| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 11 |
| Topic | Hypothesis test of Spearman’s rank correlation coefficien |
| Type | Critical region or probability |
| Difficulty | Challenging +1.2 This is a Further Maths Statistics question requiring Spearman's rank correlation hypothesis test and probability calculations. Part (i) is standard application of tables/critical values. Part (ii)(a) is straightforward probability (1/7!). Part (ii)(b) requires understanding that r_s values are constrained by the formula involving sum of squared rank differences, making it moderately challenging but still within typical Further Stats scope. |
| Spec | 5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank |
| Wine | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) |
| Judge I | 86.3 | 87.5 | 87.6 | 88.8 | 89.4 | 89.9 | 90.5 |
| Judge II | 85.3 | 88.1 | 82.7 | 87.7 | 89.0 | 89.4 | 91.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Rankings | 1 | 2 |
| 2 | 4 | 1 |
| M1 | Attempt to rank | Allow rankings in opposite order |
| M1 | ||
| \(r_s = 1 - \frac{6 \times 10}{7 \times 48} = 0.8214...\) | A1 | Or \(\frac{23}{28}\) |
| \(0.8214 < 0.8929\) | B1 | Correct comparison |
| Do not reject \(H_0\). Insufficient evidence of agreement between judges' rankings. | M1ft | FT on their TS or on 0.9286 |
| A1ft | Contextualised, not too definite | NOT "there is agreement" |
| Answer | Marks | Guidance |
|---|---|---|
| \(17!\) | M1 | |
| \(= \frac{1}{5040}\) or 0.000198... | A1 | Exact or awrt 0.000198 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - \frac{62d^2}{7 \times 48} = \frac{55}{56}\) \(\Rightarrow\) \(\Sigma d^2 = 1\) | M1 | Find \(\Sigma d^2\) |
| A1 | Justify impossible | |
| Impossible as 1 incorrect ranking implies another | A1 |
## Part (i)
$H_0$: no association between rankings, $H_1$: positive association
| Rankings | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|----------|---|---|---|---|---|---|---|
| | 2 | 4 | 1 | 3 | 5 | 6 | 7 |
| M1 | Attempt to rank | Allow rankings in opposite order
| M1 |
$r_s = 1 - \frac{6 \times 10}{7 \times 48} = 0.8214...$ | A1 | Or $\frac{23}{28}$
$0.8214 < 0.8929$ | B1 | Correct comparison
Do not reject $H_0$. Insufficient evidence of agreement between judges' rankings. | M1ft | FT on their TS or on 0.9286
| A1ft | Contextualised, not too definite | NOT "there is agreement"
## Part (ii)(a)
$17!$ | M1 |
$= \frac{1}{5040}$ or 0.000198... | A1 | Exact or awrt 0.000198
## Part (ii)(b)
$1 - \frac{62d^2}{7 \times 48} = \frac{55}{56}$ $\Rightarrow$ $\Sigma d^2 = 1$ | M1 | Find $\Sigma d^2$
| A1 | Justify impossible
Impossible as 1 incorrect ranking implies another | A1 |8 At a wine-tasting competition, two judges give marks out of 100 to 7 wines as follows.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
Wine & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ \\
\hline
Judge I & 86.3 & 87.5 & 87.6 & 88.8 & 89.4 & 89.9 & 90.5 \\
\hline
Judge II & 85.3 & 88.1 & 82.7 & 87.7 & 89.0 & 89.4 & 91.5 \\
\hline
\end{tabular}
\end{center}
(i) A spectator claims that there is a high level of agreement between the rank orders of the marks given by the two judges. Test the spectator's claim at the $1 \%$ significance level.\\
(ii) A competitor ranks the wines in a random order. The value of Spearman's rank correlation coefficient between the competitor and Judge I is $r _ { s }$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that $r _ { s } = 1$.
\item Show that $r _ { s }$ cannot take the value $\frac { 55 } { 56 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2018 Q8 [11]}}