Question 4 - A-Level Maths

OCR MEI Further Statistics Minor 2023 June — Question 4 13 marks

Exam Board	OCR MEI
Module	Further Statistics Minor (Further Statistics Minor)
Year	2023
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Poisson distribution
Type	Goodness-of-fit test for Poisson
Difficulty	Standard +0.3 This is a standard goodness-of-fit test for a Poisson distribution with straightforward calculations: estimating μ from data, computing Poisson probabilities, expected frequencies, and chi-squared contributions using given formulas. Part (c) requires standard hypothesis test procedure. While it involves multiple steps, all are routine applications of A-level Further Statistics techniques with no novel problem-solving required.
Spec	5.02i Poisson distribution: random events model 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities 5.02l Poisson conditions: for modelling 5.02m Poisson: mean = variance = lambda 5.06b Fit prescribed distribution: chi-squared test 5.06c Fit other distributions: discrete and continuous

4 Eve lives in a narrow lane in the country. She wonders whether the number of vehicles passing her house per minute can be modelled by a Poisson distribution with mean \(\mu\). She counts the number of vehicles passing her house over 100 randomly selected one-minute intervals. The results are shown in Table 4.1. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Table 4.1}

Number of vehicles	0	1	2	3	4	5	6	7	8	9	10	\(\geqslant 11\)
Frequency	36	33	14	10	4	1	0	0	1	0	1	0

\end{table}

Use the results to find an estimate for \(\mu\). The spreadsheet in Fig. 4.2 shows data for a \(\chi ^ { 2 }\) test to assess the goodness of fit of a Poisson model. The sample mean from part (a) has been used as an estimate for the population mean. Some of the values in the spreadsheet have been deliberately omitted. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Fig. 4.2}
\multirow[b]{2}{*}{1} A B C D E
Number of vehicles Observed frequency Poisson probability Expected frequency Chi-squared contribution
2 0 36 0.2725 27.2532 2.8073
3 1 33 0.3543 35.4291
4 2 14 3.5400
5 \(\geqslant 3\) 17 0.5145
6
\end{table}
Calculate the missing values in each of the following cells, giving your answers correct to 4 decimal places.
Carry out the \(\chi ^ { 2 }\) test at the 5\% significance level.
Eve checks her data and notices that the two largest numbers of vehicles per minute (8 and 10) occurred when some horses were being ridden along the lane, causing delays to the vehicles. She therefore repeats the analysis, missing out these two items of data. She finds that the value of the \(\chi ^ { 2 }\) test statistic is now 4.748. The number of degrees of freedom of the test is unchanged. Make two comments about this revised test.

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Question 4:

Answer	Marks	Guidance
4	(a)	1.3 oe
[1]	1.1	0×36+1×33+2×14+3×10+4×4+5×1+8×1+10×1 130

=

100 100

Answer	Marks	Guidance
4	(b)	C4 probability = 0.2303

D5 expected frequency = 14.2888

(33−35.4291)2

E3 contribution =

35.4291

Answer	Marks
= 0.1665	B1 FT

B1 FT

M1

A1

Answer	Marks
[4]	3.4

2.2a

1.1a

Answer	Marks
1.1	Allow 14.2911 for use of calculator probability for x=2 and

4dp values of x=1 and x=0

Answer	Marks	Guidance
4	(c)	DR

H : Poisson model is a good fit

0 H : Poisson model is not a good fit

1 X2 = 7.0(283)

Use of ν = 2

Critical value at 5% level = 5.99(1)

their 7.0(283) > their 5.99(1)

(Reject H )

0 There is sufficient evidence to suggest that the Poisson

model is not a good fit for the number of vehicles (passing

Answer	Marks
Eve’s house each minute)	B1

B1 FT

M1

A1

M1

A1

Answer	Marks
[6]	2.5

1.1

3.4

1.1

Answer	Marks
3.5a	Reference to ‘mean 1.3’ in hypotheses scores B0

Allow omission of context at this stage

FT (6.8618+𝑡ℎ𝑒𝑖𝑟E3)

or 𝜒2(7.0283) = 0.9702

5 0.9702 > 0.95 critical value must be from 𝜒2 table

Correct test statistic and critical value required

Conclusion must follow correct hypotheses, not be too

assertive and refer to context.

Answer	Marks	Guidance
4	(d)	We would now not reject H (insufficient evidence to

0 suggest that the Poisson model is not a good fit).

It is reasonable remove these two values as they are not

representative of the normal situation.

(vehicles would not be travelling independently, mean rate

Answer	Marks
would not be constant)	E1 FT

E1

Answer	Marks
[2]	3.2a
2.4	Comment on impact on conclusion. Allow ‘accept H ’

0 Allow alternative answers such as ‘If horse-riders regularly

use the lane, even if not very frequently, then the Poisson

model may not be valid.’

