| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Interpret regression line parameters |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on basic regression interpretation requiring only standard recall and simple calculations: substituting into a given equation, commenting on extrapolation vs interpolation, understanding which regression line to use, taking a square root of r², and making a basic comment on fit. All parts are routine textbook exercises with no problem-solving or novel insight required. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | A |
| x = 60, y = 19.0157 | B1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Accept 15.5 or better |
| Answer | Marks |
|---|---|
| B | x = 45 involves interpolation (and points are fairly |
| Answer | Marks |
|---|---|
| good estimate | E1 |
| Answer | Marks |
|---|---|
| [2] | 3.5a |
| Answer | Marks |
|---|---|
| (ii) | To estimate the mean value of x from a value of y, |
| Answer | Marks |
|---|---|
| then substitute y = 16 into this equation. | E1 |
| Answer | Marks |
|---|---|
| [2] | 3.5b |
| 3.5c | For indication that the use of the |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) | PMCC = √0.5748 = 0.758 | B1 |
| [1] | 1.1 | Accept 0.76, 0.7582 |
| (iv) | Because r2 = 0.5748 the fit is (moderately) good |
| Answer | Marks |
|---|---|
| line. | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.2a |
| 2.4 | Referring to 0.5748, or r2 is quite |
Question 6:
6 | (i) | A | x = 45, y = 15.5357
x = 60, y = 19.0157 | B1
B1
[2] | 1.1
1.1 | Accept 15.5 or better
Accept 19 or better
B | x = 45 involves interpolation (and points are fairly
close to line) so probably a reasonably good
estimate
x = 60 involves extrapolation so probably not a
good estimate | E1
E1
[2] | 3.5a
3.5b
(ii) | To estimate the mean value of x from a value of y,
you would need to calculate the equation of the
regression line of x on y,
then substitute y = 16 into this equation. | E1
E1
[2] | 3.5b
3.5c | For indication that the use of the
regression line of x on y is required.
(iii) | PMCC = √0.5748 = 0.758 | B1
[1] | 1.1 | Accept 0.76, 0.7582
(iv) | Because r2 = 0.5748 the fit is (moderately) good
and because the points lie (fairly) close to a straight
line. | E1
E1
[2] | 2.2a
2.4 | Referring to 0.5748, or r2 is quite
large, as supporting good fit
For reference to points being close to
a line as supporting a good fit.
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© OCR 20186 A researcher is investigating various bodily characteristics of frogs of various species. She collects data on length, $x \mathrm {~mm}$, and head width, $y \mathrm {~mm}$, of a random sample of 14 frogs of a particular species. A scatter diagram of the data is shown in Fig. 6, together with the equation of the regression line of $y$ on $x$ and also the value of $r ^ { 2 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e3ac0ba0-9692-4018-894e-2b04b07eaf32-6_949_1616_450_228}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item (A) Use the equation of the regression line to estimate the mean head width for frogs of each of the following lengths.
\begin{itemize}
\item 45 mm
\item 60 mm\\
(B) Comment briefly on each of the estimates in part (i)(A).
\item Explain how the mean length of frogs with head width 16 mm should be estimated.
\item Calculate the value of the product moment correlation coefficient.
\item In the light of the information in the scatter diagram, comment on the goodness of fit of the regression line.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2018 Q6 [9]}}