| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Calculate r from summary statistics |
| Difficulty | Moderate -0.8 This is a straightforward application of standard S1 formulas for correlation coefficient. All parts involve direct substitution into given formulas (standard deviation, S_yy, S_xy, and r) with no conceptual challenges or problem-solving required. Part (e) requires basic understanding that adding a point inconsistent with the trend weakens correlation, which is standard bookwork knowledge. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.08a Pearson correlation: calculate pmcc |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sigma_x^2 = \frac{985.88}{8} - \left(\frac{86.8}{8}\right)^2 = \frac{985.88}{8} - 10.85^2\) | M1 | Correct expression for st. dev or variance (ignore label) |
| \(\sigma_x = \sqrt{\frac{985.88}{8} - \left(\frac{86.8}{8}\right)^2} = \sqrt{123.235 - 117.7225} = \sqrt{5.5125}\) or \(\sqrt{\frac{44.1}{8}}\) | A1 | Correct expression for st. dev (must have square root) |
| \(= 2.3478\ldots = \text{awrt } \mathbf{2.35}\) | A1 | For awrt 2.35 (allow \(s = 2.5099\ldots\) or awrt 2.51; if \(\sigma^2 = 2.35\) score A0) |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{yy} = 8 \times \sigma_y^2 = 716\) (3 sf) | B1cso | Correct expression or sight of at least 716.08 (limits 716.00–716.16); no circular arguments |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{xy} = 4900.5 - 58 \times 86.8\) or \(4900.5 - \frac{86.8 \times 464}{8}\) | M1 | Correct expression for \(S_{xy}\) (NB \(\Sigma y = 464\)) |
| \(= -\mathbf{133.9}\) (Allow \(-134\)) | A1 | For \(-133.9\) or awrt \(-134\) [No fractions] |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = \frac{-133.9/8}{\sigma_x \times \sigma_y}\) or \(\frac{-133.9}{\sqrt{44.1 \times 716}}\) | M1 | Correct expression for \(r\) (ft their \(S_{xy}\), \(\sigma_x\) or \(S_{xx}\), allow ft \(S_{yy}\)) |
| \(= \text{awrt } \mathbf{-0.753}\) or \(\mathbf{-0.754}\) | A1 | For awrt \(-0.753\) or \(-0.754\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(r < 0\) means high sunshine and low rain; this is high sunshine high rain | B1 | Suitable reason contradicting \(r < 0\) |
| [not in keeping with trend so] \(r\) is closer to 0 or \( | r | \) decreases |
# Question 3:
## Part (a)
| $\sigma_x^2 = \frac{985.88}{8} - \left(\frac{86.8}{8}\right)^2 = \frac{985.88}{8} - 10.85^2$ | M1 | Correct expression for st. dev or variance (ignore label) |
| $\sigma_x = \sqrt{\frac{985.88}{8} - \left(\frac{86.8}{8}\right)^2} = \sqrt{123.235 - 117.7225} = \sqrt{5.5125}$ or $\sqrt{\frac{44.1}{8}}$ | A1 | Correct expression for st. dev (must have square root) |
| $= 2.3478\ldots = \text{awrt } \mathbf{2.35}$ | A1 | For awrt 2.35 (allow $s = 2.5099\ldots$ or awrt 2.51; if $\sigma^2 = 2.35$ score A0) |
## Part (b)
| $S_{yy} = 8 \times \sigma_y^2 = 716$ (3 sf) | B1cso | Correct expression or sight of at least 716.08 (limits 716.00–716.16); no circular arguments |
## Part (c)
| $S_{xy} = 4900.5 - 58 \times 86.8$ or $4900.5 - \frac{86.8 \times 464}{8}$ | M1 | Correct expression for $S_{xy}$ (NB $\Sigma y = 464$) |
| $= -\mathbf{133.9}$ (Allow $-134$) | A1 | For $-133.9$ or awrt $-134$ [No fractions] |
## Part (d)
| $r = \frac{-133.9/8}{\sigma_x \times \sigma_y}$ or $\frac{-133.9}{\sqrt{44.1 \times 716}}$ | M1 | Correct expression for $r$ (ft their $S_{xy}$, $\sigma_x$ or $S_{xx}$, allow ft $S_{yy}$) |
| $= \text{awrt } \mathbf{-0.753}$ or $\mathbf{-0.754}$ | A1 | For awrt $-0.753$ or $-0.754$ |
## Part (e)
| $r < 0$ means high sunshine and low rain; this is high sunshine high rain | B1 | Suitable reason contradicting $r < 0$ |
| [not in keeping with trend so] $r$ is closer to 0 or $|r|$ decreases | B1 | Correct statement about $r$ getting closer to zero; "$r$ increases" is B0 unless they say it gets closer to 0 |
---3. Before going on holiday to Seapron, Tania records the weekly rainfall ( $x \mathrm {~mm}$ ) at Seapron for 8 weeks during the summer. Her results are summarised as
$$\sum x = 86.8 \quad \sum x ^ { 2 } = 985.88$$
\begin{enumerate}[label=(\alph*)]
\item Find the standard deviation, $\sigma _ { x }$, for these data.\\
(3)
Tania also records the number of hours of sunshine ( $y$ hours) per week at Seapron for these 8 weeks and obtains the following
$$\bar { y } = 58 \quad \sigma _ { y } = 9.461 \text { (correct to } 4 \text { significant figures) } \quad \sum x y = 4900.5$$
\item Show that $\mathrm { S } _ { y y } = 716$ (correct to 3 significant figures)
\item Find $\mathrm { S } _ { x y }$
\item Calculate the product moment correlation coefficient, $r$, for these data.
During Tania's week-long holiday at Seapron there are 14 mm of rain and 70 hours of sunshine.
\item State, giving a reason, what the effect of adding this information to the above data would be on the value of the product moment correlation coefficient.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q3 [10]}}