OCR Further Statistics 2023 June — Question 8 16 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2023
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeChi-squared goodness of fit: Poisson
DifficultyChallenging +1.2 This is a comprehensive chi-squared goodness of fit question with standard calculations (mean, variance, expected frequencies) and routine hypothesis testing. Parts (a)-(f) are textbook applications requiring no novel insight. Part (g) adds mild complexity by asking students to fit a custom discrete distribution, but this is still a straightforward parameter estimation exercise using observed frequencies. The multi-part structure and length push it slightly above average difficulty, but all components are standard Further Statistics fare.
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

8 A team of researchers have reason to believe that the number of calls received in randomly chosen 10-minute intervals to a call centre can be well modelled by a Poisson distribution. To test this belief the researchers record the number of telephone calls received in 60 randomly chosen 10-minute intervals. The results, together with relevant calculations, are shown in the following table.
Total
Number of calls, \(r\)01234\(\geqslant 5\)
Observed frequency, \(f\)18131298060
rf013242732096
\(\mathrm { r } ^ { 2 } \mathrm { f }\)01348811280270
Expected frequency12.11419.38215.5068.2703.3081.42160
Contribution to test statistic2.8602.1010.7931.2326.99
  1. Calculate the mean of the observed number of calls received.
  2. Calculate the variance of the observed number of calls received.
  3. Comment on what your answers to parts (a) and (b) suggest about the proposed model.
  4. Explain why it is necessary to combine some cells in the table.
  5. Show how the values 15.506 and 0.793 in the table were obtained.
  6. Carry out the test, at the \(5 \%\) significance level. In the light of the result of the test, the team consider that a different model is appropriate. They propose the following improved model: $$P ( R = r ) = \begin{cases} \frac { 1 } { 60 } ( a + ( 2 - r ) b ) & r = 0,1,2,3,4 \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) and \(b\) are integers.
  7. Use at least three of the observed frequencies to suggest appropriate values for \(a\) and \(b\). You should consider more than one possible pair of values, and explain which pair of values you consider best. (Do not carry out a goodness-of-fit test.)

