Linear regression and prediction Problem set A One question at a time

A kiosk records the hours of sunshine x in a day and the number y of cold drinks sold on 8 days. x: 3, 5, 6, 8, 9, 11, 12, 14 y: 14, 19, 20, 27, 27, 33, 36, 41

1. (a) Find the equation of the regression line of y on x.

   • From the calculator: \(a=2.45\), \(b=6.27\)
   • \(y=2.45x+6.27\)

2. (b) Use the line to predict y when x = 10.

   • \(y=2.45\times10+6.27=30.8\)
   • 10 is inside the range of the data, so this is a reliable estimate

A taxi firm models the fare y (€) for a journey of x km. The regression line of y on x, from 30 journeys of 2 km to 25 km, is \(y=1.9x+4.5\).

1. (a) Interpret the gradient.

   • The gradient is 1.9
   • For each extra km (x up by 1) the fare y rises by €1.9 on average

2. (b) Interpret the intercept.

   • The intercept is 4.5: the value of y when x = 0
   • The model gives €4.5 for a journey of 0 km
   • x = 0 is outside the data range, so this is a modelled value

On 10 afternoons, a shop records the number of customers x and the number of items sold y. x: 12, 15, 16, 18, 19, 21, 22, 24, 25, 27 y: 40, 46, 45, 52, 49, 58, 55, 61, 95, 66

1. (a) Find the quartiles of y. Use the 1.5 × IQR rule to test y for outliers.

   • From the calculator: \(Q_1=46\), \(Q_3=61\), IQR = 15
   • Limits: \(46-1.5\times15=23.5\) and \(61+1.5\times15=83.5\)
   • 95 is above the limit: an outlier in y

2. (b) Test x for outliers.

   • \(Q_1=16\), \(Q_3=24\), IQR = 8
   • Limits: 4 and 36
   • Every x value is inside the limits: no outlier in x

A sunflower's height y (cm) is measured x weeks after planting, for x from 5 to 25. The regression line of y on x is \(y=6.4x+12\), with r = 0.97.

1. (a) Predict y when x = 15.

   • \(y=6.4\times15+12=108\)
   • 15 is inside the range 5 to 25 weeks: interpolation, so the estimate is reliable

2. (b) Predict y when x = 60. Comment on the reliability of this prediction.

   • \(y=6.4\times60+12=396\)
   • 60 is outside the range 5 to 25 weeks: extrapolation
   • The line will not hold outside the data: a sunflower stops growing, so this prediction is unreliable

For 10 students, the weekly revision time x (hours) and the exam mark y (%) are recorded. x: 2, 4, 5, 7, 8, 10, 11, 13, 14, 16 y: 38, 44, 50, 52, 35, 66, 68, 77, 80, 88

1. (a) Write down the coordinates of the point that does not fit the pattern.

   • On the diagram, (8, 35) is far below the line of the others

2. (b) Find the regression line of y on x with and without this point.

   • All 10 points: \(y=3.69x+26.6\), \(r=0.913\)
   • Without (8, 35): \(y=3.57x+30.1\), \(r=0.996\)

3. (c) Describe the effect of the point on the line and on r.

   • With the point the gradient is 3.69; without it, 3.57
   • |r| is 0.913 with the point and 0.996 without it
   • The point does not follow the pattern, so it weakens the correlation and moves the line

A trainee's typing speed y (words per minute) after x weeks on a course is: x: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 y: 3, 6, 10, 13, 16, 17, 18, 18, 20, 22

1. (a) Plot the data. Explain why one line is not a good model.

   • The points rise steeply for weeks 1 to 5, then more gently
   • One line would fit neither part well

2. (b) Find the regression line for each part.

   • Weeks 1 to 5: \(y=3.3x-0.3\)
   • Weeks 6 to 10: \(y=1.2x+9.4\)

3. (c) Predict y when x = 8.5.

   • x = 8.5 is in the part weeks 6 to 10
   • \(y=1.2\times8.5+9.4=19.6\)

A café records the midday temperature x (°C) and the daily sales y (€) on 30 days. The temperatures ranged from 9 °C to 30 °C. The regression line of y on x is \(y=18.5x+120\) and the mean temperature was 21.4 °C.

1. (a) Find the mean daily sales.

   • The mean point is on the line: \(\bar{y}=18.5(21.4)+120\)
   • \(\bar{y}=515.9\), so the mean daily sales are €516 (3 s.f.)

2. (b) The owner solves \(700=18.5x+120\) to find the temperature needed for sales of €700. Explain why this is not a reliable use of the regression line.

   • The line of y on x makes the vertical distances small: it is for estimating y from x.
   • Solving for x gives \(x=\frac{700-120}{18.5}=31.4\), which is outside the range 9 to 30: extrapolation, so it is not reliable.

In week x of a fitness programme, a client does y press-ups. x: 1, 2, 3, 4, 5, 6, 7, 8 y: 12, 15, 17, 22, 24, 27, 31, 33 In week 9 the client does 15 press-ups, after a minor injury. Add the point (9, 15).

1. (a) Find the regression line and r for the original data.

   • \(y=3.08x+8.75\), \(r=0.997\)

2. (b) Find the regression line and r when the new point is included.

   • