| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Calculate regression line equation |
| Difficulty | Moderate -0.8 This is a standard S1 regression question requiring routine application of given formulas with all summary statistics provided. Parts (i) and (iv) involve direct substitution into standard formulas, while parts (ii), (iii), and (v) test basic conceptual understanding of correlation and regression. No problem-solving insight or novel reasoning is required—this is textbook exercise material, making it easier than average A-level questions. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09a Dependent/independent variables5.09c Calculate regression line |
| \(x\) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| \(y\) | 1.2 | 1.4 | 1.2 | 0.9 | 0.8 | 0.5 | 0.5 | 0.3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{xx} = 20400 - \frac{360^2}{8} \quad (=4200)\) | ||
| \(S_{yy} = 6.88 - \frac{6.8^2}{8} \quad (=1.1)\) | ||
| \(S_{xy} = 241 - \frac{360\times6.8}{8} \quad (=-65)\) | M1 | Correct sub in a correct \(S\) formula |
| \(r = \frac{"-65"}{\sqrt{"4200"\times"1.1"}}\) | M1 | Correct sub in 3 correct \(S\) formulae and a correct \(r\) formula |
| \(= -0.956\) (3 sf) | A1 | Correct ans with no working M2A1 — Ignore comment about \(-1 < r < -0.9\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| eg As you move further away, prices drop | B1 | High prices go with short distances oe. Allow "Strong (or high or good or equiv) neg corr'n between price and distance" — Both variables must be in context; miles & £ enough. Ignore all else, even if incorrect. NOT just neg corr'n between price & dist |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| None | B1 | Ignore all else, even if incorrect |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(b = \frac{"-65"}{" 4200"} \quad (=-0.0154762)\) | M1 | ft their \(S_{xy}\) & \(S_{xx}\) from (i) for M-marks only — or fresh start correct method |
| \(Y - \frac{6.8}{8} = "-0.0154762"(x - \frac{360}{8})\) oe | M1 | or \(a = \frac{6.8}{8} + "0.0154762"\times\frac{360}{8}\) oe |
| \(y = -0.0155x + 1.55\) (3 sf) oe | A1 | allow \(y = -0.015x + 1.5\) (or figs which round to these). (NOT \(y=-0.016x+1.6\), NOT \(y=-0.02x+1.5\)). Must have "\(y=\)" — Allow figures in eqn which round to the correct figures to either 3 sf or 2 sf, even if they result from arith errors. |
| or \(y = \frac{433}{280} - \frac{13}{840}x\) oe | ||
| [3] | Correct ans with no working M2A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Values of \(x\) are chosen beforehand or \(x\) is independent or controlled | B1 | \(x\) is fixed or given or set or predetermined oe — Not "\(x\) is constant." Not just "\(y\) depends on \(x\)". Ignore all other, even if incorrect |
| [1] |
# Question 5:
## Part 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{xx} = 20400 - \frac{360^2}{8} \quad (=4200)$ | | |
| $S_{yy} = 6.88 - \frac{6.8^2}{8} \quad (=1.1)$ | | |
| $S_{xy} = 241 - \frac{360\times6.8}{8} \quad (=-65)$ | M1 | Correct sub in a correct $S$ formula |
| $r = \frac{"-65"}{\sqrt{"4200"\times"1.1"}}$ | M1 | Correct sub in 3 correct $S$ formulae and a correct $r$ formula |
| $= -0.956$ (3 sf) | A1 | Correct ans with no working M2A1 — Ignore comment about $-1 < r < -0.9$ |
| **[3]** | | |
## Part 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| eg As you move further away, prices drop | B1 | High prices go with short distances oe. Allow "Strong (or high or good or equiv) neg corr'n between price and distance" — Both variables must be in context; miles & £ enough. Ignore all else, even if incorrect. NOT just neg corr'n between price & dist |
| **[1]** | | |
## Part 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| None | B1 | Ignore all else, even if incorrect |
| **[1]** | | |
## Part 5(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $b = \frac{"-65"}{" 4200"} \quad (=-0.0154762)$ | M1 | ft their $S_{xy}$ & $S_{xx}$ from (i) for M-marks only — or fresh start correct method |
| $Y - \frac{6.8}{8} = "-0.0154762"(x - \frac{360}{8})$ oe | M1 | or $a = \frac{6.8}{8} + "0.0154762"\times\frac{360}{8}$ oe |
| $y = -0.0155x + 1.55$ (3 sf) oe | A1 | allow $y = -0.015x + 1.5$ (or figs which round to these). (NOT $y=-0.016x+1.6$, NOT $y=-0.02x+1.5$). Must have "$y=$" — Allow figures in eqn which round to the correct figures to either 3 sf or 2 sf, even if they result from arith errors. |
| or $y = \frac{433}{280} - \frac{13}{840}x$ oe | | |
| **[3]** | | Correct ans with no working M2A1 |
## Part 5(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Values of $x$ are chosen beforehand or $x$ is independent or controlled | B1 | $x$ is fixed or given or set or predetermined oe — Not "$x$ is constant." Not just "$y$ depends on $x$". Ignore all other, even if incorrect |
| **[1]** | | |
---5 Tariq collected information about typical prices, $\pounds y$ million, of four-bedroomed houses at varying distances, $x$ miles, from a large city. He chose houses at 10 -mile intervals from the city. His results are shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
$x$ & 10 & 20 & 30 & 40 & 50 & 60 & 70 & 80 \\
\hline
$y$ & 1.2 & 1.4 & 1.2 & 0.9 & 0.8 & 0.5 & 0.5 & 0.3 \\
\hline
\end{tabular}
\end{center}
$$n = 8 \quad \Sigma x = 360 \quad \Sigma x ^ { 2 } = 20400 \quad \Sigma y = 6.8 \quad \Sigma y ^ { 2 } = 6.88 \quad \Sigma x y = 241$$
(i) Use an appropriate formula to calculate the product moment correlation coefficient, $r$, showing that $- 1.0 < r < - 0.9$.\\
(ii) State what this value of $r$ shows in this context.\\
(iii) Tariq decides to recalculate the value of $r$ with the house prices measured in hundreds of thousands of pounds, instead of millions of pounds. State what effect, if any, this will have on the value of $r$.\\
(iv) Calculate the equation of the regression line of $y$ on $x$.\\
(v) Explain why the regression line of $y$ on $x$, rather than $x$ on $y$, should be used for estimating a value of $x$ from a given value of $y$.
\hfill \mbox{\textit{OCR S1 2014 Q5 [9]}}