b | M | SC2 For Wilcoxon test with
(cid:120) Cannot be sure the data are
from a Normal distribution
(cid:120) Mean ≈ median indicates
distribution is fairly
symmetrical
OR SC1 For Wilcoxon with one of
the above bullet points
7 | (ii) | H : (cid:80)(cid:32)110.2 H : (cid:80)(cid:122)110.2
0 1
Where (cid:80) is the population mean petrol price
for Yorkshire
111.6733(cid:16)110.2
Test statistic is
6.4048
15
P
S
= 0.8909
Use of t
14
Critical value = 2.145
0.8089 < 2.145 so do not reject H
0
There is insufficient evidence that the
average price in Yorkshire is different from
that in the UK | B1
B1
M1
E
A1
M1
A1
M1
A1
[8] | 1.1a
2.5
3.3
I
C
1.1
3.4
1.1
2.2b
3.5a | Both hypotheses
Correct verbal definition
N
E
M
Conclusion in context | Allow hypotheses based on median if
Wilcoxon chosen in (i)
FT method for calculation of
Wilcoxon statistic
Price -110.2 rank
116.9 6.7 11
117.9 7.7 13
123.9 13.7 15
114.9 4.7 8
112.9 2.7 4
105.7 -4.5 6
110.9 0.7 1
99.9 -10.3 14
108.9 -1.3 2
113.9 3.7 5
114.9 4.7 8
102.9 -7.3 12
114.9 4.7 8
103.9 -6.3 10
112.7 2.5 3
FT test statistic = 44
FT use of n = 15 row of Wilcoxon
tables
FT Critical value = 25
FT 44 > 25 so do not reject H
0
Price | -110.2 | rank
116.9 | 6.7 | 11
117.9 | 7.7 | 13
123.9 | 13.7 | 15
114.9 | 4.7 | 8
112.9 | 2.7 | 4
105.7 | -4.5 | 6
110.9 | 0.7 | 1
99.9 | -10.3 | 14
108.9 | -1.3 | 2
113.9 | 3.7 | 5
114.9 | 4.7 | 8
102.9 | -7.3 | 12
114.9 | 4.7 | 8
103.9 | -6.3 | 10
112.7 | 2.5 | 37 A newspaper reports that the average price of unleaded petrol in the UK is 110.2 p per litre.
The price, in pence, of a litre of unleaded petrol at a random sample of 15 petrol stations in Yorkshire is shown below together with some output from software used to analyse the data.
\begin{center}
\begin{tabular}{ | r | r | r | r | r | }
\hline
116.9 & 114.9 & 110.9 & 113.9 & 114.9 \\
\hline
117.9 & 112.9 & 99.9 & 114.9 & 103.9 \\
\hline
123.9 & 105.7 & 108.9 & 102.9 & 112.7 \\
\hline
\end{tabular}
\end{center}
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | l | }
$| l |$ & Statistics \\
\hline
n & 15 \\
\hline
Mean & 111.6733 \\
\hline
$\sigma$ & 6.1877 \\
\hline
s & 6.4048 \\
\hline
$\Sigma \mathrm { x }$ & 1675.1 \\
\hline
$\Sigma \mathrm { x } ^ { 2 }$ & 187638.31 \\
\hline
Min & 99.9 \\
\hline
Q 1 & 105.7 \\
\hline
Median & 112.9 \\
\hline
Q 3 & 114.9 \\
\hline
Max & 123.9 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{table}
\begin{center}
\begin{tabular}{ | l | l | }
\hline
$n$ & 15 \\
\hline
\begin{tabular}{ l }
Kolmogorov-Smirnov \\
test \\
\end{tabular} & $p > 0.15$ \\
\hline
Null hypothesis & \begin{tabular}{ l }
The data can be modelled \\
by a Normal distribution \\
\end{tabular} \\
\hline
Alternative hypothesis & \begin{tabular}{ l }
The data cannot be \\
modelled by a Normal \\
distribution \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
(i) Select a suitable hypothesis test to investigate whether there is any evidence that the average price of unleaded petrol in Yorkshire is different from 110.2 p. Justify your choice of test.\\
(ii) Conduct the hypothesis test at the $5 \%$ level of significance.
\hfill \mbox{\textit{OCR MEI Further Statistics Major Q7 [11]}}