| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Find means from regression lines |
| Difficulty | Standard +0.3 Part (a) requires conceptual understanding of correlation coefficients (r_s=1 means perfect rank correlation allowing non-linear relationships, while r=1 requires perfect linear relationship), which is moderately straightforward. Part (b) involves using the property that both regression lines pass through (x̄, ȳ), requiring students to find ȳ from the first regression equation then substitute into the second—a standard multi-step application of regression theory with no novel insight required. This is slightly easier than average for S1. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08e Spearman rank correlation5.09a Dependent/independent variables5.09b Least squares regression: concepts |
| Answer | Marks |
|---|---|
| B:Diag or expl based on \(r=1\Rightarrow\)pts on st line \(\Rightarrow r(s)=1\) | B1 |
| B1 3 | Diag or expl based on \(r(s)\neq 1\Rightarrow\)pts not on st line \(\Rightarrow r\neq 1\) \(r=1\Rightarrow\)pts on st line&\(r(s)\neq 1\Rightarrow\)pts not on st line B1B1 \(r=1\Rightarrow r(s)=1\) B2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{y} = 2.4 \times 4.5 + 3.7\) | M1 | Attempt to sub expression for \(y\) \(x=0.96x+1.48\)-c oe |
| \(= 14.5\) \(4.5 = 0.4 \times ^{14.5}\) - c \(c = 1.3\) | A1 M1 A1 4 | sub \(x=4.5\) and solve \(c=1.3\) |
| a'=x-b'y :-14.5 M1A1; then a'=4.5-0.4x14.5=-1.3 M1A1 | 14.5 M1A1.(y-3.7)/2.4=0.4y-c and sub14.5 M1 c=1.3 A1 |
### (a)
A: diag or explanation showing pts close to st line, always increasing
B:Diag or expl based on $r=1\Rightarrow$pts on st line $\Rightarrow r(s)=1$ | B1 |
| B1 3 | Diag or expl based on $r(s)\neq 1\Rightarrow$pts not on st line $\Rightarrow r\neq 1$ $r=1\Rightarrow$pts on st line&$r(s)\neq 1\Rightarrow$pts not on st line B1B1 $r=1\Rightarrow r(s)=1$ B2
### (b)
$\bar{y} = 2.4 \times 4.5 + 3.7$ | M1 | Attempt to sub expression for $y$ $x=0.96x+1.48$-c oe
$= 14.5$ $4.5 = 0.4 \times ^{14.5}$ - c $c = 1.3$ | A1 M1 A1 4 | sub $x=4.5$ and solve $c=1.3$
a'=x-b'y :-14.5 M1A1; then a'=4.5-0.4x14.5=-1.3 M1A1 | | 14.5 M1A1.(y-3.7)/2.4=0.4y-c and sub14.5 M1 c=1.3 A1
**Total [7]**
---\begin{enumerate}[label=(\alph*)]
\item A student calculated the values of the product moment correlation coefficient, $r$, and Spearman's rank correlation coefficient, $r_s$, for two sets of bivariate data, $A$ and $B$. His results are given below.
$$A: \quad r = 0.9 \text{ and } r_s = 1$$
$$B: \quad r = 1 \quad \text{and } r_s = 0.9$$
With the aid of a diagram where appropriate, explain why the student's results for $A$ could both be correct but his results for $B$ cannot both be correct. [3]
\item An old research paper has been partially destroyed. The surviving part of the paper contains the following incomplete information about some bivariate data from an experiment.
\includegraphics{figure_6}
The mean of $x$ is 4.5. The equation of the regression line of $y$ on $x$ is $y = 2.4x + 3.7$. The equation of the regression line of $x$ on $y$ is $x = 0.40y$ + [missing constant]
Calculate the missing constant at the end of the equation of the second regression line. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q6 [7]}}