| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Pearson’s product-moment correlation coefficient |
| Type | Calculate PMCC from summary statistics |
| Difficulty | Standard +0.3 This question tests standard formulas and procedures: (a)(i) uses the well-known relationship r = √(b_yx × b_xy), (a)(ii) is a routine hypothesis test with given critical values, and (b) requires looking up or calculating the minimum n for significance. While it requires knowledge of multiple concepts, each step follows textbook procedures without requiring problem-solving insight or novel approaches. |
| Spec | 2.05a Hypothesis testing language: null, alternative, p-value, significance5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = \sqrt{(0.38 \times 0.96)} = 0.604\) | M1* A1 | Find correlation coefficient \(r\) from \(r^2 = b_1b_2\) |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \rho = 0\), \(H_1: \rho > 0\) | B1 | State both hypotheses (B0 for \(r\)...) |
| \(r_{10,\ 5\%} = 0.549\) | *B1 | State or use correct tabular one-tail \(r\)-value |
| Reject \(H_0\) if \( | r | >\) tab. value (AEF) |
| There is positive correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |
| Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r_{16,\ 5\%} = 0.497\) (*or* \(r_{15,\ 5\%} = 0.514\)) | M1 | State or use relevant tabular two-tail \(r\)-value |
| \(n_{\min} = 16\) | A1 | Find least possible value of \(n\) |
| Total: 2 marks | SR: M1 A1 for stating 16 without explanation; B1 for stating 15 without explanation; B1 for finding or stating one-tail result 12 |
## Question 8(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \sqrt{(0.38 \times 0.96)} = 0.604$ | M1* A1 | Find correlation coefficient $r$ from $r^2 = b_1b_2$ |
| **Total: 2 marks** | | |
## Question 8(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \rho = 0$, $H_1: \rho > 0$ | B1 | State both hypotheses (B0 for $r$...) |
| $r_{10,\ 5\%} = 0.549$ | *B1 | State or use correct tabular one-tail $r$-value |
| Reject $H_0$ if $|r| >$ tab. value (AEF) | M1 | State or imply valid method for reaching conclusion |
| There is positive correlation | A1 | Correct conclusion (AEF, dep *A1, *B1) |
| **Total: 4 marks** | | |
## Question 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r_{16,\ 5\%} = 0.497$ (*or* $r_{15,\ 5\%} = 0.514$) | M1 | State or use relevant tabular two-tail $r$-value |
| $n_{\min} = 16$ | A1 | Find least possible value of $n$ |
| **Total: 2 marks** | | SR: M1 A1 for stating 16 without explanation; B1 for stating 15 without explanation; B1 for finding or stating one-tail result 12 |8
\begin{enumerate}[label=(\alph*)]
\item For a random sample of ten pairs of values of $x$ and $y$ taken from a bivariate distribution, the equations of the regression lines of $y$ on $x$ and of $x$ on $y$ are, respectively,
$$y = 0.38 x + 1.41 \quad \text { and } \quad x = 0.96 y + 7.47$$
\begin{enumerate}[label=(\roman*)]
\item Find the value of the product moment correlation coefficient for this sample.
\item Using a $5 \%$ significance level, test whether there is positive correlation between the variables.
\end{enumerate}\item For a random sample of $n$ pairs of values of $u$ and $v$ taken from another bivariate distribution, the value of the product moment correlation coefficient is 0.507 . Using a test at the $5 \%$ significance level, there is evidence of non-zero correlation between the variables. Find the least possible value of $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2015 Q8 [8]}}