| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of Pearson’s product-moment correlation coefficient |
| Type | One-tailed test for positive correlation |
| Difficulty | Standard +0.3 This is a straightforward hypothesis test question requiring standard procedure (state H₀ and H₁, compare r=0.446 to critical value from tables for n=24 at 5% level), followed by interpretation of correlation improvement and basic log manipulation. All parts are routine textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.05g Hypothesis test using Pearson's r5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \rho = 0 \quad H_1: \rho > 0\) | B1 | Both hypotheses correct in terms of \(\rho\) |
| Critical value 0.3438 | M1 | Sight of 0.3438 or any cv such that \(0.25 < |
| \(0.446 > 0.3438\) so evidence pmcc is greater than 0/positive correlation | A1 | Mentions "pmcc/correlation/relationship" and "greater than 0/positive" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Value is close(r) to 1 or there is strong(er) positive correlation | B1 | Do not allow 'association' |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_{10} y = -1.82 + 0.89(\log_{10} x)\) leading to \(y = ax^n\) form | M1 | Correct substitution for both \(c\) and \(m\) (may be implied by 2nd M1) |
| \(y = 10^{-1.82 + 0.89(\log_{10} x)}\) | M1 | Making \(y\) subject to give \(y = 10^{a+b(\log_{10} x)}\) |
| \(y = 10^{-1.82} \times 10^{0.89(\log_{10} x)}\) \([= 10^{-1.82} \times 10^{(\log_{10} x)^{0.89}}]\) | M1 | Correct multiplication: \(y = 10^a \times 10^{b(\log_{10} x)}\) |
| \(y = 0.015x^{0.89}\) | A1A1 | \(n = 0.89\) and \(a =\) awrt \(0.015\); do not award final A1 if in incorrect form e.g. \(y = 0.015 + x^{0.89}\) |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \rho = 0 \quad H_1: \rho > 0$ | B1 | Both hypotheses correct in terms of $\rho$ |
| Critical value 0.3438 | M1 | Sight of 0.3438 or any cv such that $0.25 < |cv| < 0.45$ |
| $0.446 > 0.3438$ so evidence pmcc is greater than 0/positive correlation | A1 | Mentions "pmcc/correlation/relationship" and "greater than 0/positive" |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Value is close(r) to 1 **or** there is strong(er) positive correlation | B1 | Do not allow 'association' |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_{10} y = -1.82 + 0.89(\log_{10} x)$ leading to $y = ax^n$ form | M1 | Correct substitution for both $c$ and $m$ (may be implied by 2nd M1) |
| $y = 10^{-1.82 + 0.89(\log_{10} x)}$ | M1 | Making $y$ subject to give $y = 10^{a+b(\log_{10} x)}$ |
| $y = 10^{-1.82} \times 10^{0.89(\log_{10} x)}$ $[= 10^{-1.82} \times 10^{(\log_{10} x)^{0.89}}]$ | M1 | Correct multiplication: $y = 10^a \times 10^{b(\log_{10} x)}$ |
| $y = 0.015x^{0.89}$ | A1A1 | $n = 0.89$ **and** $a =$ awrt $0.015$; do not award final A1 if in incorrect form e.g. $y = 0.015 + x^{0.89}$ |
---3. Barbara is investigating the relationship between average income (GDP per capita), $x$ US dollars, and average annual carbon dioxide ( $\mathrm { CO } _ { 2 }$ ) emissions, $y$ tonnes, for different countries.
She takes a random sample of 24 countries and finds the product moment correlation coefficient between average annual $\mathrm { CO } _ { 2 }$ emissions and average income to be 0.446
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test, at the $5 \%$ level of significance, whether or not the product moment correlation coefficient for all countries is greater than zero.
Barbara believes that a non-linear model would be a better fit to the data.\\
She codes the data using the coding $m = \log _ { 10 } x$ and $c = \log _ { 10 } y$ and obtains the model $c = - 1.82 + 0.89 m$
The product moment correlation coefficient between $c$ and $m$ is found to be 0.882
\item Explain how this value supports Barbara's belief.
\item Show that the relationship between $y$ and $x$ can be written in the form $y = a x ^ { n }$ where $a$ and $n$ are constants to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 2019 Q3 [9]}}