| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Moderate -0.8 This is a standard S1 linear regression question requiring routine application of given formulas for correlation coefficient and regression line, with straightforward interpretation. All summary statistics are provided, eliminating calculation burden. The multi-part structure is typical but each part follows textbook procedures without requiring problem-solving insight. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| \(x\) | 2.0 | 2.4 | 3.0 | 3.1 | 3.4 | 3.7 | 3.8 | 3.9 | 4.0 | 4.5 | 4.6 | 4.9 |
| \(y\) | 15.2 | 22.0 | 25.2 | 33.0 | 33.1 | 34.2 | 51.0 | 42.3 | 45.0 | 50.7 | 61.0 | 59.2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{1837.78 - \frac{43.3 \times 471.9}{12}}{\sqrt{\left(164.69 - \frac{43.3^2}{12}\right)\left(20915.75 - \frac{471.9^2}{12}\right)}}\) | M1 | For correct formula or calculator use |
| A1 | For correct value |
| Answer | Marks | Guidance |
|---|---|---|
| The value is close to +1, and the points in the diagram lie (fairly) close to a straight line with positive gradient | B1 | For relating the value to 1 |
| B1 | For a reasonable comment about linearity |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1837.78 - \frac{43.3 \times 471.9}{12}}{164.69 - \frac{43.3^2}{12}} = 15.9789\) | M1 | For correct formula or calculator use |
| A1 | For correct value for the regression coeff | |
| \(y - \frac{471.9}{12} = 15.9789\left(x - \frac{43.3}{12}\right)\) | M1 | For correct form of equn (may be implied) |
| \(y = 16.0x - 18.3\) | A1 | For correct (simplified) equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 16.0 \times 4.2 - 18.3\) | M1 | For substitution into equation from (ii) |
| A1√ | For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| High value of pmcc suggests fairly reliable | B1 | For any one reasonable comment |
| Answer | Marks | Guidance |
|---|---|---|
| As extrapolation is involved, the prediction would be (very) unreliable | M1 | For identifying extrapolation |
| A1 | For correct conclusion |
**Part (i)**
$r = \frac{1837.78 - \frac{43.3 \times 471.9}{12}}{\sqrt{\left(164.69 - \frac{43.3^2}{12}\right)\left(20915.75 - \frac{471.9^2}{12}\right)}}$ | M1 | For correct formula or calculator use
| A1 | For correct value
$= 0.956$
The value is close to +1, and the points in the diagram lie (fairly) close to a straight line with positive gradient | B1 | For relating the value to 1
| B1 | For a reasonable comment about linearity
**Part (ii)**
Gradient of regression line is
$\frac{1837.78 - \frac{43.3 \times 471.9}{12}}{164.69 - \frac{43.3^2}{12}} = 15.9789$ | M1 | For correct formula or calculator use
| A1 | For correct value for the regression coeff
$y - \frac{471.9}{12} = 15.9789\left(x - \frac{43.3}{12}\right)$ | M1 | For correct form of equn (may be implied)
$y = 16.0x - 18.3$ | A1 | For correct (simplified) equation
**Part (iii)**
$y = 16.0 \times 4.2 - 18.3$ | M1 | For substitution into equation from (ii)
| A1√ | For correct answer
Current is 48.8 cm s$^{-1}$
Comment could include:
Diagram indicates some uncertainty
High value of pmcc suggests fairly reliable | B1 | For any one reasonable comment
**Part (iv)**
As extrapolation is involved, the prediction would be (very) unreliable | M1 | For identifying extrapolation
| A1 | For correct conclusion
---8 An experiment was conducted to see whether there was any relationship between the maximum tidal current, $y \mathrm {~cm} \mathrm {~s} ^ { - 1 }$, and the tidal range, $x$ metres, at a particular marine location. [The tidal range is the difference between the height of high tide and the height of low tide.] Readings were taken over a period of 12 days, and the results are shown in the following table.
\begin{center}
\begin{tabular}{ | c | c c c c c c c c c c c c | }
\hline
$x$ & 2.0 & 2.4 & 3.0 & 3.1 & 3.4 & 3.7 & 3.8 & 3.9 & 4.0 & 4.5 & 4.6 & 4.9 \\
\hline
$y$ & 15.2 & 22.0 & 25.2 & 33.0 & 33.1 & 34.2 & 51.0 & 42.3 & 45.0 & 50.7 & 61.0 & 59.2 \\
\hline
\end{tabular}
\end{center}
$$\left[ \Sigma x = 43.3 , \Sigma y = 471.9 , \Sigma x ^ { 2 } = 164.69 , \Sigma y ^ { 2 } = 20915.75 , \Sigma x y = 1837.78 . \right]$$
The scatter diagram below illustrates the data.\\
\includegraphics[max width=\textwidth, alt={}, center]{2fb25fc5-0445-44fa-a23e-647d14b1a376-4_462_793_1464_644}\\
(i) Calculate the product moment correlation coefficient for the data, and comment briefly on your answer with reference to the appearance of the scatter diagram.\\
(ii) Calculate the equation of the regression line of maximum tidal current on tidal range.\\
(iii) Estimate the maximum tidal current on a day when the tidal range is 4.2 m , and comment briefly on how reliable you consider your estimate is likely to be.\\
(iv) It is suggested that the equation found in part (ii) could be used to predict the maximum tidal current on a day when the tidal range is 15 m . Comment briefly on the validity of this suggestion.
\hfill \mbox{\textit{OCR S1 Q8 [13]}}