2011 H2 Mathematics Paper 2 Question 8

Linear Correlation and Regression

Answers

(i)

(ii)

r=−0.992{r=-0.992}
Even though ∣r∣≈−1,{|r| \approx -1, } which suggests a strong linear correlation, there may be other models for the relationship between x{x} and y{y} that will describe the data trend even better and give a product moment correlation with absolute value even closer to 1.{1.}

(iii)

The product moment correlation coefficient with absolute value closer 1{1} will be the better model.
For the y=a+bx2{y=a+bx^2} model, the product moment correlation coefficient is −0.99998.{-0.99998. } Hence it is the better model.

(iv)

y=22.2−0.856x2{y = 22.2-0.856x^2}
y=13.5{y=13.5}

Full solutions

(i)

(ii)

Using a GC, product moment correlation coefficient between x{x} and y:{y:}
r=−0.992 (3 sf)  ■r=-0.992 \textrm{ (3 sf)} \; \blacksquare
Even though ∣r∣≈−1,{|r| \approx -1, } which suggests a strong linear correlation, there may be other models for the relationship between x{x} and y{y} that will describe the data trend even better and give a product moment correlation with absolute value even closer to 1.  ■{1. \; \blacksquare}

(iii)

The product moment correlation coefficient with absolute value closer 1{1} will be the better model. ■{\blacksquare}
For the y=a+bx2{y=a+bx^2} model, the product moment correlation coefficient is −0.99998.{-0.99998. } Hence it is the better model.

(iv)

Using a GC, least squares regression line of y{y } on x2:{x^2:}
y=22.230−0.85621x2y=22.2−0.856x2 (3 sf)  ■\begin{align*} & y = 22.230-0.85621x^2 \\ & y = 22.2-0.856x^2 \textrm{ (3 sf)} \; \blacksquare \end{align*}
When x=3.2,{x=3.2,}
y=22.230−0.85621(3.2)2=13.5 (3 sf)  ■\begin{align*} y &= 22.230 -0.85621 (3.2)^2 \\ &= 13.5 \textrm{ (3 sf)} \; \blacksquare \end{align*}