OCR Further Statistics 2024 June — Question 5 12 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2024
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeAssess model suitability before testing
DifficultyStandard +0.3 This is a standard chi-squared goodness of fit test for a Poisson distribution with straightforward parts: stating assumptions (recall), interpreting mean vs variance (basic understanding), stating hypotheses (routine), calculating one contribution (standard formula application), and completing the test with given test statistic. Part (f) requires contextual interpretation but is accessible. Slightly easier than average due to being mostly procedural with the test statistic already calculated.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.06b Fit prescribed distribution: chi-squared test

5 Some bird-watchers study the song of chaffinches in a particular wood. They investigate whether the number, \(N\), of separate bursts of song in a 5 minute period can be modelled by a Poisson distribution. They assume that a burst of song can be considered as a single event, and that bursts of song occur randomly. \section*{(a) State two further assumptions needed for \(N\) to be well modelled by a Poisson distribution.} The bird-watchers record the value of \(N\) in each of 60 periods of 5 minutes. The mean and variance of the results are 3.55 and 5.6475 respectively.
(b) Explain what this suggests about the validity of a Poisson distribution as a model in this context. The complete results are shown in the table.
\(n\)012345678\(\geqslant 9\)
Frequency103781366250
The bird-watchers carry out a \(\chi ^ { 2 }\) goodness of fit test at the \(5 \%\) significance level.
(c) State suitable hypotheses for the test.
(d) Determine the contribution to the test statistic for \(n = 3\).
(e) The total value of the test statistic, obtained by combining the cells for \(n \leqslant 1\) and also for \(n \geqslant 6\), is 9.202 , correct to 4 significant figures. Complete the goodness of fit test.
(f) It is known that chaffinches are more likely to sing in the presence of other chaffinches. Explain whether this fact affects the validity of a Poisson model for \(N\).

