Edexcel A-Level Mathematics AS S1.4 Correlation and Regression Questions

f=10.8+0.748(62.4)=57.4752
Mean final-exam mark = awrt 57.5

(a)

For each additional mark scored on the pre-test, the average mark on the final exam
increases by 0.748

B1 (1)

$\mathrm{Al}^{\mathrm{P}_{3}} \mathrm{O}_{2}$
(2)
(1)

The statement is not reliable as there is no data below 19 (extrapolation).

B1 (1)

76

$S_{pp}=15573.76$ (no change)
$\sum pf=133486-2842+9016=139660$$\sum pf$ increases by 98(92-29)=6174$\sum f=57.47\times34+(92-29)\approx2017$$\frac{\sum p\sum f}{n}$ increases by $\frac{2120(92-29)}{34}=3928.235\ldots$$S_{pf}$ increases by $6174-3928.235=2245.764\ldots$$b=\frac{11648.35+2245.764\ldots}{15573.76}=0.89\ldots$
awrt 0.9

$$S_{hw}=604135-\frac{1160\times4160}{8}$$
or
$$S_{ww}=2166600-\frac{4160^2}{8}$$$$S_{hw}=935$$$$S_{ww}=3400$$

$$r=\frac{935}{\sqrt{286\times3400}}=0.9481\ldots$$
awrt 0.948

$$b=\frac{935}{286}=3.269\ldots$$$$a=\frac{4160}{8}-3.269\ldots\times\frac{1160}{8}=45.96\ldots$$w=46.0+3.27h
M1, M1, A1*

On average, for each 1 cm increase in height, the weight increases by 3.27 kg.

The first estimate is more reliable since, for example:
- it is an interpolation, while the second estimate is an extrapolation;
- 153 cm is within the data range, while 170 cm is not;
- 153 cm is closer to the mean height than 170 cm.

(c)

= 16.946

M1

Need $\mathrm{S}_{x y}$ so use b so $\mathrm{S}_{x y}=b \times \mathrm{S}_{x x}=0.136 \times 3818$ or 519.248
$[r=] \frac{0.136 \times " 3818^{\prime \prime}}{\sqrt{" 3818^{\prime \prime} \times " 85.1 "}}$$=0.9109448 \ldots$

awrt $\underline{0.911}$
(4)

Since (new x=90 and [original or] new $\bar{x}=90$ ) the term ( $x-\bar{x}$ ) will be 0
Therefore (the $11{ }^{\text {th }}$ shop makes no change) $\mathrm{S}_{x y}$ stays the same}

M1

(2)

(e)

$\mathrm{S}_{x x}$ will be the same so b will be the same

Equation is $\underline{\boldsymbol{y}=\mathbf{5. 7 6 6}+\mathbf{0. 1 3 6 x}}$

A1

x=300 is outside the range $300 \gg 90[300 \gg 90+3 \sigma=90+3 \times 18.63.. \approx 146]$

So not suitable (since involves extrapolation) (o.e.)

(3)
(1)

Notes

(a)

M1 for substituting x=80 into the given equation

A1 for awrt $ 16900 (or better)(allow "16.9 thousand dollars"). Must have some units. Condone y=

(b)

M1 for a correct expression for either (can be implied by sight of either correct answer)

$1^{\text {st }}$ A1 for 85.1

$2^{\text {nd }}$ A1 for 3818 or accept 3820

(c)

$1^{\text {st }}$ M1 for an attempt to use gradient of regression line to find $\mathrm{S}_{x y}$$1^{\text {st }}$ A1 for awrt 519

$2^{\text {nd }}$ M1 for a correct expression using their values (M0 if $\mathrm{S}_{x y}=900 \times 183=164700$ )

$2^{\text {nd }}$ A1 for awrt 0.911

(d)

M1 for stating or showing [old or] new $\bar{x}=90$ (new x=90 implied) or stating that $(x-\bar{x})$ term =0

A1 for a fully correct argument mentioning new $x=\bar{x}=90$ and that extra $(x-\bar{x})$ term =0

Condone using $\bar{y}=18.3$ instead of 18

(e)

$1^{\text {st }}$ M1 for a correct statement about $\mathrm{S}_{x x}$ or b (may be implied by 0.136 used correctly)

$2^{\text {nd }}$ M1 for a correct value for new $\bar{y}$ (calculation may be seen in (d) scores here when 18 is used)

A1 for y=5.766 (or awrt 5.77 or awrt 5.76 ) +0.136 x (correct equation scores 3 / 3 )

B1 for suitable comparison (must see 300 vs 90 or 3000 vs 900 ) that says or implies that 300 will be
outside the range and therefore not suitable. Not sufficient to just say "larger"

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