| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Spearman’s rank correlation coefficien |
| Type | Hypothesis test for positive correlation |
| Difficulty | Standard +0.3 This is a straightforward application of Spearman's rank correlation coefficient with standard hypothesis testing. The calculation involves ranking 8 values (routine), computing the coefficient using the formula, and comparing to critical values from tables. Part (c) requires minimal conceptual understanding of how rank changes affect the coefficient. This is slightly easier than average as it's a textbook procedure with no novel problem-solving required. |
| Spec | 5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank |
| \cline { 2 - 9 } \multicolumn{1}{c|}{} | Skater | |||||||
| \cline { 2 - 9 } | \(A\) | \(B\) | \(C\) | \(D\) | \(E\) | \(F\) | \(G\) | \(H\) |
| Judge 1 | 71 | 70 | 72 | 62 | 63 | 61 | 57 | 53 |
| Judge 2 | 73 | 71 | 67 | 64 | 62 | 56 | 52 | 53 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempt to rank at least one row (at least 4 correct) | M1 | 1.1b |
| Correct \(d\) or \(d^2\) row for their ranks | dM1 | 1.1b — dep on previous M1 |
| \(r_s = 1 - \dfrac{6 \times "10"}{8(64-1)}\) | M1 | 1.1b — Allow if not ranked \(\sum d^2 = 85\) |
| \(r_s = 0.8809...\) | A1 | 1.1b — awrt 0.881 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \rho = 0 \quad H_1: \rho > 0\) | B1 | 2.5 — Both hypotheses stated in terms of \(\rho\) |
| Critical value \(\rho = 0.8333\) | B1 | 1.1b — Allow even if 2-tail test (sign must match their \(r_s\)) |
| \(r_s = 0.8809\) lies in the critical region / reject \(H_0\) / significant | M1 | 2.1 — For comparing their 0.881 with "their 0.8333" |
| There is evidence that the two judges are in agreement. | A1cso | 2.2b — Correct contextual conclusion with no contradictions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum d^2\) will decrease since the new rankings given by Judge 1 are now the same as the rankings given by Judge 2 for Skater D and E whereas previously they were different. \(\sum d^2\) decreases or \(d/d^2\) decreases for \(D\) and \(E\); idea of same rankings e.g. \(d^2\) will reduce by 2. Do not allow \(d^2\) will reduce by 1. | M1 | 2.4 |
| Therefore Spearman's rank correlation coefficient will increase | A1 | 2.2a — For a correct deduction from the information. Allow "closer to 1." |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to rank at least one row (at least 4 correct) | M1 | 1.1b |
| Correct $d$ or $d^2$ row for their ranks | dM1 | 1.1b — dep on previous M1 |
| $r_s = 1 - \dfrac{6 \times "10"}{8(64-1)}$ | M1 | 1.1b — Allow if not ranked $\sum d^2 = 85$ |
| $r_s = 0.8809...$ | A1 | 1.1b — awrt 0.881 |
$\sum d^2 = 10$
**(4 marks)**
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \rho = 0 \quad H_1: \rho > 0$ | B1 | 2.5 — Both hypotheses stated in terms of $\rho$ |
| Critical value $\rho = 0.8333$ | B1 | 1.1b — Allow even if 2-tail test (sign must match their $r_s$) |
| $r_s = 0.8809$ lies in the critical region / reject $H_0$ / significant | M1 | 2.1 — For comparing their 0.881 with "their 0.8333" |
| There is evidence that the two judges are in agreement. | A1cso | 2.2b — Correct contextual conclusion with no contradictions |
**(4 marks)**
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum d^2$ will decrease since the new rankings given by Judge 1 are now the same as the rankings given by Judge 2 for Skater D and E whereas previously they were different. $\sum d^2$ decreases or $d/d^2$ decreases for $D$ and $E$; idea of same rankings e.g. $d^2$ will reduce by 2. Do not allow $d^2$ will reduce by 1. | M1 | 2.4 |
| Therefore Spearman's rank correlation coefficient will increase | A1 | 2.2a — For a correct deduction from the information. Allow "closer to 1." |
**(2 marks)**
---\begin{enumerate}
\item Bara is investigating whether or not the two judges of a skating competition are in agreement. The two judges gave a score to each of the 8 skaters in the competition as shown in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\cline { 2 - 9 }
\multicolumn{1}{c|}{} & \multicolumn{8}{c|}{Skater} \\
\cline { 2 - 9 }
& $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Judge 1 & 71 & 70 & 72 & 62 & 63 & 61 & 57 & 53 \\
\hline
Judge 2 & 73 & 71 & 67 & 64 & 62 & 56 & 52 & 53 \\
\hline
\end{tabular}
\end{center}
Bara decided to calculate Spearman's rank correlation coefficient for these data.\\
(a) Calculate Spearman's rank correlation coefficient between the ranks of the two judges.\\
(b) Test, at the $1 \%$ level of significance, whether or not the two judges are in agreement.
Judge 1 accidentally swapped the scores for skaters $D$ and $E$. The score for skater $D$ should be 63 and the score for skater $E$ should be 62\\
(c) Without carrying out any further calculations, explain how Spearman's rank correlation coefficient will change. Give a reason for your answer.
\hfill \mbox{\textit{Edexcel FS2 AS 2019 Q1 [10]}}