OCR MEI Further Statistics A AS 2019 June — Question 6 13 marks

Exam BoardOCR MEI
ModuleFurther Statistics A AS (Further Statistics A AS)
Year2019
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear regression
TypeInterpret regression line parameters
DifficultyStandard +0.3 This is a straightforward linear regression question covering standard AS-level techniques: calculating regression line equation from summary statistics, making predictions, and computing residuals. Part (a) requires recognizing non-linearity from a graph, and part (e) involves commenting on extrapolation—both routine interpretations. The calculations are mechanical with no conceptual challenges beyond basic formula application, making this slightly easier than average.
Spec5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context
6 A meteorologist is investigating the relationship between altitude \(x\) metres and mean annual temperature \(y ^ { \circ } \mathrm { C }\) in an American state.
She selects 12 locations at various altitudes and then stations a remote monitoring device at each of them to measure the temperature over the course of a year. Fig. 6 illustrates the data which she obtains. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fd496303-10f1-450e-bbeb-421ab6f4de21-6_686_1477_486_292} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Explain why it would not be appropriate to carry out a hypothesis test for correlation based on the product moment correlation coefficient.
  2. Explain why altitude has been plotted on the horizontal axis in Fig. 6. Summary statistics for \(x\) and \(y\) are as follows. $$\sum x = 21200 \quad \sum y = 105.4 \quad \sum x ^ { 2 } = 39100000 \quad \sum y ^ { 2 } = 1004 \quad \sum x y = 176090$$
  3. Calculate the equation of the regression line of \(y\) on \(x\).
  4. Use the equation of the regression line to predict the values of the mean annual temperature at each of the following altitudes.
Question 6:
AnswerMarks Guidance
6(a) Because the test is only valid for random on
random
and the altitudes are not random variables but
AnswerMarks
controlledB1
B1
AnswerMarks
[2]3.5a
3.5b
AnswerMarks Guidance
(b)Because altitude is the independent variable B1
[1]2.4 Allow ‘controlled’, ‘selected’ for ‘independent’
(c)S 176090(21200105.4/12)
b xy 
S 39100000212002 /12
xx
= 0.00614 3 s.f.
x = 1766.666…, y = 8.783333…
hence least squares regression line is:
y – 8.7833… = 0.00614(x – 1766.666..)
AnswerMarks
y = 0.00614x + 19.637M1
A1
M1
A1
AnswerMarks
[4]1.1a
1.1
1.1
AnswerMarks
1.1For correct structure to find gradient (b) as written or
176090/12(1766.78.783)
equivalent e.g.
39100000/121766.72
For b correct to 3 s.f. or better
For equation of line with their b < 0
Values correct to 3 s.f. or better
AnswerMarks
(d)Prediction for 2000m is 7.35 C
Prediction for 3000m is 1.21 CB1
B1
AnswerMarks
[2]3.4
1.1Allow 7.36
Allow 1.22
FT their equation with b < 0
AnswerMarks
(e)Prediction for 2000m is more likely to be reliable
as interpolation
Prediction for 3000m is less likely to be reliable as
AnswerMarks
extrapolationB1
B1
AnswerMarks
[2]3.5a
3.5bIf ‘interpolation’ not stated then alternative wording must
be convincing - e.g. ‘within domain of validity’
If ‘extrapolation’ not stated then alternative wording must
be convincing. e.g. allow ‘outside the range of the data
(values)’
AnswerMarks
(f)Residual = 8.1  (0.00614×1600 + 19.637)
= 1.71 3 s.f.M1
A1
AnswerMarks
[2]1.1
1.1For difference between 8.1 and their prediction for 1600
For – 1.71 FT their b < 0
PPMMTT
Y412/01 Mark Scheme June 2019
Additional notes RE:Non-assertive conclusions relating to H in hypothesis tests
1
There is insufficient evidence to believe/suggest/indicate that there is an association between…
There is insufficient evidence to believe/suggest/indicate that… …are not independent.
There is sufficient evidence to believe/suggest/indicate that there is an association between…
There is sufficient evidence to believe/suggest/indicate that… …are not independent.
9
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
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 (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
Question 6:
6 | (a) | Because the test is only valid for random on
random
and the altitudes are not random variables but
controlled | B1
B1
[2] | 3.5a
3.5b
(b) | Because altitude is the independent variable | B1
[1] | 2.4 | Allow ‘controlled’, ‘selected’ for ‘independent’
(c) | S 176090(21200105.4/12)
b xy 
S 39100000212002 /12
xx
= 0.00614 3 s.f.
x = 1766.666…, y = 8.783333…
hence least squares regression line is:
y – 8.7833… = 0.00614(x – 1766.666..)
y = 0.00614x + 19.637 | M1
A1
M1
A1
[4] | 1.1a
1.1
1.1
1.1 | For correct structure to find gradient (b) as written or
176090/12(1766.78.783)
equivalent e.g.
39100000/121766.72
For b correct to 3 s.f. or better
For equation of line with their b < 0
Values correct to 3 s.f. or better
(d) | Prediction for 2000m is 7.35 C
Prediction for 3000m is 1.21 C | B1
B1
[2] | 3.4
1.1 | Allow 7.36
Allow 1.22
FT their equation with b < 0
(e) | Prediction for 2000m is more likely to be reliable
as interpolation
Prediction for 3000m is less likely to be reliable as
extrapolation | B1
B1
[2] | 3.5a
3.5b | If ‘interpolation’ not stated then alternative wording must
be convincing - e.g. ‘within domain of validity’
If ‘extrapolation’ not stated then alternative wording must
be convincing. e.g. allow ‘outside the range of the data
(values)’
(f) | Residual = 8.1  (0.00614×1600 + 19.637)
= 1.71 3 s.f. | M1
A1
[2] | 1.1
1.1 | For difference between 8.1 and their prediction for 1600
For – 1.71 FT their b < 0
PPMMTT
Y412/01 Mark Scheme June 2019
Additional notes RE:Non-assertive conclusions relating to H in hypothesis tests
1
There is insufficient evidence to believe/suggest/indicate that there is an association between…
There is insufficient evidence to believe/suggest/indicate that… …are not independent.
There is sufficient evidence to believe/suggest/indicate that there is an association between…
There is sufficient evidence to believe/suggest/indicate that… …are not independent.
9
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
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 (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2019
6 A meteorologist is investigating the relationship between altitude $x$ metres and mean annual temperature $y ^ { \circ } \mathrm { C }$ in an American state.\\
She selects 12 locations at various altitudes and then stations a remote monitoring device at each of them to measure the temperature over the course of a year. Fig. 6 illustrates the data which she obtains.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fd496303-10f1-450e-bbeb-421ab6f4de21-6_686_1477_486_292}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Explain why it would not be appropriate to carry out a hypothesis test for correlation based on the product moment correlation coefficient.
\item Explain why altitude has been plotted on the horizontal axis in Fig. 6.

Summary statistics for $x$ and $y$ are as follows.

$$\sum x = 21200 \quad \sum y = 105.4 \quad \sum x ^ { 2 } = 39100000 \quad \sum y ^ { 2 } = 1004 \quad \sum x y = 176090$$
\item Calculate the equation of the regression line of $y$ on $x$.
\item Use the equation of the regression line to predict the values of the mean annual temperature at each of the following altitudes.

\begin{itemize}
  \item 2000 metres
  \item 3000 metres
\item Comment on the reliability of your predictions in part (d).
\item Calculate the value of the residual for the data point ( $1600,8.1$ ).
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2019 Q6 [13]}}
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