Correlation & Regression: PMCC; scatter diagram — HCI 2025 H2 Math Prelim Paper 2
What this question tests
Question
An ice-cream seller records the monthly ice cream sales, \(s\) thousands dollars for different temperature, \(t\) degrees Celsius during the winter season. The recorded values are shown in the table below.
| \(t\) | 1 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|
| \(s\) | 14 | 15 | 15 | 16 | 18 | 21 | 23 |
(a) It is given that the regression line of \(s\) on \(t\) is \(s=1.125t+11\). Using this regression line, find the sum of the squares of the residuals.
(b) State the coordinates of an additional data point such that, with all 8 data points, the regression line remains the same as in part (a).
(c) Sketch a scatter diagram of \(s\) against \(t\) for the data given in the table.
The following three models are proposed, where \(a\), \(b\), \(c\), \(d\), \(f\) and \(h\) are positive constants.
(A) \(s=at^{2}+b\) (B) \(s=-c\mathrm{e}^{t}+d\) (C) \(s=f\ln(t+h)\)
(d) Explain which of these models give the best fit to the data. State the values of the constants for the chosen model.
A temperature of \(F\) degrees Fahrenheit is equivalent to a temperature of \(C\) degrees Celsius, where \(F=\dfrac{9}{5}C+32\).
(e) Using the model you chose in part (d), re-write the equation so that it can be used to estimate the monthly sales when the temperature, \(T\), is given in degrees Fahrenheit.
Show full worked solution▾
(a) Required sum of square of residuals \(= 14.75\).
(b) Required point \(=(\bar{t},\bar{s})=(5.71,17.4)\).
OR
Any points on the given regression line such as \((0,11)\).
(c) Scatter diagram of \(s\) against \(t\):
Figure 6 — Scatter diagram of \(s\) against \(t\).
(d) Since the data points lie close to a curve with positive gradient and concave upwards, model (A) (\(s=at^{2}+b\)) is most appropriate.
Or
The scatter diagram shows that as \(t\) increases, \(s\) increases at an increasing rate. Thus, model (A) is most appropriate.
\[ s=0.118586994\,t^{2}+12.82061957 \] \(\therefore\ a\approx 0.119\) (to 3sf) and \(b\approx 12.8\) (to 3sf).
(e) Let \(T\) be the temperature given in Fahrenheit. \[ T=\frac{9}{5}t+32 \quad\Rightarrow\quad t=\frac{5}{9}(T-32). \] \[ s=0.118586994\left[\frac{5}{9}(T-32)\right]^{2}+12.82061957 \] \[ s\approx 0.0366(T-32)^{2}+12.8\ (\text{to 3 s.f.}). \]