Question 8:
AnswerMarks Guidance
8(a) 1.6
[1]1.1
(b)1.94 B1
[1]1.1
(c)These are fairly but not very close, so the Poisson
model may be good but there is some doubt.B1ft
[1]3.5a Nuanced conclusion, not just yes/no, based on whether mean and variance
are close, not just 1.6  1.94. FT on their s but must be nuanced
AnswerMarks Guidance
(d)The expected frequencies for 4 and  5 are both < 5 B1
[1]2.4 State “two cells have expected frequencies < 5”, or specify at least one cell
with expected frequency < 5. Not general statement, nor “some are ...”
AnswerMarks
(e)Po(1.6)
1 2 .6
P(R = 2) = e −1.6
2 !
Expected frequency is 60  0.25843 = 15.506 AG
AnswerMarks
(15.506 – 12)2/15.506 = 0.793 AGB1
M1
A1
B1
AnswerMarks
[4]3.3
1.1
3.4
AnswerMarks
1.1Stated, or implied by formula
Correct formula used, or at least 0.25843 or 15.5056… seen,
SC: 60  0.258 or 60  0.2584 without working for prob: M1A0
Correctly obtain 15.506, e.g. 15.5056 seen AG so full working necessary
Independent, but need to see formula used AG
AnswerMarks
(f)H : Population of number of calls has a Poisson
0
distribution. H : it doesn’t
1
 = 2
CV is 5.991
(Their) 6.99 > 5.991 so reject H (or Accept H )
0 1
There is significant evidence that the data is not
AnswerMarks
well modelled by a Poisson distributionB1
M1
A1
M1ft
A1ft
AnswerMarks
[5]1.1
1.1
1.1
1.1
AnswerMarks
2.2bNeeds general Poisson and not Po(1.6). Not “evidence that ...”, not “data
follows Poisson distribution”, allow “data fits Poisson distribution”
Allow  = 3 for M1
5.991, allow either 5.991 or 7.815 if hypotheses mention 1.6
Allow if 6.99 wrong or not explicit; FT on 7.815 but no other CVs.
Contextualised, not over-assertive, needs 5.99(1) seen. If 7.815 used can
get A1 with conclusion reversed: insufficient evidence that it doesn’t fit.
Needs hypotheses right way round for A1 but can be poorly stated
AnswerMarks
(g)a + 2b = 18, a + b = 13, a = 12, a – b = 9, a – 2b = 8
e.g. (a, b) = (12, 3) [→ 18, 15, 12, 9, 6];
(11, 3) [→ 18, 13, 12, 9, 4], and (12, 3) is better
E.g. a must be 12 and b = 3 gives a close
AnswerMarks
approximation to the frequenciesM1
B1
A1
AnswerMarks
[3]3.1b
3.4
AnswerMarks
3.5cConsider at least three of these (e.g. a = 12 and two other equations),
allow from 60 (e.g. a + 2b = 1080), allow sum of probs = 1 for one eqn
Consider at least two different possibilities and select one, allow from 60
(or, if equations used are a + 2b = 18, a = 12, a – 2b = 8, check third eqn)
a = 12 only, b = 1 or 2 or 3, with any relevant reason.
SC: if M0B0, give SC B1 for (12, 3) or (12, 2)
AnswerMarks Guidance
(f)A H : Number of calls has a Poisson distribution. H : it doesn’t 
0 1
 = 3 
CV is 7.815 X
6.99 < 7.815 so accept H FT 
0
There is significant evidence that the data has a Poisson distribution X
AnswerMarks
(could get this last A1 if conclusion was “insufficient evidence that it doesn’t have a Poisson distribution”)B1
M1
A0
M1ft
A0
[3/5]
AnswerMarks
BH : Number of calls fits Po(1.6). H : it doesn’t X
0 1
 = 3 
CV is 7.815 FT 
6.99 < 7.815 so do not reject H FT 
0
AnswerMarks
There is insufficient evidence that the data does not have a Poisson distribution  BOD (ignoring the 1.6)B0
M1
A1
M1ft
A1ft
[4/5]
AnswerMarks
CH : Number of calls fits Po(1.6). H : it doesn’t X
0 1
 = 2 
CV is 5.991  (allow this even though it’s not consistent with their hypotheses)
6.99 > 5.991 so accept H 
1
AnswerMarks
There is significant evidence that the data does not fit Po(1.6)  BOD (again ignoring the 1.6)B0
M1
A1
M1ft
A1
[4/5]
AnswerMarks
DH : Data doesn’t fit Poisson distribution. H : it does X
0 1
 = 2 
CV is 5.99 
6.99 > 5.99 so reject H 
0
There is insufficient evidence that the data does not have a Poisson distribution X
AnswerMarks
(can’t get final A1 if hypotheses wrong way round)B0
M1
A1
M1ft
A0
[3/5]
PMT
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Question 8:
8 | (a) | 1.6 | B1
[1] | 1.1
(b) | 1.94 | B1
[1] | 1.1
(c) | These are fairly but not very close, so the Poisson
model may be good but there is some doubt. | B1ft
[1] | 3.5a | Nuanced conclusion, not just yes/no, based on whether mean and variance
are close, not just 1.6  1.94. FT on their s but must be nuanced
(d) | The expected frequencies for 4 and  5 are both < 5 | B1
[1] | 2.4 | State “two cells have expected frequencies < 5”, or specify at least one cell
with expected frequency < 5. Not general statement, nor “some are ...”
(e) | Po(1.6)
1 2 .6
P(R = 2) = e −1.6
2 !
Expected frequency is 60  0.25843 = 15.506 AG
(15.506 – 12)2/15.506 = 0.793 AG | B1
M1
A1
B1
[4] | 3.3
1.1
3.4
1.1 | Stated, or implied by formula
Correct formula used, or at least 0.25843 or 15.5056… seen,
SC: 60  0.258 or 60  0.2584 without working for prob: M1A0
Correctly obtain 15.506, e.g. 15.5056 seen AG so full working necessary