Question 4:
4 | (a) | 1.3 oe | B1
[1] | 1.1 | 0×36+1×33+2×14+3×10+4×4+5×1+8×1+10×1 130
=
100 100
4 | (b) | C4 probability = 0.2303
D5 expected frequency = 14.2888
(33−35.4291)2
E3 contribution =
35.4291
= 0.1665 | B1 FT
B1 FT
M1
A1
[4] | 3.4
2.2a
1.1a
1.1 | Allow 14.2911 for use of calculator probability for x=2 and
4dp values of x=1 and x=0
4 | (c) | DR
H : Poisson model is a good fit
0
H : Poisson model is not a good fit
1
X2 = 7.0(283)
Use of ν = 2
Critical value at 5% level = 5.99(1)
their 7.0(283) > their 5.99(1)
(Reject H )
0
There is sufficient evidence to suggest that the Poisson
model is not a good fit for the number of vehicles (passing
Eve’s house each minute) | B1
B1 FT
M1
A1
M1
A1
[6] | 2.5
1.1
3.4
1.1
1.1
3.5a | Reference to ‘mean 1.3’ in hypotheses scores B0
Allow omission of context at this stage
FT (6.8618+𝑡ℎ𝑒𝑖𝑟E3)
or 𝜒2(7.0283) = 0.9702
5
0.9702 > 0.95 critical value must be from 𝜒2 table
Correct test statistic and critical value required
Conclusion must follow correct hypotheses, not be too
assertive and refer to context.
4 | (d) | We would now not reject H (insufficient evidence to
0
suggest that the Poisson model is not a good fit).
It is reasonable remove these two values as they are not
representative of the normal situation.
(vehicles would not be travelling independently, mean rate
would not be constant) | E1 FT
E1
[2] | 3.2a
2.4 | Comment on impact on conclusion. Allow ‘accept H ’
0
Allow alternative answers such as ‘If horse-riders regularly
use the lane, even if not very frequently, then the Poisson
model may not be valid.’

Show LaTeX source

4 Eve lives in a narrow lane in the country. She wonders whether the number of vehicles passing her house per minute can be modelled by a Poisson distribution with mean $\mu$. She counts the number of vehicles passing her house over 100 randomly selected one-minute intervals. The results are shown in Table 4.1.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Table 4.1}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Number of vehicles & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & $\geqslant 11$ \\
\hline
Frequency & 36 & 33 & 14 & 10 & 4 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\
\hline
\end{tabular}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use the results to find an estimate for $\mu$.

The spreadsheet in Fig. 4.2 shows data for a $\chi ^ { 2 }$ test to assess the goodness of fit of a Poisson model. The sample mean from part (a) has been used as an estimate for the population mean. Some of the values in the spreadsheet have been deliberately omitted.

\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multirow[b]{2}{*}{1} & A & B & C & D & E \\
\hline
 & Number of vehicles & Observed frequency & Poisson probability & Expected frequency & Chi-squared contribution \\
\hline
2 & 0 & 36 & 0.2725 & 27.2532 & 2.8073 \\
\hline
3 & 1 & 33 & 0.3543 & 35.4291 &  \\
\hline
4 & 2 & 14 &  &  & 3.5400 \\
\hline
5 & $\geqslant 3$ & 17 &  &  & 0.5145 \\
\hline
6 &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}
\end{table}
\item Calculate the missing values in each of the following cells, giving your answers correct to 4 decimal places.

\begin{itemize}
  \item C4
  \item D5
  \item E3
\item In this question you must show detailed reasoning.
\end{itemize}

Carry out the $\chi ^ { 2 }$ test at the 5\% significance level.
\item Eve checks her data and notices that the two largest numbers of vehicles per minute (8 and 10) occurred when some horses were being ridden along the lane, causing delays to the vehicles. She therefore repeats the analysis, missing out these two items of data. She finds that the value of the $\chi ^ { 2 }$ test statistic is now 4.748. The number of degrees of freedom of the test is unchanged.

Make two comments about this revised test.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2023 Q4 [13]}}

This paper (7 questions)

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Q1 8 Q2 5 Q3 10 Q4 13 Q5 8 Q6 10 Q7 6

\multirow[b]{2}{*}{1}	A	B	C	D	E
	Number of vehicles	Observed frequency	Poisson probability	Expected frequency	Chi-squared contribution
2	0	36	0.2725	27.2532	2.8073
3	1	33	0.3543	35.4291
4	2	14			3.5400
5	\(\geqslant 3\)	17			0.5145
6