| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate regression line then predict |
| Difficulty | Moderate -0.8 This is a routine S1 regression question requiring standard formula application (calculating Stt, correlation coefficient, regression line) with straightforward arithmetic. The only slight challenge is recognizing the validity issue in part (f) about extrapolation/different location, but this is a common exam trope. Overall, this is easier than average A-level maths as it's pure formula recall with no problem-solving or conceptual insight required. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(S_n = 1828 - \frac{(136)^2}{12} = 286.6......\) o.e. awrt 287 | M1A1 (2) | Correct expression for \(S_n\). Awrt 287 allows exact fractions e.g. \(\frac{860}{3}\) or \(286\frac{2}{3}\) |
| (b) \(r = \frac{S_n}{\sqrt{S_n S_{hh}}} = \frac{-236}{\sqrt{286.6...} \times 297}\) or \(\frac{-236}{\sqrt{85140}} := -0.8088...\) awrt \(-0.809\) | M1A1 (2) | For attempt at correct formula, values must be substituted. Allow \(\frac{-236}{\sqrt{287 \times 297}}\). Awrt \(-0.809\) (allow \(-0.808\) from a correct expression with 287 used) |
| (c) Temperature decreases as height increases. | B1ft (1) | For a comment in context. Must see "height" (or h) and "temperature" (or t) mentioned. Allow "as the temperature increases the height above sea level decreases" (o.e.). If \( |
| (d) \(b = \frac{S_{nh}}{S_{hh}} = \frac{-236}{297} (= -0.7946...)\) | M1 | Correct expression for \(b\). Allow \(11.3... -\) "their \(b \times 9.33...\)" [\(a = \frac{16706}{891}\) scores M1 but A0] |
| \(a = \bar{t} - b\bar{h} = 11.3... + 0.7946 \times 9.33... = 18.7497...\) | M1 | |
| \(t = 18.7 - 0.795h\) | A1 | \(t = (18.75\) or awrt \(18.7) - (\) awrt \(0.795)h\) [No fractions and no x, y] |
| (3) | ||
| (e) \(t = 18.7 - 0.795 \times 5 = 14.7\) | M1 A1 (2) | Substitute \(h = 5\) or 500 into their regression line. Answer in range [14.7, 14.8] (condone coming from y, x equation) |
| (f) Unreliable as the data is from France not South Africa | B1 (1) | Unreliable with a reason. [Use of 500 in (e) and stating "out of range" is B0]. Must mention France or (S) Africa and at least imply the other |
| Total 11 |
**(a)** $S_n = 1828 - \frac{(136)^2}{12} = 286.6......$ o.e. awrt 287 | M1A1 (2) | Correct expression for $S_n$. Awrt 287 allows exact fractions e.g. $\frac{860}{3}$ or $286\frac{2}{3}$
**(b)** $r = \frac{S_n}{\sqrt{S_n S_{hh}}} = \frac{-236}{\sqrt{286.6...} \times 297}$ or $\frac{-236}{\sqrt{85140}} := -0.8088...$ awrt $-0.809$ | M1A1 (2) | For attempt at correct formula, values must be substituted. Allow $\frac{-236}{\sqrt{287 \times 297}}$. Awrt $-0.809$ (allow $-0.808$ from a correct expression with 287 used)
**(c)** Temperature decreases as height increases. | B1ft (1) | For a comment in context. Must see "height" (or h) and "temperature" (or t) mentioned. Allow "as the temperature increases the height above sea level decreases" (o.e.). If $|r| > 1$ score B0 in (c). Saying "sea level increases" (o.e.) is B0
**(d)** $b = \frac{S_{nh}}{S_{hh}} = \frac{-236}{297} (= -0.7946...)$ | M1 | Correct expression for $b$. Allow $11.3... -$ "their $b \times 9.33...$" [$a = \frac{16706}{891}$ scores M1 but A0]
| $a = \bar{t} - b\bar{h} = 11.3... + 0.7946 \times 9.33... = 18.7497...$ | M1 |
| $t = 18.7 - 0.795h$ | A1 | $t = (18.75$ or awrt $18.7) - ($ awrt $0.795)h$ [No fractions and no x, y]
| | (3) |
**(e)** $t = 18.7 - 0.795 \times 5 = 14.7$ | M1 A1 (2) | Substitute $h = 5$ or 500 into their regression line. Answer in range [14.7, 14.8] (condone coming from y, x equation)
**(f)** Unreliable as the data is from France not South Africa | B1 (1) | Unreliable with a reason. [Use of 500 in (e) and stating "out of range" is B0]. Must mention France or (S) Africa and at least imply the other
| | **Total 11** |
---\begin{enumerate}
\item The heights above sea level ( $h$ hundred metres) and the temperatures ( $t ^ { \circ } \mathrm { C }$ ) at 12 randomly selected places in France, at 7 am on July 31st, were recorded.\\
The data are summarised as follows\\
(a) Find the value of $S _ { t t }$\\
(b) Calculate the product moment correlation coefficient for these data.\\
(c) Interpret the relationship between $t$ and $h$.\\
(d) Find an equation of the regression line of $t$ on $h$.
\end{enumerate}
At 7 am on July 31st Yinka is on holiday in South Africa. He uses the regression equation to estimate the temperature when the height above sea level is 500 m .\\
(e) Find the estimated temperature Yinka calculates.\\
(f) Comment on the validity of your answer in part (e).
$$\sum h = 112 \quad \sum t = 136 \quad \sum t ^ { 2 } = 1828 \quad S _ { h t } = - 236 \quad S _ { h h } = 297$$
(a) Find the value of $S$ (2)
\hfill \mbox{\textit{Edexcel S1 2018 Q1 [11]}}