Unit 17 Review Exercises

Correlation and Regression

These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 A survey of 50 cities finds a correlation coefficient of \(r = 0.88\) between the number of libraries and the number of restaurants in each city.

(a) Describe the correlation.

(b) A journalist writes: “Building more libraries will bring more restaurants.” Comment.

Solution

(a) Strong positive linear correlation.

(b) Correlation does not imply causation. Both numbers probably depend on a third variable, the population of the city: large cities have more of both. See 17.1.

2 Paper 2 The number of hours of sunshine (\(x\)) and the number of visitors to a beach (\(y\) hundred) are recorded on nine days.

\(x\)	2.5	3.8	4.6	5.2	6.0	6.9	7.5	8.3	9.1
\(y\)	12	15	19	18	24	26	27	31	33

(a) Find \(r\). \(\qquad\) (b) Find the regression line of \(y\) on \(x\).

(c) Estimate the number of visitors on a day with 5.5 hours of sunshine.

Solution

(a) \(r = 0.990\) (3 s.f.).

(b) \(y = 3.29x + 3.09\) (3 s.f.).

(c) \(y = 21.2\) (3 s.f.), so about \(2120\) visitors. This is interpolation with strong correlation, so it is reliable. See 17.2 and 17.3.

3 Paper 1 For 8 pairs of data, \(\sum x = 64\) and \(\sum y = 296\). The regression line of \(y\) on \(x\) is \(y = ax + 12\).

(a) Write down the mean point.

(b) Find \(a\).

(c) Estimate \(y\) when \(x = 9\).

Solution

(a) \(\bar{x} = \dfrac{64}{8} = 8\), \(\bar{y} = \dfrac{296}{8} = 37\). Mean point \((8, 37)\).

(b) \(37 = 8a + 12 \Rightarrow a = \dfrac{25}{8} = 3.125\).

(c) \(y = 3.125 \times 9 + 12 = 40.125\). See 17.2.

4 Paper 1 The regression lines for a set of data are \(y = 0.8x + 6\) (\(y\) on \(x\)) and \(x = 0.9y + 1.2\) (\(x\) on \(y\)).

(a) Find the mean point. \(\qquad\) (b) Estimate \(x\) when \(y = 40\).

Solution

(a) \(x = 0.9(0.8x + 6) + 1.2 = 0.72x + 6.6 \Rightarrow 0.28x = 6.6 \Rightarrow x = \dfrac{165}{7} \approx 23.6\).

\(y = 0.8 \times \dfrac{165}{7} + 6 = \dfrac{174}{7} \approx 24.9\). Mean point \(\left(\dfrac{165}{7}, \dfrac{174}{7}\right)\).

(b) Use the \(x\) on \(y\) line: \(x = 0.9 \times 40 + 1.2 = 37.2\). See 17.2 and 17.3.

5 Paper 1 For a set of data, \(r_{xy} = -0.74\). Write down the correlation coefficient between

(a) \(u = 5x + 2\) and \(v = 3y - 7\),

(b) \(u = 5x + 2\) and \(w = -y\).

Solution

(a) Multipliers \(5\) and \(3\) have the same sign: \(r_{uv} = -0.74\).

(b) Multipliers \(5\) and \(-1\) have opposite signs: \(r_{uw} = 0.74\). See 17.1.

6 Paper 1 The regression line of \(y\) on \(x\) for the data below is \(y = 2.41x + 13.5\). Find the value of \(k\), to the nearest integer.

\(x\)	3	5	7	9	11	13
\(y\)	20	26	\(k\)	35	41	44

Solution

\(\bar{x} = \dfrac{48}{6} = 8\), so \(\bar{y} = 2.41 \times 8 + 13.5 = 32.78\) and \(\sum y = 6 \times 32.78 = 196.68\).

\(k = 196.68 - (20 + 26 + 35 + 41 + 44) = 196.68 - 166 = 30.68\), so \(k = 31\). See 17.3.

7 Paper 2 A scientist sets the concentration of a fertiliser (\(y\) mg per litre) and measures the growth of a plant (\(x\) cm) after two weeks.

\(y\)	0	5	10	15	20	25
\(x\)	3.2	4.0	5.1	5.7	6.9	7.4

(a) Explain why the regression line of \(x\) on \(y\) should be used.

(b) Find its equation.

(c) Estimate the growth when the concentration is \(12\) mg per litre.

Solution

(a) The concentration \(y\) is set by the scientist, so \(y\) is the independent variable and \(x\) depends on it.

(b) GDC with X List \(y\), Y List \(x\): \(\;x = 0.173y + 3.22\) (3 s.f.).

(c) \(x = 0.17314\ldots \times 12 + 3.2190\ldots = 5.30\) cm (3 s.f.). See 17.2.

8 Paper 1 For children aged \(4\) to \(12\) years, the regression line of reading time (\(y\) minutes per day) on age (\(x\) years) is \(y = 3.5x + 2\).

(a) Interpret the value \(3.5\).

(b) Use the line to estimate the reading time of a 40-year-old adult, and comment.

Solution

(a) For each extra year of age, children read for about \(3.5\) more minutes per day, on average.

(b) \(y = 3.5 \times 40 + 2 = 142\) minutes. This is extrapolation far outside the ages \(4\) to \(12\); adult reading habits are very different, so it is not reliable. See 17.2 and 17.3.

9 Paper 2 Seven pairs of data are shown.

\(x\)	10	20	30	40	50	60	70
\(y\)	15	24	36	44	13	63	75

(a) Find \(r\) for all seven points.

(b) Find \(r\) without the point \((50, 13)\).

(c) Comment.

Solution

(a) \(r = 0.764\) (3 s.f.). \(\qquad\) (b) \(r = 0.999\) (3 s.f.).

(c) The single outlier \((50, 13)\) makes the correlation look much weaker than it really is. See 17.1.

10 Paper 2 The age (\(x\) years) and price (\(y\) thousand dollars) of eight used cars of the same model are shown.

\(x\)	1	2	2	3	4	5	6	7
\(y\)	27.5	24.8	25.6	22.1	20.3	17.9	16.2	13.4

(a) Find \(r\) and describe the correlation.

(b) Find the regression line of \(y\) on \(x\) and interpret the gradient.

(c) Estimate the price of a car that is 4.5 years old.

(d) A car of this model costs $19 000. Use a suitable regression line to estimate its age.

Solution

(a) \(r = -0.997\): very strong negative linear correlation.

(b) \(y = -2.32x + 29.7\) (3 s.f.). On average, the price falls by about $2320 each year.

(c) \(y = 19.2\), about $19 200 (3 s.f.).

(d) Given \(y\), use the regression line of \(x\) on \(y\): \(\;x = -0.429y + 12.8\) (3 s.f.). \(x = -0.42923\ldots \times 19 + 12.753\ldots = 4.60\) years (3 s.f.). See 17.2 and 17.3.