OCR MEI Further Statistics A AS 2022 June — Question 6 10 marks

Exam BoardOCR MEI
ModuleFurther Statistics A AS (Further Statistics A AS)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeNon-geometric distribution identification
DifficultyModerate -0.8 This is a straightforward regression question requiring standard formula application for the regression line equation, simple substitution for predictions, and recall of the property that both regression lines pass through (x̄, ȳ). All calculations are routine with no conceptual challenges or novel problem-solving required, making it easier than average for A-level.
Spec5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context

6 Tom has read in a newspaper that you can tell the air temperature by counting how often a cricket chirps in a period of 20 seconds. (A cricket is a type of insect.) He wants to know exactly how the temperature can be predicted. On 8 randomly selected days, when Tom can hear crickets chirping, he records the number of chirps, \(x\), made by a cricket in a 20-second interval, and also the temperature, \(y ^ { \circ } \mathrm { C }\), at that time. The data are summarised as follows. \(n = 8 \quad \sum x = 268 \quad \sum y = 141.9 \quad \sum x ^ { 2 } = 9618 \quad \sum y ^ { 2 } = 2630.55 \quad \sum \mathrm { xy } = 5009.1\) These data are illustrated below. \includegraphics[max width=\textwidth, alt={}, center]{8f1e0c68-a334-4657-823e-386ab0994c02-5_661_1035_699_242}
  1. Determine the equation of the regression line of \(y\) on \(x\). Give your answer in the form \(\mathrm { y } = \mathrm { ax } + \mathrm { b }\), giving the values of \(a\) and \(b\) correct to \(\mathbf { 3 }\) significant figures.
  2. Use the equation of the regression line to predict the temperature for the following values of \(x\).

Question 6:
AnswerMarks Guidance
6(a) 1
𝑆 = 5009.1− ×268×141.9 = 255.45
𝑥𝑦
8
S = 9 6 1 8 − 18  2 6 8 2 = 6 4 0
x x
255.45
b= =0.3991
640
141.9 268
a= −0.3991  =4.3662
8 8
AnswerMarks
y = 0.399x + 4.37M1
A1
M1
A1
AnswerMarks
[4]1.1a
1.1
1.1
AnswerMarks
1.1For correct substitution into S or S
x y x x
For b
For method for a. FT their b and means.
cwo Allow final mark only if equation explicitly stated.
Awrt 0.399 and 4.37 in final answer.
AnswerMarks Guidance
6(b) Prediction for x = 35 is y = 18.3
Prediction for x = 10 is y = 8.4B1
B1
AnswerMarks
[2]1.1
1.1Allow 18.34. FT their final equation if correct substitution
seen and given as a decimal correct to no more than 2 d.p.
Allow 8.36. FT their final equation if correct substitution
seen and given as a decimal correct to no more than 2 d.p.
SCB1 if B0B0 awarded due to over-specification
AnswerMarks Guidance
6(c) e.g. The prediction for x = 35 is likely to be fairly
reliable since this is interpolation
and the points appear to follow a linear trend.
The prediction for x = 10 is less likely to be reliable
since this is extrapolation, well below any data
AnswerMarks
values.B1
E1
B1
AnswerMarks
[3]2.2b
2.2b
AnswerMarks
3.5aMust have some element of doubt to get 2 marks for
prediction for x = 35
AnswerMarks Guidance
6(d) Coordinates are (33.5, 17.7(375))
[1]1.1
Question 6:
6 | (a) | 1
𝑆 = 5009.1− ×268×141.9 = 255.45
𝑥𝑦
8
S = 9 6 1 8 − 18  2 6 8 2 = 6 4 0
x x
255.45
b= =0.3991
640
141.9 268
a= −0.3991  =4.3662
8 8
y = 0.399x + 4.37 | M1
A1
M1
A1
[4] | 1.1a
1.1
1.1
1.1 | For correct substitution into S or S
x y x x
For b
For method for a. FT their b and means.
cwo Allow final mark only if equation explicitly stated.
Awrt 0.399 and 4.37 in final answer.
6 | (b) | Prediction for x = 35 is y = 18.3
Prediction for x = 10 is y = 8.4 | B1
B1
[2] | 1.1
1.1 | Allow 18.34. FT their final equation if correct substitution
seen and given as a decimal correct to no more than 2 d.p.
Allow 8.36. FT their final equation if correct substitution
seen and given as a decimal correct to no more than 2 d.p.
SCB1 if B0B0 awarded due to over-specification
6 | (c) | e.g. The prediction for x = 35 is likely to be fairly
reliable since this is interpolation
and the points appear to follow a linear trend.
The prediction for x = 10 is less likely to be reliable
since this is extrapolation, well below any data
values. | B1
E1
B1
[3] | 2.2b
2.2b
3.5a | Must have some element of doubt to get 2 marks for
prediction for x = 35
6 | (d) | Coordinates are (33.5, 17.7(375)) | B1
[1] | 1.1
6 Tom has read in a newspaper that you can tell the air temperature by counting how often a cricket chirps in a period of 20 seconds. (A cricket is a type of insect.) He wants to know exactly how the temperature can be predicted. On 8 randomly selected days, when Tom can hear crickets chirping, he records the number of chirps, $x$, made by a cricket in a 20-second interval, and also the temperature, $y ^ { \circ } \mathrm { C }$, at that time. The data are summarised as follows.\\
$n = 8 \quad \sum x = 268 \quad \sum y = 141.9 \quad \sum x ^ { 2 } = 9618 \quad \sum y ^ { 2 } = 2630.55 \quad \sum \mathrm { xy } = 5009.1$\\
These data are illustrated below.\\
\includegraphics[max width=\textwidth, alt={}, center]{8f1e0c68-a334-4657-823e-386ab0994c02-5_661_1035_699_242}
\begin{enumerate}[label=(\alph*)]
\item Determine the equation of the regression line of $y$ on $x$. Give your answer in the form $\mathrm { y } = \mathrm { ax } + \mathrm { b }$, giving the values of $a$ and $b$ correct to $\mathbf { 3 }$ significant figures.
\item Use the equation of the regression line to predict the temperature for the following values of $x$.

\begin{itemize}
  \item 35
  \item 10
\item Comment on the reliability of your predictions in part (b).
\item State the coordinates of the point of intersection of the line whose equation you have calculated with the regression line of $x$ on $y$.
\end{itemize}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2022 Q6 [10]}}