| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate from summary statistics |
| Difficulty | Moderate -0.3 This is a standard S1 linear regression question requiring routine application of formulas (Sxx, Syy, correlation coefficient) and interpretation. Part (d) requires algebraic manipulation but follows a structured approach. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(y = 6.066 + 0.136\times 80 = 16.946\) so annual rent is \(\\)16\,946\( | M1, A1 | M1 substituting \)x=80\(; A1 for awrt \\)16,900 or better, must have units |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(S_{yy} = 3434 - \frac{183^2}{10}\) or \(S_{xx} = 84818 - \frac{900^2}{10}\) | M1 | Correct expression for either |
| \(S_{yy} = \mathbf{85.1}\) | A1 | |
| \(S_{xx} = \mathbf{3818}\) | A1 | Accept 3820 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(S_{xy} = b \times S_{xx} = 0.136\times 3818 = 519.248\) | M1; A1 | M1 attempt to use gradient to find \(S_{xy}\); A1 for awrt 519 |
| \(r = \frac{0.136\times``3818"}{\sqrt{``3818"\times``85.1"}}\) | M1 | Correct expression using their values |
| \(= 0.9109448...\) awrt 0.911 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Since new \(x=90\) and new \(\bar{x}=90\), the term \((x-\bar{x})\) will be 0 | M1 | Stating/showing \(\bar{x}=90\) or that \((x-\bar{x})\) term \(=0\) |
| Therefore \(S_{xy}\) stays the same | A1 | Fully correct argument |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(S_{xx}\) will be the same so \(b\) will be the same | M1 | Correct statement about \(S_{xx}\) or \(b\) |
| New \(\bar{y} = \frac{183+15}{11} = 18\) (or \(a\) is reduced by 0.3) | M1 | Correct new \(\bar{y}\) |
| Equation is \(y = 5.766 + 0.136x\) | A1 | For \(y=5.766\) (or awrt 5.77 or 5.76) \(+0.136x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(x=300\) is outside the range \(300 \gg 90\) [\(300 \gg 90+3\sigma = 90+3\times18.63\approx146\)] | B1 | Must compare 300 vs 90 or 3000 vs 900 and conclude outside range |
| So not suitable (involves extrapolation) | (o.e.) | Not sufficient to just say "larger" |
# Question 5:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $y = 6.066 + 0.136\times 80 = 16.946$ so annual rent is **$\$16\,946$** | M1, A1 | M1 substituting $x=80$; A1 for awrt \$16,900 or better, must have units |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $S_{yy} = 3434 - \frac{183^2}{10}$ or $S_{xx} = 84818 - \frac{900^2}{10}$ | M1 | Correct expression for either |
| $S_{yy} = \mathbf{85.1}$ | A1 | |
| $S_{xx} = \mathbf{3818}$ | A1 | Accept 3820 |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| $S_{xy} = b \times S_{xx} = 0.136\times 3818 = 519.248$ | M1; A1 | M1 attempt to use gradient to find $S_{xy}$; A1 for awrt 519 |
| $r = \frac{0.136\times``3818"}{\sqrt{``3818"\times``85.1"}}$ | M1 | Correct expression using their values |
| $= 0.9109448...$ awrt **0.911** | A1 | |
## Part (d):
| Working | Marks | Notes |
|---------|-------|-------|
| Since new $x=90$ and new $\bar{x}=90$, the term $(x-\bar{x})$ will be 0 | M1 | Stating/showing $\bar{x}=90$ or that $(x-\bar{x})$ term $=0$ |
| Therefore $S_{xy}$ stays the same | A1 | Fully correct argument |
## Part (e):
| Working | Marks | Notes |
|---------|-------|-------|
| $S_{xx}$ will be the same so $b$ will be the same | M1 | Correct statement about $S_{xx}$ or $b$ |
| New $\bar{y} = \frac{183+15}{11} = 18$ (or $a$ is reduced by 0.3) | M1 | Correct new $\bar{y}$ |
| Equation is $y = 5.766 + 0.136x$ | A1 | For $y=5.766$ (or awrt 5.77 or 5.76) $+0.136x$ |
## Part (f):
| Working | Marks | Notes |
|---------|-------|-------|
| $x=300$ is outside the range $300 \gg 90$ [$300 \gg 90+3\sigma = 90+3\times18.63\approx146$] | B1 | Must compare 300 vs 90 or 3000 vs 900 and conclude outside range |
| So not suitable (involves extrapolation) | (o.e.) | Not sufficient to just say "larger" |
---\begin{enumerate}
\item A large company rents shops in different parts of the country. A random sample of 10 shops was taken and the floor area, $x$ in $10 \mathrm {~m} ^ { 2 }$, and the annual rent, $y$ in thousands of dollars, were recorded.\\
The data are summarised by the following statistics
\end{enumerate}
$$\sum x = 900 \quad \sum x ^ { 2 } = 84818 \quad \sum y = 183 \quad \sum y ^ { 2 } = 3434$$
and the regression line of $y$ on $x$ has equation $y = 6.066 + 0.136 x$\\
(a) Use the regression line to estimate the annual rent in dollars for a shop with a floor area of $800 \mathrm {~m} ^ { 2 }$\\
(b) Find $\mathrm { S } _ { y y }$ and $\mathrm { S } _ { x x }$\\
(c) Find the product moment correlation coefficient between $y$ and $x$.
An 11th shop is added to the sample. The floor area is $900 \mathrm {~m} ^ { 2 }$ and the annual rent is 15000 dollars.\\
(d) Use the formula $\mathrm { S } _ { x y } = \sum ( x - \bar { x } ) ( y - \bar { y } )$ to show that the value of $\mathrm { S } _ { x y }$ for the 11 shops will be the same as it was for the original 10 shops.\\
(e) Find the new equation of the regression line of $y$ on $x$ for the 11 shops.
The company is considering renting a larger shop with area of $3000 \mathrm {~m} ^ { 2 }$\\
(f) Comment on the suitability of using the new regression line to estimate the annual rent. Give a reason for your answer.
\hfill \mbox{\textit{Edexcel S1 2020 Q5 [15]}}