Question 5:
AnswerMarks Guidance
5(a) Bursts occur independently of one another ...
and at constant average rate or at a uniform rateB1
B1
AnswerMarks
[2]3.3
3.3Allow “probabilities of bursts must be independent”.
Not “number in one interval independent of number in another.”
Not “constant probability”, not “constant rate”, not “average
constant rate”, not “mean  variance”.
B0B0 unless at least one contextualised.
Ignore “singly”, but otherwise if more than 2 given, max B1
AnswerMarks
(b)These are not very close together
... so a Poisson distribution may not be valid
AnswerMarks
(Ignore “mean = variance for (exact) Poisson”)M1
A1
AnswerMarks
[2]2.2b
1.2Comment on closeness (not on equality)
Inference, not too definite (i.e., not “not valid”).
“In Poisson they’re the same but here they’re different” M0
AnswerMarks
(c)H : N has (follows) a Poisson distribution
0
H : N does not have a Poisson distribution
AnswerMarks Guidance
1B1
[1]1.1 Allow H : “Poisson is a good model”, “data can be modelled by
0
…”, “data consistent with Poisson”, “data follows/fits Poisson”,
“supports”, but not “there is evidence that …” Not Poisson(3.55).
AnswerMarks
(d)E (3) = 60P(3) = 12.85
f
(8 – 12.85)2/12.85
AnswerMarks
= 1.83(1207)B1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1E in range [12.8, 12.9], can be inferred from answer
f
(8 – their E)2/their E needed if answer not correct
f f
Awrt 1.83
AnswerMarks
(e)9.202 < 9.488
Do not reject H .
0
There is insufficient evidence that the number
of bursts of song in 5-minute periods does not
AnswerMarks
have a Poisson distributionB1
M1ft
A1ft
AnswerMarks
[3]1.1
1.1
AnswerMarks
2.2bCompare their 9.202 with 9.488 (wrong 9.202 loses no marks)
Allow for comparison with 11.07 or 12.59, or, with reversed
conclusion, 7.815. FT on their 9.202
Contextualised or N, not too assertive, FT on their 9.202. Needs
double negative, not “can be modelled by …” etc.
Condone “Po(3.55)” here.
AnswerMarks
(f)It affects the validity as it suggests that bursts
are not independent (as if one burst is heard it is
AnswerMarks Guidance
more likely that another will be)B1
[1]3.2a “Yes” or “less valid” or “not valid” oe, and reason in context.
Allow “bursts are not independent” or “not random” .
Not “N is not independent”, nor “not singly”.
AnswerMarks Guidance
QuestionAnswer Marks
Question 5:
5 | (a) | Bursts occur independently of one another ...
and at constant average rate or at a uniform rate | B1
B1
[2] | 3.3
3.3 | Allow “probabilities of bursts must be independent”.
Not “number in one interval independent of number in another.”
Not “constant probability”, not “constant rate”, not “average
constant rate”, not “mean  variance”.
B0B0 unless at least one contextualised.
Ignore “singly”, but otherwise if more than 2 given, max B1
(b) | These are not very close together
... so a Poisson distribution may not be valid
(Ignore “mean = variance for (exact) Poisson”) | M1
A1
[2] | 2.2b
1.2 | Comment on closeness (not on equality)
Inference, not too definite (i.e., not “not valid”).
“In Poisson they’re the same but here they’re different” M0
(c) | H : N has (follows) a Poisson distribution
0
H : N does not have a Poisson distribution
1 | B1
[1] | 1.1 | Allow H : “Poisson is a good model”, “data can be modelled by
0
…”, “data consistent with Poisson”, “data follows/fits Poisson”,
“supports”, but not “there is evidence that …” Not Poisson(3.55).
(d) | E (3) = 60P(3) = 12.85
f
(8 – 12.85)2/12.85
= 1.83(1207) | B1
M1
A1
[3] | 3.1a
1.1
1.1 | E in range [12.8, 12.9], can be inferred from answer
f
(8 – their E)2/their E needed if answer not correct
f f
Awrt 1.83
(e) | 9.202 < 9.488
Do not reject H .
0
There is insufficient evidence that the number
of bursts of song in 5-minute periods does not
have a Poisson distribution | B1
M1ft
A1ft
[3] | 1.1
1.1
2.2b | Compare their 9.202 with 9.488 (wrong 9.202 loses no marks)
Allow for comparison with 11.07 or 12.59, or, with reversed
conclusion, 7.815. FT on their 9.202
Contextualised or N, not too assertive, FT on their 9.202. Needs
double negative, not “can be modelled by …” etc.
Condone “Po(3.55)” here.
(f) | It affects the validity as it suggests that bursts
are not independent (as if one burst is heard it is
more likely that another will be) | B1
[1] | 3.2a | “Yes” or “less valid” or “not valid” oe, and reason in context.
Allow “bursts are not independent” or “not random” .
Not “N is not independent”, nor “not singly”.
Question | Answer | Marks | AO | Guidance
5 Some bird-watchers study the song of chaffinches in a particular wood. They investigate whether the number, $N$, of separate bursts of song in a 5 minute period can be modelled by a Poisson distribution. They assume that a burst of song can be considered as a single event, and that bursts of song occur randomly.

\section*{(a) State two further assumptions needed for $N$ to be well modelled by a Poisson distribution.}
The bird-watchers record the value of $N$ in each of 60 periods of 5 minutes. The mean and variance of the results are 3.55 and 5.6475 respectively.\\
(b) Explain what this suggests about the validity of a Poisson distribution as a model in this context.

The complete results are shown in the table.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | c | }
\hline
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & $\geqslant 9$ \\
\hline
Frequency & 10 & 3 & 7 & 8 & 13 & 6 & 6 & 2 & 5 & 0 \\
\hline
\end{tabular}
\end{center}

The bird-watchers carry out a $\chi ^ { 2 }$ goodness of fit test at the $5 \%$ significance level.\\
(c) State suitable hypotheses for the test.\\
(d) Determine the contribution to the test statistic for $n = 3$.\\
(e) The total value of the test statistic, obtained by combining the cells for $n \leqslant 1$ and also for $n \geqslant 6$, is 9.202 , correct to 4 significant figures.

Complete the goodness of fit test.\\
(f) It is known that chaffinches are more likely to sing in the presence of other chaffinches.

Explain whether this fact affects the validity of a Poisson model for $N$.

\hfill \mbox{\textit{OCR Further Statistics 2024 Q5 [12]}}