| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate PMCC from summary statistics |
| Difficulty | Standard +0.3 This is a straightforward application of standard formulas from linear regression. Part (a) requires substituting given summary statistics into the PMCC formula r = S_sh/√(S_hh × S_ss), part (b) is direct substitution into the regression equation, and part (c) requires understanding that a point on the regression line doesn't change the gradient. All techniques are routine for Further Statistics students with no problem-solving or novel insight required. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{ss} = \frac{S_{sh}}{b} = \frac{0.352}{0.919} = 0.383\ldots\) or \(\frac{352}{919}\) | M1 | Realising the need to find \(S_{ss}\). Must have \(0.919 = \frac{0.352}{S_{ss}}\) or better |
| \(r = \frac{0.352}{\sqrt{0.377 \times 0.383}}\) | M1 | Attempt to find \(r\) ft their \(S_{ss}\) provided \(S_{ss} > 0.33\) |
| \(r = 0.9263\ldots\) awrt 0.926 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h - 1.68 = 0.919(1.79 - 1.70)\) | M1 | Using model with means and 1.79 to find \(h\). May see: \(0.1177 + 0.919 \times 1.79\); allow \(0.1177 + 0.919 \times 1.75\) as MR |
| \(h = 1.76271\ldots\) awrt 1.76 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_{ss}\) would remain the same since \(s_{25} = \bar{s}\) | M1 | Explaining the effect on \(S_{ss}\) |
| \(S_{sh}\) would remain the same; additional \((s - \bar{s})(h - \bar{h}) = 0\) | M1 | Considering the effect on \(S_{sh}\) |
| So the gradient would be unchanged. | A1 | Correct deduction. ALT: 1st M1 explaining \(s_{25} = \bar{s}\) so point lies on \(s = \bar{s}\); 2nd M1 \(h_{25} > \bar{h}\) so line would move upwards (parallel to original regression line) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{ss} = \frac{S_{sh}}{b} = \frac{0.352}{0.919} = 0.383\ldots$ or $\frac{352}{919}$ | M1 | Realising the need to find $S_{ss}$. Must have $0.919 = \frac{0.352}{S_{ss}}$ or better |
| $r = \frac{0.352}{\sqrt{0.377 \times 0.383}}$ | M1 | Attempt to find $r$ ft their $S_{ss}$ provided $S_{ss} > 0.33$ |
| $r = 0.9263\ldots$ awrt **0.926** | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h - 1.68 = 0.919(1.79 - 1.70)$ | M1 | Using model with means and 1.79 to find $h$. May see: $0.1177 + 0.919 \times 1.79$; allow $0.1177 + 0.919 \times 1.75$ as MR |
| $h = 1.76271\ldots$ awrt **1.76** | A1 | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{ss}$ would remain the same since $s_{25} = \bar{s}$ | M1 | Explaining the effect on $S_{ss}$ |
| $S_{sh}$ would remain the same; additional $(s - \bar{s})(h - \bar{h}) = 0$ | M1 | Considering the effect on $S_{sh}$ |
| So the gradient would be unchanged. | A1 | Correct deduction. ALT: 1st M1 explaining $s_{25} = \bar{s}$ so point lies on $s = \bar{s}$; 2nd M1 $h_{25} > \bar{h}$ so line would move upwards (parallel to original regression line) |\begin{enumerate}
\item A random sample of 24 adults is taken. The height, $h$ metres, and the arm span, $s$ metres, for each adult are recorded.
\end{enumerate}
These data are summarised below.
$$\mathrm { S } _ { h h } = 0.377 \quad \mathrm {~S} _ { s h } = 0.352 \quad \bar { s } = 1.70 \quad \bar { h } = 1.68$$
The least squares regression line of $h$ on $s$ is
$$h = a + 0.919 s$$
where $a$ is a constant.\\
(a) Calculate the product moment correlation coefficient.
A doctor uses the least squares regression line of $h$ on $s$ as a model to predict a person's height based on their arm span.\\
(b) Use the model to predict the height of an adult with arm span 1.79 metres.
Ewan has an arm span of 1.70 metres and a height of 1.75 metres. His information is added to the sample as the 25th adult.\\
(c) Explain how the gradient of the regression line for the sample of 25 adults compares with the gradient of the regression line for the original sample of 24 adults.\\
Give a reason for your answer.
\hfill \mbox{\textit{Edexcel FS2 AS 2024 Q5 [8]}}