| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Pearson’s product-moment correlation coefficient |
| Type | Two-tailed test for any correlation |
| Difficulty | Standard +0.3 Part (i) is a standard two-tailed hypothesis test requiring table lookup with n=15 and comparison to 0.430. Part (ii) requires working backwards from critical values to find minimum sample size, which adds modest problem-solving but remains a routine Further Maths Statistics exercise with clear methodology. |
| Spec | 5.08d Hypothesis test: Pearson correlation |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): \(\rho = 0\), \(H_1\): \(\rho \neq 0\) | B1 | State both hypotheses (B0 for \(r\ldots\)) |
| \(r_{15,\,10\%} = 0.441\) | *B1 | State or use correct tabular two-tail \(r\)-value |
| Accept \(H_0\) if \(0.430 <\) tab. \(r\)-value (AEF) | M1 | State or imply valid method for conclusion |
| OR: \(t_r = r\sqrt{(n-2)/(1-r^2)} = 1.72\), \(t_{13,\,0.95} = 1.771\) | (*B1) | (Rarely seen) |
| Accept \(H_0\) if \( | t_r | <\) tab. \(T\)-value (AEF) |
| No [non-zero] correlation (AEF) | A1 | Correct conclusion (dep *B1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r_{8,\,5\%} = 0.621\), \(r_{9,\,5\%} = 0.582\) so \(N_{min} = 9\) (or equivalent argument) | M1 A1 | Find \(N_{min}\) from relevant [one-tail] tabular values. SC: Award B1 for stating 9 with inadequate justification |
## Question 6:
### Part 6(i):
$H_0$: $\rho = 0$, $H_1$: $\rho \neq 0$ | B1 | State both hypotheses (B0 for $r\ldots$)
$r_{15,\,10\%} = 0.441$ | *B1 | State or use correct tabular two-tail $r$-value
Accept $H_0$ if $0.430 <$ tab. $r$-value (AEF) | M1 | State or imply valid method for conclusion
OR: $t_r = r\sqrt{(n-2)/(1-r^2)} = 1.72$, $t_{13,\,0.95} = 1.771$ | (*B1) | (Rarely seen)
Accept $H_0$ if $|t_r| <$ tab. $T$-value (AEF) | (M1) |
No [non-zero] correlation (AEF) | A1 | Correct conclusion (dep *B1)
### Part 6(ii):
$r_{8,\,5\%} = 0.621$, $r_{9,\,5\%} = 0.582$ so $N_{min} = 9$ (or equivalent argument) | M1 A1 | Find $N_{min}$ from relevant [one-tail] tabular values. **SC:** Award B1 for stating 9 with inadequate justification
---6 A random sample of 15 observations of pairs of values of two variables gives a product moment correlation coefficient of 0.430 .\\
(i) Test at the $10 \%$ significance level whether there is evidence of non-zero correlation between the variables.\\
A second random sample of $N$ observations gives a product moment correlation coefficient of 0.615 . Using a 5\% significance level, there is evidence of positive correlation between the variables.\\
(ii) Find the least possible value of $N$, justifying your answer.\\
\hfill \mbox{\textit{CAIE FP2 2018 Q6 [6]}}