2024 H2 Mathematics Paper 2 Question 11

Linear Correlation and Regression

Answers

The scatter diagram shows that, as $m$ increases, $p$ increases at a decreasing rate on average. This indicate a likely non-linear relationship between $m$ and $p .$

(ii)

$p = d + e \sqrt{m}$
$r = 0.991.$

(iii)

$510 cents .$

Since $m = 750$ is within the given data range $24 \leq m \leq 5000,$ and the $r -value = 0.991$ is close to $1,$ our estimate is reliable.

(b)

(i)

Sketch.

(ii)

The residuals can be negative or positive depending on whether the point is above or below the regression line. Hence, using the sum of the squares of the residuals ensure the error values used in assessing the fit of the model are positive.
Sum of squares = $\frac{107}{4} .$

(iii)

For a line that is a better fit for the data, the sum of the squares of the residuals will be smaller.

(iv)

$- \frac{6}{5} < c < 1.$

Full Solutions

(a)

(i)

The scatter diagram shows that, as $m$ increases, $p$ increases at a decreasing rate on average. This indicate a likely non-linear relationship between $m$ and $p$ $∎$

(ii)

The $p = d + e \sqrt{m}$ model gives a better fit to the data as it has an $r -value$ of $0.991 ∎$ which is closer to $1$ than the $r -value$ of $0.929$ for the $p = a + b \ln m$ model

p = 32.953 + 17.270 \sqrt{m}

\begin{aligned} d & = 33.0 (3 s.f.) ∎ \\ e & = 17.3 (3 s.f.) ∎ \end{aligned}

(iii)

When $m = 750,$

\begin{aligned} p & = 32.953 + 17.270 \sqrt{750} \\ = 505.91 \\ = 510 cents (nearest 10 cents) ∎ \end{aligned}

Since $m = 750$ is within the given data range $24 \leq m \leq 5000,$ and the $r -value = 0.991$ is close to $1,$ our estimate is reliable $∎$

(b)

(i)

scatter-plot

(ii)

The residuals can be negative or positive depending on whether the point is above or below the regression line. Hence, using the sum of the squares of the residuals ensure the error values used in assessing the fit of the model are positive $∎$

$n$	2	5	6	7	11
$y$	6	10	16	20	25
$f (n) = \frac{5}{2} n + 1$	$6$	$\frac{27}{2}$	$16$	$\frac{37}{2}$	$\frac{57}{2}$
$Residuals = y - f (n)$	$0$	$- \frac{7}{2}$	$0$	$\frac{3}{2}$	$- \frac{7}{2}$

\begin{aligned} Sum of squares \\ = 0^{2} + {(- \frac{7}{2})}^{2} + 0^{2} + {\frac{3}{2}}^{2} + {(- \frac{7}{2})}^{2} \\ = \frac{107}{4} ∎ \end{aligned}

(iii)

For a line that is a better fit for the data, the sum of the squares of the residuals will be smaller $∎$

(iv)

$n$	2	5	6	7	11
$y$	6	10	16	20	25
$f (n) = \frac{5}{2} n + c$	$5 + c$	$\frac{25}{2} + c$	$15 + c$	$\frac{35}{2} + c$	$\frac{55}{2} + c$
$Residuals = y - f (n)$	$1 - c$	$- \frac{5}{2} - c$	$1 - c$	$\frac{5}{2} - c$	$- \frac{5}{2} - c$

Considering the sum of squares,

\begin{matrix} {(1 - c)}^{2} + {(- \frac{5}{2} - c)}^{2} + {(1 - c)}^{2} + {(\frac{5}{2} - c)}^{2} + {(- \frac{5}{2} - c)}^{2} < \frac{107}{4} \\ 5 c^{2} + c + \frac{83}{4} < \frac{107}{4} \\ 5 c^{2} + c - 6 < 0 \\ (5 c + 6) (c - 1) < 0 \\ - \frac{6}{5} < c < 1 ∎ \end{matrix}