| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from summary statistics |
| Difficulty | Moderate -0.5 This is a straightforward S1 regression question requiring standard formula application (Sxp, PMCC, regression line) with clear summary statistics provided. Part (e) adds a simple percentage calculation but no conceptual challenge. Slightly easier than average due to being entirely procedural with no interpretation complexity or problem-solving required. |
| Spec | 2.02g Calculate mean and standard deviation2.02h Recognize outliers2.05f Pearson correlation coefficient2.05g Hypothesis test using Pearson's r5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09d Linear coding: effect on regression |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xp} = 2347 - \frac{93 \times 273}{12}\) or \(2347 - \frac{25389}{12}\) \([= 231.25]\) | B1cso | Correct expression; don't need = 231.25 |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{pp} = 6602.72 - \frac{273^2}{12} = [391.97]\) | M1 | Attempt at correct expression for \(S_{pp}\); allow one transcription error; may be seen in part (a) |
| \(r = \frac{231.25}{\sqrt{148.25 \times 391.97}}\) | M1 | Correct expression for \(r\), ft their \(S_{pp}\) |
| \(= 0.959307...\) awrt 0.959 | A1 | For awrt 0.959 |
| Answer | Marks | Guidance |
|---|---|---|
| \(b = \frac{S_{xp}}{S_{xx}} = \frac{231.25}{148.25} [= 1.559865...]\) | M1 | Correct expression for \(b\); may be implied by awrt 1.56 |
| \(a = \frac{273}{12} - \text{"1.56"} \times \frac{93}{12}\) or \(22.75 - \text{"1.56"} \times 7.75\) \([= 10.66...]\) | M1 | Correct expression for \(a\) ft their \(b\); may be implied by awrt 10.7 |
| \(b =\) awrt 1.6 or \(a =\) awrt 11 | A1 | For \(b =\) awrt 1.6 or \(a =\) awrt 11 |
| \(p = 10.7 + 1.56x\) | A1 | Correct equation in \(p\) and \(x\) with \(b =\) awrt 1.56 and \(a =\) awrt 10.7 |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. each extra employee costs the company (on average) \(\\)156\( a year in paper | B1 | Suitable contextual comment mentioning value of \)b\(; allow multiples e.g. every extra 100 employees costs "\\)15600"; condone missing \\(; do not allow "\\)1.56" for 1 person unless indicates in 100s |
| Answer | Marks | Guidance |
|---|---|---|
| New \(p = 0.8 \times \text{"10.66..."} + \frac{\text{"1.559..."}}{2} \times \frac{93}{12}\) \([= 14.573...]\) | M1 | Correct expression for average value of \(p\) using new model [ft their \(a\) and \(b\)] |
| compared with \(\bar{p} = 22.75\), percentage saving is \(\frac{22.75 - 14.573...}{22.75} \times 100\) | M1 | Correct percentage saving calculation using 22.75 |
| \(= 35.94...\) awrt 36[%] | A1 | For awrt 36 |
# Question 2:
## Part (a)
| $S_{xp} = 2347 - \frac{93 \times 273}{12}$ or $2347 - \frac{25389}{12}$ $[= 231.25]$ | B1cso | Correct expression; don't need = 231.25 |
## Part (b)
| $S_{pp} = 6602.72 - \frac{273^2}{12} = [391.97]$ | M1 | Attempt at correct expression for $S_{pp}$; allow one transcription error; may be seen in part (a) |
| $r = \frac{231.25}{\sqrt{148.25 \times 391.97}}$ | M1 | Correct expression for $r$, ft their $S_{pp}$ |
| $= 0.959307...$ awrt **0.959** | A1 | For awrt 0.959 |
## Part (c)
| $b = \frac{S_{xp}}{S_{xx}} = \frac{231.25}{148.25} [= 1.559865...]$ | M1 | Correct expression for $b$; may be implied by awrt 1.56 |
| $a = \frac{273}{12} - \text{"1.56"} \times \frac{93}{12}$ or $22.75 - \text{"1.56"} \times 7.75$ $[= 10.66...]$ | M1 | Correct expression for $a$ ft their $b$; may be implied by awrt 10.7 |
| $b =$ awrt 1.6 **or** $a =$ awrt 11 | A1 | For $b =$ awrt 1.6 or $a =$ awrt 11 |
| $p = 10.7 + 1.56x$ | A1 | Correct equation in $p$ and $x$ with $b =$ awrt 1.56 and $a =$ awrt 10.7 |
## Part (d)
| e.g. each extra **employee** costs the company (on average) $\$156$ a year in **paper** | B1 | Suitable contextual comment mentioning value of $b$; allow multiples e.g. every extra 100 employees costs "\$15600"; condone missing \$; do not allow "\$1.56" for 1 person unless indicates in 100s |
## Part (e)
| New $p = 0.8 \times \text{"10.66..."} + \frac{\text{"1.559..."}}{2} \times \frac{93}{12}$ $[= 14.573...]$ | M1 | Correct expression for average value of $p$ using new model [ft their $a$ and $b$] |
| compared with $\bar{p} = 22.75$, percentage saving is $\frac{22.75 - 14.573...}{22.75} \times 100$ | M1 | Correct percentage saving calculation using 22.75 |
| $= 35.94...$ awrt **36[%]** | A1 | For awrt 36 |
---2. A large company is analysing how much money it spends on paper in its offices each year. The number of employees in the office, $x$, and the amount spent on paper in a year, $p$ (\$ hundreds), in each of 12 randomly selected offices were recorded.
The results are summarised in the following statistics.
$$\sum x = 93 \quad \mathrm {~S} _ { x x } = 148.25 \quad \sum p = 273 \quad \sum p ^ { 2 } = 6602.72 \quad \sum x p = 2347$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { S } _ { x p } = 231.25$
\item Find the product moment correlation coefficient for these data.
\item Find the equation of the regression line of $p$ on $x$ in the form $p = a + b x$
\item Give an interpretation of the gradient of your regression line.
The director of the company wants to reduce the amount spent on paper each year.
He wants each office to aim for a model of the form $p = \frac { 4 } { 5 } a + \frac { 1 } { 2 } b x$, where $a$ and $b$ are the values found in part (c).
Using the data for the 93 employees from the 12 offices,
\item estimate the percentage saving in the amount spent on paper each year by the company using the director's model.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2021 Q2 [12]}}