| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Spearman’s rank correlation coefficien |
| Type | Compare with PMCC |
| Difficulty | Standard +0.3 Part (i) requires substituting the pattern into Spearman's formula and solving algebraically for n, which is straightforward but involves some algebraic manipulation. Parts (ii)(a) and (ii)(b) test conceptual understanding of correlation coefficients at a basic level—these are standard textbook concepts requiring minimal problem-solving. Overall slightly easier than average for A-level. |
| Spec | 5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank5.08g Compare: Pearson vs Spearman |
| Competitor | \(C_1\) | \(C_2\) | \(C_3\) | \(C_4\) | \(C_5\) | \(C_6\) | \(\ldots\) | \(C_{n-1}\) | \(C_n\) |
| Judge 1 rank | 1 | 2 | 3 | 4 | 5 | 6 | \(\ldots\) | \(n-1\) | \(n\) |
| Judge 2 rank | 2 | 1 | 4 | 3 | 6 | 5 | \(\ldots\) | \(n\) | \(n-1\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Sigma d^2 = n\) seen or implied; \(1 - \frac{6×\text{anything}}{n(n^2-1)}\) = \(\frac{63}{65}\) or \(\frac{6×\text{anything}}{n(n^2-1)}\) = \(\frac{2}{65}\); or eg 390 = \(2(n^2 − 1)\) | M1, M1, A1 depM2 | eg \(1 - \frac{6×y^2}{6}\) = \(\frac{63}{65}\) or \(1 - \frac{6×n^2}{n(n^2-1)}\) = \(\frac{63}{65}\); \(\frac{6}{(n^2-1)}\) = \(\frac{2}{65}\); Any correct eqn after cancelling \(n\) or take out factor of \(n\); can be implied by \(n = 14\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 1 \Rightarrow\) st line, hence true (or \(r_s = 1\)) oe; Explanation essential. Must state or imply "true" | B1, [1] | \(r = 1 \Rightarrow y\) incr as \(x\) incr, so \(r_s = 1\) oe; Allow "True because \(r = 1\) means pts ranked in order so \(r_s = 1\)" or "r = 1 means all d's are 0, hence \(r_s = 1−0 = 1\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Diag. > 3 pts, not on st line but with \(x_{n+1}>x_n\) & \(y_{n+1}>y_n\); Zig zag line or curve, moving up & right; so \(r_s\) can still be 1 | B1, B1dep, [2] | Ignore explan if correct diag given; Ignore any st line drawn; Allow numerical example for which \(r ≠ 1\) but \(r_s = 1\). If expl'n contradicts diag, mark diag; For 2ⁿᵈ B1 must state or imply "false" |
## (i)
$\Sigma d^2 = n$ seen or implied; $1 - \frac{6×\text{anything}}{n(n^2-1)}$ = $\frac{63}{65}$ or $\frac{6×\text{anything}}{n(n^2-1)}$ = $\frac{2}{65}$; or eg 390 = $2(n^2 − 1)$ | M1, M1, A1 depM2 | eg $1 - \frac{6×y^2}{6}$ = $\frac{63}{65}$ or $1 - \frac{6×n^2}{n(n^2-1)}$ = $\frac{63}{65}$; $\frac{6}{(n^2-1)}$ = $\frac{2}{65}$; Any correct eqn after cancelling $n$ or take out factor of $n$; can be implied by $n = 14$ | $\frac{6}{(n^2-1)}$ = $\frac{2}{65}$ or 390 = $2(n^2 − 1)$; A1 depM2 | But A0 if $n = 14$ clearly follows from incorrect working; If no working or unclear working, but $n = 14$, M1M1A1A1 | n = 14; NOT $n = ±14$; A1; Conclusion needed | [4] | Trial method: eg $\Sigma d^2 = 14$; M1; $1 - \frac{6×14}{14(14^2-1)}$ oe M1; = $\frac{63}{65}$ (0.969 : A0)
## (ii)(a)
$r = 1 \Rightarrow$ st line, hence true (or $r_s = 1$) oe; Explanation essential. Must state or imply "true" | B1, [1] | $r = 1 \Rightarrow y$ incr as $x$ incr, so $r_s = 1$ oe; Allow "True because $r = 1$ means pts ranked in order so $r_s = 1$" or "r = 1 means all d's are 0, hence $r_s = 1−0 = 1$" | NOT "r incr so ranks incr"; NOT "$r_s = r$ for ranks so true"; NOT "True because strong corr'n"
## (ii)(b)
Diag. > 3 pts, not on st line but with $x_{n+1}>x_n$ & $y_{n+1}>y_n$; Zig zag line or curve, moving up & right; so $r_s$ can still be 1 | B1, B1dep, [2] | Ignore explan if correct diag given; Ignore any st line drawn; Allow numerical example for which $r ≠ 1$ but $r_s = 1$. If expl'n contradicts diag, mark diag; For 2ⁿᵈ B1 must state or imply "false" | eg "expon'l curve gives $r ≠ 1$ but $r_s = 1$"
---\begin{enumerate}[label=(\roman*)]
\item Two judges rank $n$ competitors, where $n$ is an even number. Judge 2 reverses each consecutive pair of ranks given by Judge 1, as shown.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
Competitor & $C_1$ & $C_2$ & $C_3$ & $C_4$ & $C_5$ & $C_6$ & $\ldots$ & $C_{n-1}$ & $C_n$ \\
\hline
Judge 1 rank & 1 & 2 & 3 & 4 & 5 & 6 & $\ldots$ & $n-1$ & $n$ \\
\hline
Judge 2 rank & 2 & 1 & 4 & 3 & 6 & 5 & $\ldots$ & $n$ & $n-1$ \\
\hline
\end{tabular}
\end{center}
Given that the value of Spearman's coefficient of rank correlation is $\frac{63}{65}$, find $n$. [4]
\item An experiment produced some data from a bivariate distribution. The product moment correlation coefficient is denoted by $r$, and Spearman's rank correlation coefficient is denoted by $r_s$.
\begin{enumerate}[label=(\alph*)]
\item Explain whether the statement
$$r = 1 \Rightarrow r_s = 1$$
is true or false. [1]
\item Use a diagram to explain whether the statement
$$r \neq 1 \Rightarrow r_s \neq 1$$
is true or false. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2013 Q7 [7]}}