| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2018 |
| Session | September |
| Marks | 11 |
| Topic | Bivariate data |
| Type | Calculate r from summary statistics |
| Difficulty | Standard +0.3 This is a straightforward Further Statistics question requiring standard application of the correlation coefficient formula with given summary statistics, followed by routine interpretation and hypothesis testing. While it's Further Maths content (making it slightly harder on an absolute scale), the calculations are mechanical with no conceptual challenges or novel problem-solving required. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.08e Spearman rank correlation |
| \(x\) | 4.6 | 5.9 | 6.5 | 7.8 | 8.3 |
| \(y\) | 15.6 | 10.8 | 10.4 | 10.1 | 9.7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-0.85462\) | B2 | 1.1 |
| \([S_{xx} = 1.7656, S_{yy} = 5.7096, S_{xy} = -2.4644\) or \(8.828, 23.548, -12.322]\) |
| Answer | Marks | Guidance |
|---|---|---|
| Points lie (quite) close to a straight line with negative gradient | B1 | 2.2b |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \rho = 0, H_1: \rho \neq 0\), where \(\rho\) is the population product moment correlation coefficient | B2 | 1.1a, 2.5 |
| CV \(= (-) 0.8783\) | B1 | 1.1 |
| \(-0.855 > -0.8783\) so do not reject \(H_0\) | M1 | 2.2b |
| Insufficient evidence of association | A1 [5] | 3.5a |
| Answer | Marks | Guidance |
|---|---|---|
| \(-1\) | B1 [1] | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Neither is changed | B1 [1] | 2.2a |
| (b) \(r\) is changed but \(r_s\) is not | B1 [1] | 2.2a |
## (i)
$-0.85462$ | B2 | 1.1 | BC $-0.855$ or better | SC: If B0 give M1 for correct use of Ss
$[S_{xx} = 1.7656, S_{yy} = 5.7096, S_{xy} = -2.4644$ or $8.828, 23.548, -12.322]$ | | |
## (iii)
Points lie (quite) close to a straight line with negative gradient | B1 | 2.2b | Both nearness and sign needed
## (iii)
$H_0: \rho = 0, H_1: \rho \neq 0$, where $\rho$ is the population product moment correlation coefficient | B2 | 1.1a, 2.5 | $\rho$ not properly defined in terms of population: B1
CV $= (-) 0.8783$ | B1 | 1.1 | Correct CV
$-0.855 > -0.8783$ so do not reject $H_0$ | M1 | 2.2b | Comparison and first conclusion
Insufficient evidence of association | A1 [5] | 3.5a | Not too definite, e.g. not "there is no association" | No context needed
## (iv)
$-1$ | B1 [1] | 2.2a | Not "nearly $-1$"
## (v)
(a) Neither is changed | B1 [1] | 2.2a |
(b) $r$ is changed but $r_s$ is not | B1 [1] | 2.2a |
---7 The table shows the values of 5 observations of bivariate data $( x , y )$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 4.6 & 5.9 & 6.5 & 7.8 & 8.3 \\
\hline
$y$ & 15.6 & 10.8 & 10.4 & 10.1 & 9.7 \\
\hline
\end{tabular}
\end{center}
$$n = 5 , \Sigma x = 33.1 , \Sigma y = 56.6 , \Sigma x ^ { 2 } = 227.95 , \Sigma y ^ { 2 } = 664.26 , \Sigma x y = 362.37$$
(i) Calculate Pearson's product-moment correlation coefficient $r$ for the data.\\
(ii) State what this value of $r$ tells you about a scatter diagram illustrating the data.\\
(iii) Test at the $5 \%$ significance level whether there is association between $x$ and $y$.\\
(iv) State the value of Spearman's rank correlation coefficient $r _ { s }$ for the data.\\
(v) State whether $r , r _ { s }$, or both or neither is changed when the values of $x$ are replaced by
\begin{enumerate}[label=(\alph*)]
\item $3 x - 2$,
\item $\sqrt { x }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2018 Q7 [11]}}