Independent, but need to see formula used AG
(f) | H : Population of number of calls has a Poisson
0
distribution. H : it doesn’t
1
 = 2
CV is 5.991
(Their) 6.99 > 5.991 so reject H (or Accept H )
0 1
There is significant evidence that the data is not
well modelled by a Poisson distribution | B1
M1
A1
M1ft
A1ft
[5] | 1.1
1.1
1.1
1.1
2.2b | Needs general Poisson and not Po(1.6). Not “evidence that ...”, not “data
follows Poisson distribution”, allow “data fits Poisson distribution”
Allow  = 3 for M1
5.991, allow either 5.991 or 7.815 if hypotheses mention 1.6
Allow if 6.99 wrong or not explicit; FT on 7.815 but no other CVs.
Contextualised, not over-assertive, needs 5.99(1) seen. If 7.815 used can
get A1 with conclusion reversed: insufficient evidence that it doesn’t fit.
Needs hypotheses right way round for A1 but can be poorly stated
(g) | a + 2b = 18, a + b = 13, a = 12, a – b = 9, a – 2b = 8
e.g. (a, b) = (12, 3) [→ 18, 15, 12, 9, 6];
(11, 3) [→ 18, 13, 12, 9, 4], and (12, 3) is better
E.g. a must be 12 and b = 3 gives a close
approximation to the frequencies | M1
B1
A1
[3] | 3.1b
3.4
3.5c | Consider at least three of these (e.g. a = 12 and two other equations),
allow from 60 (e.g. a + 2b = 1080), allow sum of probs = 1 for one eqn
Consider at least two different possibilities and select one, allow from 60
(or, if equations used are a + 2b = 18, a = 12, a – 2b = 8, check third eqn)
a = 12 only, b = 1 or 2 or 3, with any relevant reason.
SC: if M0B0, give SC B1 for (12, 3) or (12, 2)
(f) | A | H : Number of calls has a Poisson distribution. H : it doesn’t 
0 1
 = 3 
CV is 7.815 X
6.99 < 7.815 so accept H FT 
0
There is significant evidence that the data has a Poisson distribution X
(could get this last A1 if conclusion was “insufficient evidence that it doesn’t have a Poisson distribution”) | B1
M1
A0
M1ft
A0
[3/5]
B | H : Number of calls fits Po(1.6). H : it doesn’t X
0 1
 = 3 
CV is 7.815 FT 
6.99 < 7.815 so do not reject H FT 
0
There is insufficient evidence that the data does not have a Poisson distribution  BOD (ignoring the 1.6) | B0
M1
A1
M1ft
A1ft
[4/5]
C | H : Number of calls fits Po(1.6). H : it doesn’t X
0 1
 = 2 
CV is 5.991  (allow this even though it’s not consistent with their hypotheses)
6.99 > 5.991 so accept H 
1
There is significant evidence that the data does not fit Po(1.6)  BOD (again ignoring the 1.6) | B0
M1
A1
M1ft
A1
[4/5]
D | H : Data doesn’t fit Poisson distribution. H : it does X
0 1
 = 2 
CV is 5.99 
6.99 > 5.99 so reject H 
0
There is insufficient evidence that the data does not have a Poisson distribution X
(can’t get final A1 if hypotheses wrong way round) | B0
M1
A1
M1ft
A0
[3/5]
PMT
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If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
8 A team of researchers have reason to believe that the number of calls received in randomly chosen 10-minute intervals to a call centre can be well modelled by a Poisson distribution. To test this belief the researchers record the number of telephone calls received in 60 randomly chosen 10-minute intervals. The results, together with relevant calculations, are shown in the following table.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
 &  &  &  &  &  &  & Total \\
\hline
Number of calls, $r$ & 0 & 1 & 2 & 3 & 4 & $\geqslant 5$ &  \\
\hline
Observed frequency, $f$ & 18 & 13 & 12 & 9 & 8 & 0 & 60 \\
\hline
rf & 0 & 13 & 24 & 27 & 32 & 0 & 96 \\
\hline
$\mathrm { r } ^ { 2 } \mathrm { f }$ & 0 & 13 & 48 & 81 & 128 & 0 & 270 \\
\hline
Expected frequency & 12.114 & 19.382 & 15.506 & 8.270 & 3.308 & 1.421 & 60 \\
\hline
Contribution to test statistic & 2.860 & 2.101 & 0.793 & \multicolumn{3}{|c|}{1.232} & 6.99 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Calculate the mean of the observed number of calls received.
\item Calculate the variance of the observed number of calls received.
\item Comment on what your answers to parts (a) and (b) suggest about the proposed model.
\item Explain why it is necessary to combine some cells in the table.
\item Show how the values 15.506 and 0.793 in the table were obtained.
\item Carry out the test, at the $5 \%$ significance level.

In the light of the result of the test, the team consider that a different model is appropriate. They propose the following improved model:

$$P ( R = r ) = \begin{cases} \frac { 1 } { 60 } ( a + ( 2 - r ) b ) & r = 0,1,2,3,4 \\ 0 & \text { otherwise } \end{cases}$$

where $a$ and $b$ are integers.
\item Use at least three of the observed frequencies to suggest appropriate values for $a$ and $b$. You should consider more than one possible pair of values, and explain which pair of values you consider best. (Do not carry out a goodness-of-fit test.)
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2023 Q8 [16]}}