| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Topic | Bivariate data |
| Type | Calculate r from summary statistics |
| Difficulty | Moderate -0.8 This is a straightforward application of standard formulas for correlation coefficient and regression line using given summary statistics. Part (i) requires substituting into the r formula, part (ii) is a 'show that' verification of regression coefficients, and part (iii) involves simple substitution with routine commentary on extrapolation. All steps are mechanical with no problem-solving or conceptual insight required. |
| Spec | 5.08a Pearson correlation: calculate pmcc5.09b Least squares regression: concepts5.09c Calculate regression line |
**Question 3**
**(i)**
$S_{xy} = 77532 - \frac{1002 \times 1865}{25} = 2782.8$ — M1 (Use of formula for numerator)
$S_{xx} = 43508 - \frac{1002^2}{25} = 3347.84$ — M1 (Use of formula for either term in denominator)
$S_{yy} = 142749 - \frac{1865^2}{25} = 3620$
$r = \frac{2782.8}{\sqrt{3347.84 \times 3620}} = 0.799(36...)$ — M1, A1 **[4]**
Use of formula for $r$. c.a.o.
**(ii)** Form $y = ax + b$
$a = \frac{S_{xy}}{S_{xx}} = \frac{2782.8}{3347.84} = 0.83(122...)$ — M1, A1
$S_{xy}$ and $S_{xx}$ from above. **AG.**
$b = \bar{y} - a\bar{x}$
$\therefore b = \frac{1865}{25} - 0.83122... \times \frac{1002}{25}$ — M1
$= 74.6 - 0.83122... \times 40.08 = 41.28(46...)$ — A1 **[4]**
**AG.** Must be convincing.
Allow M1 for use of $a = 0.83$ to find $b$ ($= 41.33...$), or $b = 41.28$ to find $a$ ($= 0.83133…$), but not both, but do not award the corresponding A mark.
**(iii)**
When $x = 50$, $y = 82.78... \approx 82.8$ — B1 (Accept a.w.r.t. 82.8)
This is ok; it is within the range of the data. — B1
When $x = 65$, $y = 95.23... \approx 95.2$ — B1 (Accept a.w.r.t. 95.2)
This is not ok; it is beyond the range of the data. — B1 **[4]**
At least one of the comments must refer to within/beyond the range of the data. (o.e.)
**Total: [12]**3 At a local athletics club, data on the ages of the members and their times to run a 10 km course are recorded. For a random sample of 25 club members aged between 20 and 60, their ages ( $x$ years) and times ( $y$ minutes) are summarised as follows.
$$n = 25 \quad \Sigma x = 1002 \quad \Sigma x ^ { 2 } = 43508 \quad \Sigma y = 1865 \quad \Sigma y ^ { 2 } = 142749 \quad \Sigma x y = 77532$$
(i) Calculate the product moment correlation coefficient for these data.\\
(ii) Show that the equation of the least squares regression line of $y$ on $x$ is $y = 0.83 x + 41.28$, where the coefficients are given correct to 2 decimal places.\\
(iii) Use the equation given in part (ii) to estimate the time taken by someone who is
\begin{enumerate}[label=(\alph*)]
\item 50 years old,
\item 65 years old.
Comment on the validity of each of these estimates.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2013 Q3 [12]}}