2008 H2 Mathematics Paper 2 Question 8

Linear Correlation and Regression

Answers

(i)

r=0.970{r = 0.970}
As r≈1,{r\approx 1,} this indicates a strong, positive linear correlation between t{t} and x,{x, } suggesting that a linear model is appropriate.

(ii)

(iii)

The remaining points may be consistent with a model of the form t=a+bln⁡x{t=a+b\ln x} because, as x{x} increases, t{t} is increasing at a decreasing rate.

(iv)

a=1.42{a = 1.42}
b=4.40{b = 4.40}

(v)

t=8.32{t = 8.32}

(vi)

The prediction will be unreliable as x=8.0{x=8.0} is outside of the given data range 1.2≤x≤6.9{1.2 \leq x \leq 6.9 } as the observed data trend may not hold outside of the range.

Full solutions

(i)

Using a GC, product moment correlation coefficient,
r=0.970 (3 sf)  ■r = 0.970 \textrm{ (3 sf)} \; \blacksquare
As r≈1,{r\approx 1,} this indicates a strong, positive linear correlation between t{t} and x,{x, } suggesting that a linear model is appropriate. ■{\blacksquare}

(ii)

(iii)

The remaining points may be consistent with a model of the form t=a+bln⁡x{t=a+b\ln x} because, as x{x} increases, t{t} is increasing at a decreasing rate. ■{\blacksquare}

(iv)

Using a GC, least squares estimates of
a=1.4247=1.42 (3 sf)  ■b=4.3966=4.40 (3 sf)  ■\begin{align*} a &= 1.4247 \\ & = 1.42 \textrm{ (3 sf)} \; \blacksquare \\ b &= 4.3966 \\ & = 4.40 \textrm{ (3 sf)} \; \blacksquare \\ \end{align*}

(v)

t=1.4247+4.3966ln⁡4.8=8.32 (3 sf)  ■\begin{align*} t & = 1.4247 + 4.3966 \ln 4.8 \\ & = 8.32 \textrm{ (3 sf)} \; \blacksquare \\ \end{align*}

(vi)

The prediction will be unreliable as x=8.0{x=8.0} is outside of the given data range 1.2≤x≤6.9{1.2 \leq x \leq 6.9 } as the observed data trend may not hold outside of the range. ■{\blacksquare}