2007 H2 Mathematics Paper 2 Question 11

Linear Correlation and Regression

Answers

x=66.20.260t{x = 66.2-0.260t}
x=11.7{x=-11.7}
The linear model is unsuitable as it predicts a negative concentration, which is not realistic. Moreover, the scatter diagram suggests that the rate of change of x{x} decreases on average as t{t} increases, which is not consistent with a linear model.
r=0.994{r = -0.994}
As r1,{r\approx -1,} this indicates a strong, negative linear correlation between y{y} and t.{t.}
As r{r} is closer to 1{-1} compared to 0.912,{-0.912, } the lnx=a+bt{\ln x = a + bt} model is a better fit for the data compared to the linear model.
t=155 min{t = 155 \textrm{ min}}

Full solutions

Using a GC, equation of the regression line of x{x} on t:{t:}
x=66.1940.25970tx=66.20.260t (3 sf)  \begin{align*} & x = 66.194 -0.25970 t \\ & x = 66.2-0.260t \textrm{ (3 sf)} \; \blacksquare \\ \end{align*}
When t=300,{t=300, } estimated value of
x=66.1940.25970(300)=11.7 (3 sf)  \begin{align*} x &= 66.194 -0.25970 (300) \\ &= -11.7 \textrm{ (3 sf)} \; \blacksquare \end{align*}
The linear model is unsuitable as it predicts a negative concentration, which is not realistic. Moreover, the scatter diagram suggests that the rate of change of x{x} decreases on average as t{t} increases, which is not consistent with a linear model. {\blacksquare}

(i)

Using a GC, product moment correlation coefficient,
r=0.994 (3 sf)  r = -0.994 \textrm{ (3 sf)} \; \blacksquare
As r1,{r\approx -1,} this indicates a strong, negative linear correlation between y{y} and t.{t.}
As r{r} is closer to 1{-1} compared to 0.912,{-0.912, } the lnx=a+bt{\ln x = a + bt} model is a better fit for the data compared to the linear model. {\blacksquare}

(ii)

Using a GC, regression line of y{y} on t:{t:}
y=4.62060.012343ty = 4.6206-0.012343t
When x=15,{x=15,}
lnx=4.62060.012343t0.012343t=4.6206ln15t=155 min (3 sf)  \begin{align*} \ln x &= 4.6206 -0.012343 t \\ 0.012343t &= 4.6206 - \ln 15 \\ t &= 155 \textrm{ min (3 sf)} \; \blacksquare \end{align*}