2012 H2 Mathematics Paper 2 Question 8

Linear Correlation and Regression

Answers

(i)

(ii)

The practice paper on week 5 may be more difficult that usual

(iii)

A linear model will predict that her percentage mark will increase at the same rate every week. This is inappropriate because it will eventually predict a percentage mark greater than 100%{100\%} on future papers which is not realistic.
A quadratic model will predict a maximum percentage mark, after which the percentage mark on future papers will start to decrease. This is not appropriate as her percentage mark should continue to increase as she practices more each week.

(iv)

r=−0.929744{r=-0.929744}

(v)

92{92} is the most appropriate value of L{L} because the product moment correlation coefficient is closest to −1,{-1, } suggesting the strongest linear correlation between ln⁡(L−y){\ln(L-y)} and x{x} when L=92{L=92}

(vi)

a=4.10{a=4.10}
b=−0.280{b=-0.280}
Week 13{13}

(vii)

The value of L{L} represent Amy's maximum percentage mark she will approach eventually in the long term

Full solutions

(i)

(ii)

The practice paper on week 5 may be more difficult that usual ■{\blacksquare}

(iii)

A linear model will predict that her percentage mark will increase at the same rate every week. This is inappropriate because it will eventually predict a percentage mark greater than 100%{100\%} on future papers which is not realistic. ■{\blacksquare}
A quadratic model will predict a maximum percentage mark, after which the percentage mark on future papers will start to decrease. This is not appropriate as her percentage mark should continue to increase as she practices more each week. ■{\blacksquare}

(iv)

r=−0.929744 (6 dp)  ■r=-0.929744 \textrm{ (6 dp)} \; \blacksquare

(v)

92{92} is the most appropriate value of L{L} because the product moment correlation coefficient is closest to −1,{-1, } suggesting the strongest linear correlation between ln⁡(L−y){\ln(L-y)} and x{x} when L=92  ■{L=92 \; \blacksquare}

(vi)

Using a GC, least squares regression line of ln⁡(92−y){\ln (92-y) } on x:{x:}
ln⁡(92−y)=4.1045−0.27960x\ln (92 - y) = 4.1045-0.27960x
a=4.1045=4.10 (3 sf)  ■b=−0.27960=−0.280 (3 sf)  ■\begin{align*} a &= 4.1045 \\ &= 4.10 \textrm{ (3 sf)} \; \blacksquare \\ b &= -0.27960 \\ &= -0.280 \textrm{ (3 sf)} \; \blacksquare \end{align*}
When y=90,{y=90,}
ln⁡(92−90)=4.1045−0.27960xx=12.2\begin{align*} \ln (92-90) &= 4.1045 -0.27960 x \\ x &= 12.2 \end{align*}
Hence Amy is predicted to obtain her first mark of at least 90%{90\%} on week 13  ■{13 \; \blacksquare}

(vii)

The value of L{L} represent Amy's maximum percentage mark she will approach eventually in the long term ■{\blacksquare}