Specimen 2017 H2 Mathematics Paper 2 Question 6

Linear Correlation and Regression

Answers

As d{d} increases, m{m} increases at an increasing rate so they should not be modelled by a linear equation y=ax+b.{y=ax+b.}
For m=ed2+f,{m=ed^2 + f,} r=0.98889.{r=0.98889.}
For m=gd3+h,{m=gd^3+h,} r=0.99951.{r=0.99951.}
Since the r-value{|r|\textrm{-value}} is closer to 1,{1,} m=gd3+h{m=gd^3+h} is the better model.

Least square regression line of m{m} on d3:{d^3:} m=3.74+0.572d3.{m = 3.74+0.572d^3.}

(iiia)
127 kg.{127 \textrm{ kg}.}
(iiib)
992 kg.{992 \textrm{ kg}.}
The answer for (a) is more reliable as the top length of 6 m{6 \textrm{ m}} lies within the given data range 2.31d9.17{2.31 \leq d \leq 9.17} while the top length of 12 m{12 \textrm{ m}} lies outside the given data range.

Full solutions

(i)

As d{d} increases, m{m} increases at an increasing rate so they should not be modeled by a linear equation y=ax+b  {y=ax+b \; \blacksquare}

(ii)

For m=ed2+f,{m=ed^2 + f,} r=0.98889.{r=0.98889.}
For m=gd3+h,{m=gd^3+h,} r=0.99951.{r=0.99951.}

Since the r-value{|r|\textrm{-value}} is closer to 1,{1,} the better model is m=gd3+h  {m=gd^3+h \; \blacksquare}

Least square regression line of m{m} on d3:{d^3:}

m=3.74+0.572d3  m = 3.74+0.572d^3 \; \blacksquare
(iiia)
m=0.57165(6)3+3.7431=127 (3 sf) kg  \begin{align*} m &= 0.57165(6)^3 + 3.7431 \\ &= 127 \textrm{ (3 sf) kg} \; \blacksquare \end{align*}
(iiib)
m=0.57165(12)3+3.7431=992 (3 sf) kg  \begin{align*} m &= 0.57165(12)^3 + 3.7431 \\ &= 992 \textrm{ (3 sf) kg} \; \blacksquare \end{align*}
The answer for (a) is more reliable as the top length of 6 m{6 \textrm{ m}} lies within the given data range 2.31d9.17{2.31 \leq d \leq 9.17} while the top length of 12 m{12 \textrm{ m}} lies outside the given data range.
d{d}
m{m}
2.31{2.31}
9.17{9.17}
11{11}
449{449}