Edexcel S3 2012 June — Question 1 12 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Year2012
SessionJune
Marks12
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TopicHypothesis test of Spearman’s rank correlation coefficien
TypeHypothesis test for association
DifficultyStandard +0.3 This is a straightforward application of Spearman's rank correlation with standard hypothesis testing. Part (a) requires ranking data and applying a formula (routine calculation), part (b) is a standard hypothesis test with critical value lookup, and part (c) tests conceptual understanding of tied ranks. The question is slightly easier than average because it's a direct application of a standard procedure with no novel problem-solving required, though the tied ranks consideration in part (c) adds minor complexity.
Spec5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank

  1. Interviews for a job are carried out by two managers. Candidates are given a score by each manager and the results for a random sample of 8 candidates are shown in the table below.
Candidate\(A\)\(B\)\(C\)\(D\)\(E\)\(F\)\(G\)\(H\)
Manager \(X\)6256875465151210
Manager \(Y\)5447715049253044
  1. Calculate Spearman's rank correlation coefficient for these data.
  2. Test, at the \(5 \%\) level of significance, whether there is agreement between the rankings awarded by each manager. State your hypotheses clearly. Manager \(Y\) later discovered he had miscopied his score for candidate \(D\) and it should be 54 .
  3. Without carrying out any further calculations, explain how you would calculate Spearman's rank correlation in this case.

Question 1:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Attempt to rank score \(X\) and score \(Y\)M1 For an attempt to rank both variables
Attempting \(d^2\) for their ranks, must be using ranksM1
\(\sum d^2 = 18\)A1 For sum of 18
\(r_s = 1 - \frac{6 \times 18}{8(64-1)} = 0.7857...\)M1A1 awrt 0.786
(5 marks)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(H_0 : \rho = 0\)B1 Null hypothesis in terms of \(\rho\) or \(\rho_s\)
\(H_1 : \rho > 0\)B1 Alt hypothesis as given
Critical region \(r_s > 0.6429\)B1 cv of \(+0.6429\) (or 0.7381 if two-tailed from hyp)
\((0.7857 > 0.6429)\) sufficient evidence to reject \(H_0\)M1 Correct statement relating \(r_s\) with cv, cv must be such that \(\
There is evidence of agreement between the scores awarded by each managerA1ft Correct contextualised comment. Must mention "scores/rankings" and "manager". Use of "association" is A0
(5 marks)
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
(\(A\) and \(D\) are now) tied ranks (for Manager \(Y\))B1 Tied ranks implied by 2.5, 6.5 or both 2 or 6 or description
Average rank awarded to \(A\) and \(D\), and use \(r_s = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\)B1 Average rank implied by 2.5 or 6.5 or description and 'use of pmcc'
(2 marks)
# Question 1:

## Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to rank score $X$ and score $Y$ | M1 | For an attempt to rank both variables |
| Attempting $d^2$ for their ranks, must be using ranks | M1 | |
| $\sum d^2 = 18$ | A1 | For sum of 18 |
| $r_s = 1 - \frac{6 \times 18}{8(64-1)} = 0.7857...$ | M1A1 | awrt 0.786 |

**(5 marks)**

## Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0 : \rho = 0$ | B1 | Null hypothesis in terms of $\rho$ or $\rho_s$ |
| $H_1 : \rho > 0$ | B1 | Alt hypothesis as given |
| Critical region $r_s > 0.6429$ | B1 | cv of $+0.6429$ (or 0.7381 if two-tailed from hyp) |
| $(0.7857 > 0.6429)$ sufficient evidence to reject $H_0$ | M1 | Correct statement relating $r_s$ with cv, cv must be such that $\|cv\| < 1$ |
| There is evidence of agreement between the scores awarded by each manager | A1ft | Correct contextualised comment. Must mention "scores/rankings" and "manager". Use of "association" is A0 |

**(5 marks)**

## Part (c)

| Answer/Working | Marks | Guidance |
|---|---|---|
| ($A$ and $D$ are now) tied ranks (for Manager $Y$) | B1 | Tied ranks implied by 2.5, 6.5 or **both** 2 or 6 or description |
| Average rank awarded to $A$ and $D$, **and** use $r_s = \frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}$ | B1 | Average rank implied by 2.5 or 6.5 or description and 'use of pmcc' |

**(2 marks)**

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\begin{enumerate}
  \item Interviews for a job are carried out by two managers. Candidates are given a score by each manager and the results for a random sample of 8 candidates are shown in the table below.
\end{enumerate}

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Candidate & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Manager $X$ & 62 & 56 & 87 & 54 & 65 & 15 & 12 & 10 \\
\hline
Manager $Y$ & 54 & 47 & 71 & 50 & 49 & 25 & 30 & 44 \\
\hline
\end{tabular}
\end{center}

(a) Calculate Spearman's rank correlation coefficient for these data.\\
(b) Test, at the $5 \%$ level of significance, whether there is agreement between the rankings awarded by each manager. State your hypotheses clearly.

Manager $Y$ later discovered he had miscopied his score for candidate $D$ and it should be 54 .\\
(c) Without carrying out any further calculations, explain how you would calculate Spearman's rank correlation in this case.\\

\hfill \mbox{\textit{Edexcel S3 2012 Q1 [12]}}