| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate from raw data |
| Difficulty | Moderate -0.8 This is a straightforward application of standard S1 formulas for regression statistics. Students need to recall and apply memorized formulas (S_ww, S_wp, PMCC, regression line) with no conceptual challenges or problem-solving required—purely computational with given summary statistics. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09c Calculate regression line |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(S_{nw} = 41252 - \frac{640^2}{10} = 292\) | M1A1 |
| \(S_{wp} = 27557.8 - \frac{640 \times 431}{10} = -26.2\) | A1 | 2nd A1 for both \(S_{nw} = 292\) and \(S_{wp} = -26.2\) |
| (b) | \(r = \frac{-26.2}{\sqrt{292 \times 2.72}} = -0.9297\) | M1 A1 |
| awrt −0.930 | (Answer only: awrt −0.930 scores 2/2 but answer only −0.93 is M1A0) | |
| (c) | As weight increases the percentage of oil content decreases o.e. | B1 |
| (d) | \(b = \frac{-26.2}{292} = -0.0897...\) | M1 A1 |
| \(a = \frac{431}{10} - \left(\frac{-26.2}{292}\right) \times \left(\frac{640}{10}\right) = 48.842...\) | M1 | 2nd M1 for correct method for \(a\) ft their value of \(b\) (Allow \(a = 43.1 + b \times 64\)) |
| \(p = 48.8 - 0.0897w\) | A1 | 2nd A1 for correct equation for \(p\) and \(w\) with \(a =\) awrt 48.8 and \(b =\) awrt −0.0897. No fractions. Equation in x and y is A0 |
| (e) | \(p = 48.8 - 0.0897 \times 60 = 43.4/43.5\) | M1 A1 |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $S_{nw} = 41252 - \frac{640^2}{10} = 292$ | M1A1 | M1 for correct expression for $S_{nw}$ (may be implied). 1st A1 for either $S_{nw} = 292$ or $S_{wp} = -26.2$ |
| | $S_{wp} = 27557.8 - \frac{640 \times 431}{10} = -26.2$ | A1 | 2nd A1 for both $S_{nw} = 292$ and $S_{wp} = -26.2$ |
| (b) | $r = \frac{-26.2}{\sqrt{292 \times 2.72}} = -0.9297$ | M1 A1 | M1 for correct expression (Allow ft of their $S_{nw}$ or $S_{wp}$ provided $S_{nw} \neq 41252$ and $S_{wp}\neq 27557.8$). Condone missing "−". A1 for awrt −0.930 (Condone −0.93 for M1A1 if correct expression is seen) |
| | awrt −0.930 | | (Answer only: awrt −0.930 scores 2/2 but answer only −0.93 is M1A0) |
| (c) | As weight increases the percentage of oil content decreases o.e. | B1 | B1 for correct contextual description of negative correlation which must include weight and oil (but w increases as p decreases is not sufficient) |
| (d) | $b = \frac{-26.2}{292} = -0.0897...$ | M1 A1 | 1st M1 for correct expression for $b$ (Allow ft). 1st A1 for awrt −0.09 |
| | $a = \frac{431}{10} - \left(\frac{-26.2}{292}\right) \times \left(\frac{640}{10}\right) = 48.842...$ | M1 | 2nd M1 for correct method for $a$ ft their value of $b$ (Allow $a = 43.1 + b \times 64$) |
| | $p = 48.8 - 0.0897w$ | A1 | 2nd A1 for correct equation for $p$ and $w$ with $a =$ awrt 48.8 and $b =$ awrt −0.0897. No fractions. Equation in x and y is A0 |
| (e) | $p = 48.8 - 0.0897 \times 60 = 43.4/43.5$ | M1 A1 | M1 substituting $w = 60$ into their equation. A1 awrt 43.4 or 43.5 (Answer only scores 2/2) |
**Total: 12 marks**
---\begin{enumerate}
\item The percentage oil content, $p$, and the weight, $w$ milligrams, of each of 10 randomly selected sunflower seeds were recorded. These data are summarised below.
\end{enumerate}
$$\sum w ^ { 2 } = 41252 \quad \sum w p = 27557.8 \quad \sum w = 640 \quad \sum p = 431 \quad \mathrm {~S} _ { p p } = 2.72$$
(a) Find the value of $\mathrm { S } _ { w w }$ and the value of $\mathrm { S } _ { w p }$\\
(b) Calculate the product moment correlation coefficient between $p$ and $w$\\
(c) Give an interpretation of your product moment correlation coefficient.
The equation of the regression line of $p$ on $w$ is given in the form $p = a + b w$\\
(d) Find the equation of the regression line of $p$ on $w$\\
(e) Hence estimate the percentage oil content of a sunflower seed which weighs 60 milligrams.\\
\hfill \mbox{\textit{Edexcel S1 2016 Q1 [12]}}