20N.1.AHL.TZ0.F_13

pestleMathematicsAIHLPaper 120N· ahl-4-17-poisson-distributionsource ↗

Observations on $12$ pairs of values of the random variables $X, Y$ yielded the following results.

$\sum x = 76.3, \sum x^{2} = 563.7, \sum y = 72.2, \sum y^{2} = 460.1, \sum x y = 495.4$

Calculate the value of $r$ , the product moment correlation coefficient of the sample.

[3]

a.i.

Assuming that the distribution of $X, Y$ is bivariate normal with product moment correlation coefficient $ρ$ , calculate the $p$ -value of your result when testing the hypotheses $H_{0} : ρ = 0; H_{1} : ρ > 0$ .

[3]

a.ii.

State whether your $p$ -value suggests that $X$ and $Y$ are independent.

[1]

a.iii.

Given a further value $x = 5.2$ from the distribution of $X$ , $Y$ , predict the corresponding value of $y$ . Give your answer to one decimal place.

[3]

Markscheme / solution

use of

$r = \frac{\sum x y - n \bar{x} \bar{y}}{\sqrt{(\sum x^{2} - n {\bar{x}}^{2}) (\sum y^{2} - n {\bar{y}}^{2})}}$ M1

$= \frac{495.4 - 12 \times \frac{76.3}{12} \times \frac{72.2}{12}}{\sqrt{(563.7 - \frac{12 \times 76 . 3^{2}}{12^{2}}) (460.1 - \frac{12 \times 72 . 2^{2}}{12^{2}})}}$ A1

$= 0.809$ A1

Note: Accept any answer that rounds to $0.81$ .

[3 marks]

a.i.

$t = 0.80856 \dots \sqrt{\frac{10}{1 - 0.80856 \dots^{2}}}$ (M1)

$= 4.345 \dots$ A1

$p$ -value $= 7.27 \times 10^{- 4}$ A1

Note: Accept any answer that rounds to $7.2$ or $7.3 \times 10^{- 4}$ .

Note: Follow through their $p$ -value

[3 marks]

a.ii.

this value indicates that $X, Y$ are not independent A1

[1 mark]

a.iii.

use of

$y - \bar{y} = \frac{\sum x y - n \bar{x} \bar{y}}{\sum x^{2} - n {\bar{x}}^{2}} (x - \bar{x})$ M1

$y - \frac{72.2}{12} = (\frac{495.4 - 12 \times \frac{76.3}{12} \times \frac{72.2}{12}}{563.7 - 12 \times \frac{76 . 3^{2}}{12^{2}}}) (x - \frac{76.3}{12})$ A1

putting $x = 5.2$ gives $y = 5.5$ A1

[3 marks]

Examiners’ report

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a.i.

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a.ii.

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a.iii.

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