2019 H2 Mathematics Paper 2 Question 10

Linear Correlation and Regression

Answers

(ia)
(ib)
(ic)
1.33{1.33}
(id)
(a,f(a)){(a,f(a))} may be higher or lower than (a,b),{(a,b),} resulting in positive or negative residuals. Using the sum of residuals directly will then result in positive and negative values canceling each other out.
Using the sum of squares of the residuals ensure that all quantities are positive.

(ii)

Bhani's model gives a better fit as the sum of the squares of the residuals is smaller.

(iii)

(50,18.6){( 50, 18.6 )}

(iv)

y=33.6−0.300x{y = 33.6-0.300x}
r=−0.985{r=-0.985}

(v)

y=24.6{y=24.6}
We do not expect this estimate to be reliable because x=30{x=30} lies outside the given data range 40≤x≤60.{40 \leq x \leq 60.} The observed data trend may no longer hold outside of the range.

(vi)

The data points in Cerie's experiment are collinear and all lie on her line.

Full solutions

(ia)
(ib)
(ic)
Sum of squares of residuals=(22−(35−13(40)))2+…+(16−(35−13(60)))2=1.33 (3 sf)  ■\begin{align*} & \textrm{Sum of squares of residuals} \\ & = \left( 22- \left( 35-{\textstyle \frac{1}{3}}(40) \right) \right)^2 + \ldots + \left( 16- \left( 35-{\textstyle \frac{1}{3}}(60) \right) \right)^2 \\ & = 1.33 \textrm{ (3 sf)} \; \blacksquare \end{align*}
(id)
(a,f(a)){(a,f(a))} may be higher or lower than (a,b),{(a,b),} resulting in positive or negative residuals. Using the sum of residuals directly will then result in positive and negative values canceling each other out.
Using the sum of squares of the residuals ensure that all quantities are positive. ■{\blacksquare}

(ii)

Bhani's model gives a better fit as the sum of the squares of the residuals is smaller. ■{\blacksquare}

(iii)

The least squares regression line of y{y} on x{x} must pass through (x‾,y‾):{(\overline{x}, \overline{y} ):}
(50,18.6)  ■( 50, 18.6 ) \; \blacksquare

(iv)

Using a GC, least squares regression line of y{y} on x:{x:}
y=33.600−0.30000xy=33.6−0.300x (3 sf)  ■\begin{align*} & y = 33.600-0.30000x \\ & y = 33.6-0.300x \textrm{ (3 sf)} \; \blacksquare \end{align*}
Product moment correlation coefficient:
r=−0.985 (3 sf)  ■r=-0.985 \textrm{ (3 sf)} \; \blacksquare

(v)

When x=30,{x=30,}
Fuel consumption=33.600−0.30000(30)=24.6 km/L (3 sf)  ■\begin{align*} & \textrm{Fuel consumption} \\ &= 33.600 -0.30000 (30) \\ &= 24.6 \textrm{ km/L (3 sf)} \; \blacksquare \end{align*}
The estimate is not reliable because x=30{x=30} lies outside the given data range 40≤x≤60.{40 \leq x \leq 60.} The observed data trend may no longer hold outside of the range. ■{\blacksquare}

(vi)

The data points in Cerie's experiment are collinear and all lie on her line. ■{\blacksquare}