AQA S3 2008 June — Question 1 7 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2008
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicHypothesis test of Pearson’s product-moment correlation coefficient
TypeOne-tailed test for positive correlation
DifficultyModerate -0.3 This is a straightforward application of standard formulas for correlation coefficient and hypothesis testing. Part (a) requires direct substitution into r = S_xy/√(S_xx·S_yy), part (b) is a routine one-tailed test comparing r to critical values from tables, and part (c) asks for basic interpretation. While it requires knowledge of hypothesis testing procedures, there are no conceptual challenges or problem-solving elements beyond following the standard algorithm.
Spec5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation

1 The best performances of a random sample of 20 junior athletes in the long jump, \(x\) metres, and in the high jump, \(y\) metres, were recorded. The following statistics were calculated from the results. $$S _ { x x } = 7.0036 \quad S _ { y y } = 0.8464 \quad S _ { x y } = 1.3781$$
  1. Calculate the value of the product moment correlation coefficient between \(x\) and \(y\).
    (2 marks)
  2. Assuming that these data come from a bivariate normal distribution, investigate, at the \(1 \%\) level of significance, the claim that for junior athletes there is a positive correlation between \(x\) and \(y\).
  3. Interpret your conclusion in the context of this question.

AnswerMarks Guidance
Part (a)
\(r = \frac{1.3781}{\sqrt{7.0036 \times 0.8464}} = 0.56 \text{ to } 0.57\)M1; A1 AWFW (0.56602)
Part (b)
\(H_0: \rho = 0\); \(H_1: \rho > 0\)B1 Both
\(\text{SL } \alpha = 0.01 \text{ (1%)}\); \(\text{CV } r = 0.515 \text{ to } 0.516\)B1 AWFW (0.5155)
Calculated \(r >\) Tabulated \(r\)M1 Comparison
Evidence, at 1% level, of a positive correlation between \(x\) and \(y\)A1↑ ft on \(r\) and CV; 4 marks
Part (b) Special Case
\(\text{CV } t_{n-2}(0.99) = 2.552\)(B1)
\(r\sqrt{\frac{n-2}{1-r^2}} = 2.913\)(B1)
Part (c)
(Strong) evidence of a positive correlation between best performances of junior athletes in the long jump and in the high jumpB1↑ ft on (b); or equivalent; 1 mark
Total: 7 marks
| **Part (a)** |
|---|
| $r = \frac{1.3781}{\sqrt{7.0036 \times 0.8464}} = 0.56 \text{ to } 0.57$ | M1; A1 | AWFW (0.56602) |

| **Part (b)** |
|---|
| $H_0: \rho = 0$; $H_1: \rho > 0$ | B1 | Both |
| $\text{SL } \alpha = 0.01 \text{ (1%)}$; $\text{CV } r = 0.515 \text{ to } 0.516$ | B1 | AWFW (0.5155) |
| Calculated $r >$ Tabulated $r$ | M1 | Comparison |
| Evidence, at 1% level, of a positive correlation between $x$ and $y$ | A1↑ | ft on $r$ and CV; 4 marks |

| **Part (b) Special Case** |
|---|
| $\text{CV } t_{n-2}(0.99) = 2.552$ | (B1) | |
| $r\sqrt{\frac{n-2}{1-r^2}} = 2.913$ | (B1) | |

| **Part (c)** |
|---|
| (Strong) evidence of a positive correlation between best performances of junior athletes in the long jump and in the high jump | B1↑ | ft on (b); or equivalent; 1 mark |

**Total: 7 marks**

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1 The best performances of a random sample of 20 junior athletes in the long jump, $x$ metres, and in the high jump, $y$ metres, were recorded. The following statistics were calculated from the results.

$$S _ { x x } = 7.0036 \quad S _ { y y } = 0.8464 \quad S _ { x y } = 1.3781$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the value of the product moment correlation coefficient between $x$ and $y$.\\
(2 marks)
\item Assuming that these data come from a bivariate normal distribution, investigate, at the $1 \%$ level of significance, the claim that for junior athletes there is a positive correlation between $x$ and $y$.
\item Interpret your conclusion in the context of this question.
\end{enumerate}

\hfill \mbox{\textit{AQA S3 2008 Q1 [7]}}