| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Bivariate data |
| Type | Calculate r from summary statistics |
| Difficulty | Moderate -0.8 This is a routine S1 bivariate data question requiring standard formula application: calculating Syy using the computational formula, finding PMCC using given summary statistics, and performing simple linear regression. All steps are direct substitutions into well-practiced formulas with no problem-solving or conceptual challenges beyond basic interpretation. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(S_{xy} = 81938.5 - \frac{2015^2}{50} \left[= 734\right]\) * | B1* | Answer is given so a correct numerical expression and no incorrect working seen |
| (a)(ii) \(r = \frac{219.55}{\sqrt{734 \times 72.25}} = 0.95338...\) | M1 A1 | For use of \(\frac{S_{xy}}{\sqrt{S_{xx} \times S_{yy}}}\) May be implied by awrt 0.953 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) e.g. [In general] the longer the rabbit the greater the weight | B1ft | A correct interpretation ft their \(r\) value (provided that \( |
| (c) Consistent/Yes as r/PMCC is close to 1 | B1ft | A correct statement with a correct reason ft their \(r\) value (provided that \( |
| (d) \(b = \frac{219.55}{734} = 0.2991...\) | M1 A1 | A correct method to find the gradient (May be implied by awrt 0.299 or \(\frac{4016}{150}\)) |
| \(a = \left(\frac{125}{50}\right) - 'b' \left(\frac{2015}{50}\right) \left[= -9.554...\right]\) | M1 | A correct method to find the intercept ft their \(b\) (May be implied by awrt –9.55) |
| \(w = -9.55 + 0.299y\) | A1 | For \(w = \)(awrt) \(-9.55 + \)(awrt)\(0.299y\) Must be seen in part (d) |
| (e) '-9.55'+0.299'×'45 = 3.905 | M1 A1ft | For substitution of 45 into their regression equation |
| awrt 3.91 | For awrt 3.91 or ft their regression line, provided their final answer is > 0. If regression line is not correct, then you will need to check their answer |
**(a)(i)** $S_{xy} = 81938.5 - \frac{2015^2}{50} \left[= 734\right]$ * | B1* | Answer is given so a correct numerical expression and no incorrect working seen
**(a)(ii)** $r = \frac{219.55}{\sqrt{734 \times 72.25}} = 0.95338...$ | M1 A1 | For use of $\frac{S_{xy}}{\sqrt{S_{xx} \times S_{yy}}}$ May be implied by awrt 0.953
awrt 0.953
**(b)** e.g. [In general] the longer the rabbit the greater the weight | B1ft | A correct interpretation ft their $r$ value (provided that $|r| < 1$) e.g length/y increases the weight/w increases Ignore any figures quoted. Do not accept comments about correlation on their own
**(c)** Consistent/Yes as r/PMCC is close to 1 | B1ft | A correct statement with a correct reason ft their $r$ value (provided that $|r| < 1$) Allow 0.953 ≈ 1
**(d)** $b = \frac{219.55}{734} = 0.2991...$ | M1 A1 | A correct method to find the gradient (May be implied by awrt 0.299 or $\frac{4016}{150}$)
$a = \left(\frac{125}{50}\right) - 'b' \left(\frac{2015}{50}\right) \left[= -9.554...\right]$ | M1 | A correct method to find the intercept ft their $b$ (May be implied by awrt –9.55)
$w = -9.55 + 0.299y$ | A1 | For $w = $(awrt) $-9.55 + $(awrt)$0.299y$ Must be seen in part (d)
**(e)** '-9.55'+0.299'×'45 = 3.905 | M1 A1ft | For substitution of 45 into their regression equation
awrt 3.91 | | For awrt 3.91 or ft their regression line, provided their final answer is > 0. If regression line is not correct, then you will need to check their answer
**Total: 11 marks**\begin{enumerate}
\item A biologist records the length, $y \mathrm {~cm}$, and the weight, $w \mathrm {~kg}$, of 50 rabbits. The following summary statistics are calculated from these data.
\end{enumerate}
$$\sum y = 2015 \quad \sum y ^ { 2 } = 81938.5 \quad \sum w = 125 \quad \mathrm {~S} _ { w w } = 72.25 \quad \mathrm {~S} _ { y w } = 219.55$$
(a) (i) Show that $\mathrm { S } _ { y y } = 734$\\
(ii) Calculate the product moment correlation coefficient for these data. Give your answer to 3 decimal places.\\
(b) Interpret your value of the product moment correlation coefficient.
The biologist believes that a linear regression model may be appropriate to describe these data.\\
(c) State, with a reason, whether or not your value of the product moment correlation coefficient is consistent with the biologist’s belief.\\
(d) Find the equation of the regression line of $w$ on $y$, giving your answer in the form $w = a + b y$
Jeff has a pet rabbit of length 45 cm .\\
(e) Use your regression equation to estimate the weight of Jeff's rabbit.
\hfill \mbox{\textit{Edexcel S1 2024 Q2}}