| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate PMCC from summary statistics |
| Difficulty | Moderate -0.8 This is a straightforward application of standard S1 formulas for PMCC and linear regression. Part (a) requires substituting into Sth and Stt formulas (routine calculation), parts (b) and (d) test interpretation (standard bookwork), part (c) applies the regression formula directly, and part (e) is a simple substitution. No problem-solving or novel insight required—purely procedural with given summary statistics. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08b Linear coding: effect on pmcc5.09c Calculate regression line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((S_{th}) = 31070 - \frac{61\times6370}{8}\) or \(31070-48571.25\); \((S_{tt}) = 693 - \frac{61^2}{8}\) or \(693-465.125\) | M1; M1 | For a correct expression for \(S_{th}\). For a correct expression for \(S_{tt}\). Allow 1 slip e.g. 6730 or \(61^2\) or 3721. Consistent use of \(n\neq8\) M0M1 |
| \((S_{th}) = -17501.25\) and \((S_{tt}) = 227.875\) | A1cso | For both answers correct and both Ms scored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \( | r | \) close to 1 or \(r\) is close to \(-1\) therefore it does support the linear model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left[r = \frac{S_{yx}}{\sqrt{S_{yy}\times S_{xx}}}\right]\) so \(r = \frac{S_{th}}{\sqrt{S_{tt}\times S_{hh}}}\) or \(r^2 = \frac{(S_{th})^2}{S_{tt}\times S_{hh}}\) or \(S_{hh} = \frac{(S_{th})^2}{r^2\times S_{tt}}\) or substitute 1 value | M1 | For the sight of the formula for \(r\) and an attempt to do something useful with it. In (c) condone \(x\) for \(h\) and \(y\) for \(t\) except in 4th A1 |
| e.g. \(\pm0.985 = \frac{\pm17501.25}{\sqrt{227.875\times S_{hh}}}\) or \(S_{hh} = \frac{(\pm17501.25)^2}{(\pm0.985)^2\times227.875}\) o.e. \((= 1\ 385\ 380.258)\) | A1 | For a correct numerical expression in \(S_{hh}\) or \(\sqrt{S_{hh}}\). Accept with 3sf values (ignore signs) |
| \(= \text{awrt } \mathbf{1\ 390\ 000}\) | A1 | For awrt 1 390 000 (3sf gives 1 384 422.948 but scores 1st A1 and 2nd A0) |
| \(b = \frac{-17501.25}{1385380.258} = -0.0126328..., = \text{awrt} -0.013\) | M1, A1 | For a correct expression for \(b\) seen (ft their values to 3sf). Use of \(S_{tt} \to -76.8\) is M0. For awrt \(-0.013\) (candidates using 3sf for \(S_{hh}\) should therefore get this) |
| \([\text{NB}\ \bar{t}=7.625,\ \bar{h}=796.25]\ a = \frac{61}{8} - \text{``}-0.0126...\text{''}\times\frac{6370}{8}\ [=17.6838...]\) | M1 | For a correct use of \(\bar{t}\) and \(\bar{h}\) to find \(a\) ft their \(b\) (allow letter \(b\) or even \(b=-0.985\)) |
| So \(\mathbf{t = 17.7 - 0.0126h}\) | A1 | For a correct equation with \(a=\) awrt 17.7 and \(b=\) awrt \(-0.0126\) [No \(y\) and \(x\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a\) is an estimate of the temperature at sea level is \((17.7\ ^\circ\text{C})\) | B1 | For stating or implying that it is the temperature (value not needed) at sea level |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((\mp)150\times b\) (o.e. e.g. \([17.7-0.0126h]-[17.7-0.0126(h+150)]\)) | M1 | For a correct expression equivalent to \((\mp)150b\). Can use letter \(b\) or ft their value(s) |
| \(= 1.89\) awrt \(\mathbf{2\ (^\circ C)}\) | A1 | For awrt 2 (°C not required). Allow \(\pm\) can give if "\(a\)" incorrect or "\(b\)" from M0A0 in (c). Common wrong answer of 11520 can score M1A0 even if no working seen |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(S_{th}) = 31070 - \frac{61\times6370}{8}$ or $31070-48571.25$; $(S_{tt}) = 693 - \frac{61^2}{8}$ or $693-465.125$ | M1; M1 | For a correct expression for $S_{th}$. For a correct expression for $S_{tt}$. Allow 1 slip e.g. 6730 or $61^2$ or 3721. Consistent use of $n\neq8$ M0M1 |
