· H2 Mathematics — Ten-Year Series, worked ← All 11 questions

2021 A-Level · Paper 2 · Question 9

2021 H2 Maths Paper 2, Question 9

Correlation and regression 7 marks transformed models
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Part (a)

1 mark

No — a linear model is not likely to fit well.

Explanation: although r=0.9803r = 0.9803 is close to 11, indicating a strong linear association, the scatter diagram shows the points bending upward as tt increases, a curved pattern rather than a straight line. A high value of rr measures only the strength of a linear relationship; it does not rule out a non-linear model fitting even better, so the curvature visible in the diagram means a straight line is not a good model here.

∴\therefore a linear model is not likely to be a good model for xx against tt.

Part (b)(i)

1 mark

We compute ln⁡x\ln x for each value of xx in the table and plot it against tt.

Sketch for question 9

∴\therefore the points now lie close to a straight line with positive gradient, with none of the curvature seen in the original xx against tt diagram.

Part (b)(ii)

3 marks

By GC, performing linear regression of ln⁡x\ln x on tt,

ln⁡x=c+dt\ln x = c + dt
c≈2.29,d≈0.0573 (3 s.f., by GC)c \approx 2.29,\quad d \approx 0.0573\ (3\ \text{s.f., by GC)}

and the product moment correlation coefficient for this model is

r≈0.999 (3 s.f., by GC).r \approx 0.999\ (3\ \text{s.f., by GC)}.
To check your answer: at t=24t = 24, ln⁡x≈2.293+0.0573(24)=3.68\ln x \approx 2.293 + 0.0573(24) = 3.68, so x≈e3.68≈39.6x \approx e^{3.68} \approx 39.6, close to the given x=39.3x = 39.3. ✓

∴\therefore c≈2.29c \approx 2.29, d≈0.0573d \approx 0.0573, and r≈0.999r \approx 0.999 (3 s.f.).

Part (c)

2 marks

The model ln⁡x=c+dt\ln x = c + dt has r≈0.999r \approx 0.999, closer to 11 than the linear model’s r=0.9803r = 0.9803, and its scatter diagram in (b)(i) shows the points lying close to a straight line with no visible curvature, unlike the original diagram. Since ln⁡x=c+dt\ln x = c + dt is equivalent to x=ecedt=Aedtx = e^{c}e^{dt} = Ae^{dt} (an exponential growth model, where A=ecA = e^{c}), Toby’s second model fits the data much better than his first, straight-line model for xx against tt.

∴\therefore the exponential model x=Aedtx = Ae^{dt}, obtained from ln⁡x=c+dt\ln x = c + dt, is a much better fit than the straight-line model xx against tt.

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