2018 H2 Mathematics Paper 2 Question 9

Linear Correlation and Regression

Answers

(i)

The scatter diagram shows that, as R{R} increases, P{P} increases at a increasing rate, which is not consistent with a linear model.

(ii)

0.969,0.993{0.969, 0.993}
The relationship is modelled better by P=aR2+b{P=aR^2 + b} because the product moment correlation between R2{R^2} and P{P} is closer to 1{1} (0.993{0.993} vs 0.969{0.969}).
P=(2.85×10−8)R2−0.283{P = (2.85 \times 10^{-8})R^2 -0.283}

(iii)

R=6,450{R = 6,450}
The estimate is reliable because
  • P=0.9{P=0.9} lies within the given data range 0.22≤P≤2.64.{0.22 \leq P \leq 2.64.}
  • ∣r∣≈1{|r|\approx 1} which suggest a strong linear correlation between R2{R^2} and P.{P.}

(iv)

P=0.0273{P=0.0273}
The estimate is not reliable because R=3300{R=3300} lies outside the given data range 3600≤R≤9900.{3600 \leq R \leq 9900.} The observed data trend may no longer hold outside of the range.

(v)

P=−0.283+0.000102Rs2{P = -0.283+0.000102R_s^2}

Full solutions

(i)

The scatter diagram shows that, as R{R} increases, P{P} increases at a increasing rate, which is not consistent with a linear model. ■{\blacksquare}

(ii)

Product moment correlation coefficient between R{R} and P:{P:}
r1=0.969 (3 sf)  ■r_1 = 0.969 \textrm{ (3 sf)} \; \blacksquare
Product moment correlation coefficient between R2{R^2} and P:{P:}
r2=0.993 (3 sf)  ■r_2 = 0.993 \textrm{ (3 sf)} \; \blacksquare
Since 0.993{0.993} is closer to 1,{1,} the relationship is modelled better by P=aR2+b.  ■{P=aR^2 + b. \; \blacksquare}
P=(2.8457×10−8)R2−0.28261P=(2.85×10−8)R2−0.283 (3 sf)  ■\begin{align*} & P = (2.8457 \times 10^{-8})R^2 -0.28261 \\ & P = (2.85 \times 10^{-8})R^2 -0.283 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

When P=0.9,{P=0.9,}
0.9=(2.8457×10−8)R2−0.28261R=6,450 rpm (3 sf)  ■\begin{align*} 0.9 &= (2.8457 \times 10^{-8}) R^2 -0.28261 \\ R &= 6,450 \textrm{ rpm (3 sf)} \; \blacksquare \end{align*}
The estimate is reliable because
  • P=0.9{P=0.9} lies within the given data range 0.22≤P≤2.64.  ■{0.22 \leq P \leq 2.64. \; \blacksquare}
  • ∣r∣≈1{|r|\approx 1} which suggest a strong linear correlation between R2{R^2} and P.  ■{P. \; \blacksquare}

(iv)

When R=3300,{R=3300,}
P=(2.8457×10−8)(3300)2−0.28261=0.0273 watts (3 sf)  ■\begin{align*} P &= (2.8457 \times 10^{-8}) (3300)^2 -0.28261 \\ &= 0.0273 \textrm{ watts (3 sf)} \; \blacksquare \end{align*}
The estimate is not reliable because R=3300{R=3300} lies outside the given data range 3600≤R≤9900.{3600 \leq R \leq 9900.} The observed data trend may no longer hold outside of the range. ■{\blacksquare}

(v)

Let Rs{R_s} denote the speed of the fan, in revolutions per second.
1 rps=60 rpmRs rps=60Rs rpmR=60Rs\begin{align*} 1 \textrm{ rps} &= 60 \textrm{ rpm} \\ R_s \textrm{ rps} &= 60R_s \textrm{ rpm} \\ R &= 60 R_s \end{align*}
P=(2.8457×10−8)R2−0.28261P=(2.8457×10−8)(60Rs)2−0.28261P=−0.283+0.000102Rs2  ■\begin{align*} & P = (2.8457 \times 10^{-8})R^2 -0.28261 \\ & P = (2.8457 \times 10^{-8})(60R_s)^2 -0.28261 \\ & P = -0.283+0.000102R_s^2 \; \blacksquare \end{align*}