| $(S_{th}) = -17501.25$ and $(S_{tt}) = 227.875$ | A1cso | For both answers correct and both Ms scored |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|r|$ close to 1 or $r$ is close to $-1$ therefore it does support the linear model | B1 | For correct and relevant comment about the value of $r$ and saying it does support or "yes". Allow "it is...""strong" or "near perfect" correlation BUT B0 for "perfect" or "highly negative" |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[r = \frac{S_{yx}}{\sqrt{S_{yy}\times S_{xx}}}\right]$ so $r = \frac{S_{th}}{\sqrt{S_{tt}\times S_{hh}}}$ or $r^2 = \frac{(S_{th})^2}{S_{tt}\times S_{hh}}$ or $S_{hh} = \frac{(S_{th})^2}{r^2\times S_{tt}}$ or substitute 1 value | M1 | For the sight of the formula for $r$ and an attempt to do something useful with it. In (c) condone $x$ for $h$ and $y$ for $t$ except in 4th A1 |
| e.g. $\pm0.985 = \frac{\pm17501.25}{\sqrt{227.875\times S_{hh}}}$ or $S_{hh} = \frac{(\pm17501.25)^2}{(\pm0.985)^2\times227.875}$ o.e. $(= 1\ 385\ 380.258)$ | A1 | For a correct numerical expression in $S_{hh}$ or $\sqrt{S_{hh}}$. Accept with 3sf values (ignore signs) |
| $= \text{awrt } \mathbf{1\ 390\ 000}$ | A1 | For awrt 1 390 000 (3sf gives 1 384 422.948 but scores 1st A1 and 2nd A0) |
| $b = \frac{-17501.25}{1385380.258} = -0.0126328..., = \text{awrt} -0.013$ | M1, A1 | For a correct expression for $b$ seen (ft their values to 3sf). Use of $S_{tt} \to -76.8$ is M0. For awrt $-0.013$ (candidates using 3sf for $S_{hh}$ should therefore get this) |
| $[\text{NB}\ \bar{t}=7.625,\ \bar{h}=796.25]\ a = \frac{61}{8} - \text{``}-0.0126...\text{''}\times\frac{6370}{8}\ [=17.6838...]$ | M1 | For a correct use of $\bar{t}$ and $\bar{h}$ to find $a$ ft their $b$ (allow letter $b$ or even $b=-0.985$) |
| So $\mathbf{t = 17.7 - 0.0126h}$ | A1 | For a correct equation with $a=$ awrt 17.7 and $b=$ awrt $-0.0126$ [No $y$ and $x$] |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a$ is an estimate of the temperature at sea level is $(17.7\ ^\circ\text{C})$ | B1 | For stating or implying that it is the temperature (value not needed) at sea level |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(\mp)150\times b$ (o.e. e.g. $[17.7-0.0126h]-[17.7-0.0126(h+150)]$) | M1 | For a correct expression equivalent to $(\mp)150b$. Can use letter $b$ or ft their value(s) |
| $= 1.89$ awrt $\mathbf{2\ (^\circ C)}$ | A1 | For awrt 2 (°C not required). Allow $\pm$ can give if "$a$" incorrect or "$b$" from M0A0 in (c). Common wrong answer of 11520 can score M1A0 even if no working seen |
---6. A group of climbers collected information about the height above sea level, $h$ metres, and the air temperature, $t ^ { \circ } \mathrm { C }$, at the same time at 8 different points on the same mountain.
The data are summarised by
$$\sum h = 6370 \quad \sum t = 61 \quad \sum t h = 31070 \quad \sum t ^ { 2 } = 693$$
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { S } _ { \text {th } } = - 17501.25$ and $\mathrm { S } _ { \text {tt } } = 227.875$
The product moment correlation coefficient for these data is - 0.985
\item State, giving a reason, whether or not this value supports the use of a regression equation to predict the air temperature at different heights on this mountain.
\item Find the equation of the regression line of $t$ on $h$, giving your answer in the form $t = a + b h$. Give the value of your coefficients to 3 significant figures.
\item Give an interpretation of your value of $a$.
One of the climbers has just stopped for a short break before climbing the next 150 metres.
\item Estimate the drop in temperature over this 150 metre climb.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2018 Q6 [14]}}