Edexcel S3 2014 June — Question 1 11 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Year2014
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHypothesis test of Spearman’s rank correlation coefficien
TypeHypothesis test for positive correlation
DifficultyStandard +0.3 This is a standard textbook application of Spearman's rank correlation coefficient with straightforward ranking, hypothesis testing using tables, and basic interpretation. The calculations are routine (8 data points, no ties), the hypothesis test follows a standard template, and parts (c)-(d) require only recall of assumptions rather than deep statistical reasoning. Slightly easier than average due to the mechanical nature of the procedure.
Spec5.08e Spearman rank correlation5.08f Hypothesis test: Spearman rank5.08g Compare: Pearson vs Spearman

  1. A journalist is investigating factors which influence people when they buy a new car. One possible factor is fuel efficiency. The journalist randomly selects 8 car models. Each model's annual sales and fuel efficiency, in km/litre, are shown in the table below.
Car model\(A\)\(B\)\(C\)\(D\)\(E\)\(F\)\(G\)\(H\)
Annual sales18005400181007100930048001220010700
Fuel efficiency5.218.614.813.218.311.916.517.7
  1. Calculate Spearman's rank correlation coefficient for these data. The journalist believes that car models with higher fuel efficiency will achieve higher sales.
  2. Stating your hypotheses clearly, test whether or not the data support the journalist's belief. Use a \(5 \%\) level of significance.
  3. State the assumption necessary for a product moment correlation coefficient to be valid in this case.
  4. The mean and median fuel efficiencies of the car models in the random sample are 14.5 km /litre and 15.65 km /litre respectively. Considering these statistics, as well as the distribution of the fuel efficiency data, state whether or not the data suggest that the assumption in part (c) might be true in this case. Give a reason for your answer. (No further calculations are required.)

1(a)
AnswerMarks Guidance
AnswerMarks Guidance
\(\Sigma d^2 = 50\)M1A1 M1 for attempting to rank at least one set of data; A1 for at least one set of data ranked correctly (NB this mark comes after 2nd M1)
\(r_s = 1 - \frac{6\Sigma d^2}{8(64-1)} = 1 - \frac{6 \times 50}{8 \times 63}\)M1 M1 for correct use of formula for \(r_s\)
\(r_s = \frac{204}{504} = 0.40476...\)A1 awrt 0.405
(5)
1(b)
AnswerMarks Guidance
AnswerMarks Guidance
\(H_0: \rho = 0\) and \(H_1: \rho > 0\) (accept \(\rho_s\) or \(\rho\))B1
1 tail critical value \(\rho = 0.6429\)B1
Test value is not in critical region so insufficient evidence to reject \(H_0\)M1A1ft M1 for correct statement relating their test statistic and critical value; A1ft their test statistic, \(H_1\) and critical value but must be in context
No significant evidence at 5% level to support journalist's belief (4)
1(c)
AnswerMarks Guidance
AnswerMarks Guidance
Underlying (bivariate) Normal distributionB1 (1)
1(d)
AnswerMarks Guidance
AnswerMarks Guidance
Evidence does not support Normal distribution since mean < median or (negative) skew, oeB1 (1)
(11 marks)
**1(a)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Sigma d^2 = 50$ | M1A1 | M1 for attempting to rank at least one set of data; A1 for at least one set of data ranked correctly (NB this mark comes after 2nd M1) |
| $r_s = 1 - \frac{6\Sigma d^2}{8(64-1)} = 1 - \frac{6 \times 50}{8 \times 63}$ | M1 | M1 for correct use of formula for $r_s$ |
| $r_s = \frac{204}{504} = 0.40476...$ | A1 | awrt 0.405 |
| | (5) | |

**1(b)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \rho = 0$ and $H_1: \rho > 0$ (accept $\rho_s$ or $\rho$) | B1 | |
| 1 tail critical value $\rho = 0.6429$ | B1 | |
| Test value is not in critical region so insufficient evidence to reject $H_0$ | M1A1ft | M1 for correct statement relating their test statistic and critical value; A1ft their test statistic, $H_1$ and critical value but must be in context |
| No significant evidence at 5% level to support journalist's belief | | (4) |

**1(c)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Underlying (bivariate) Normal distribution | B1 | (1) |

**1(d)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Evidence does not support Normal distribution since mean < median or (negative) skew, oe | B1 | (1) |
| | (11 marks) | |

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\begin{enumerate}
  \item A journalist is investigating factors which influence people when they buy a new car. One possible factor is fuel efficiency. The journalist randomly selects 8 car models. Each model's annual sales and fuel efficiency, in km/litre, are shown in the table below.
\end{enumerate}

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Car model & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & $G$ & $H$ \\
\hline
Annual sales & 1800 & 5400 & 18100 & 7100 & 9300 & 4800 & 12200 & 10700 \\
\hline
Fuel efficiency & 5.2 & 18.6 & 14.8 & 13.2 & 18.3 & 11.9 & 16.5 & 17.7 \\
\hline
\end{tabular}
\end{center}

(a) Calculate Spearman's rank correlation coefficient for these data.

The journalist believes that car models with higher fuel efficiency will achieve higher sales.\\
(b) Stating your hypotheses clearly, test whether or not the data support the journalist's belief. Use a $5 \%$ level of significance.\\
(c) State the assumption necessary for a product moment correlation coefficient to be valid in this case.\\
(d) The mean and median fuel efficiencies of the car models in the random sample are 14.5 km /litre and 15.65 km /litre respectively. Considering these statistics, as well as the distribution of the fuel efficiency data, state whether or not the data suggest that the assumption in part (c) might be true in this case. Give a reason for your answer. (No further calculations are required.)\\

\hfill \mbox{\textit{Edexcel S3 2014 Q1 [11]}}