| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Topic | Bivariate data |
| Type | Calculate r from summary statistics |
| Difficulty | Moderate -0.8 This is a straightforward application of the product-moment correlation coefficient formula using given summary statistics. Students need only substitute values into the standard formula r = [nΣxy - ΣxΣy]/√[(nΣx² - (Σx)²)(nΣy² - (Σy)²)], requiring careful arithmetic but no conceptual insight or problem-solving beyond direct recall. |
| Spec | 2.05f Pearson correlation coefficient |
**Question 1** [5 marks]
$S_{xx} = 804.34 - \frac{87.6^2}{10} = 36.964$ — M1: Correct use of formula or equivalent form.
$S_{yy} = 596 - \frac{76.4^2}{10} = 12.304$ — M1: As above.
$S_{xy} = 684.02 - \frac{87.6 \times 76.4}{10} = 14.756$ — M1: As above.
$r = \frac{S_{xy}}{\sqrt{S_{xx} \times S_{yy}}} = 0.69192... \approx 0.692$ (3sf) — M1: As above. A1: c.a.o.
**Total: [5]**1 The information below summarises the percentages of males unemployed ( $x$ ) and the percentages of females unemployed ( $y$ ) in 10 different locations in the UK.
$$n = 10 \quad \Sigma x = 87.6 \quad \Sigma x ^ { 2 } = 804.34 \quad \Sigma y = 76.4 \quad \Sigma y ^ { 2 } = 596 \quad \Sigma x y = 684.02$$
Find the product-moment correlation coefficient for these data.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2015 Q1 [5]}}