2020 H2 Mathematics Paper 2 Question 7

Linear Correlation and Regression

Answers

(i)

r1=−0.848{r_1 = -0.848}

(ii)

a=145{a=145}
b=9,460{b=9,460}
r=0.941{r=0.941}

(iii)

Lim's model does not fit this additional data as the model predicts that d=−35.2{d=-35.2} when t=86{t=86}

(iv)

d=−145+47,3009T+160{d = -145 + \frac{47,300}{9T+160}}

Full solutions

(i)

Using a GC, product moment correlation coefficient between d{d} and t:{t:}
r1=−0.848 (3 sf)  ■r_1 = -0.848 \textrm{ (3 sf)} \; \blacksquare

(ii)

Using a GC, least square regression line of d{d} on u:{u:}
d=−145.13+9456.4u−145+9,460u (3 sf)\begin{align*} & d = -145.13+9456.4u \\ & -145 + 9,460u \textrm{ (3 sf)} \end{align*}
a=145  ■b=9,460  ■\begin{align*} a &= 145 \; \blacksquare \\ b &= 9,460 \; \blacksquare \end{align*}
Product moment correlation coefficient between d{d} and u:{u:}
r=0.941 (3 sf)  ■r = 0.941 \textrm{ (3 sf)} \; \blacksquare

(iii)

When t=86,{t=86,}
d=−145.13+9456.4(86)=−35.2 days (3 sf)\begin{align*} d &= -145.13 + 9456.4 (86) \\ &= -35.2 \textrm{ days (3 sf)} \end{align*}
Hence Lim's model does not fit this additional data. ■{\blacksquare}

(iv)

C=59(F−32)F=95C+32t=95T+32\begin{gather*} C = \frac{5}{9}(F-32) \\ F = \frac{9}{5}C + 32 \\ t = \frac{9}{5}T + 32 \end{gather*}
d=−145.13+9456.4ud=−145.13+9456.4td=−145.13+9456.495T+32d=−145+47,3009T+160 (3 sf)  ■\begin{align*} & d = -145.13+9456.4u \\ & d = -145.13 + \frac{9456.4}{t} \\ & d = -145.13 + \frac{9456.4}{\frac{9}{5}T+32} \\ & d = -145 + \frac{47,300}{9T+160} \textrm{ (3 sf)} \; \blacksquare \end{align*}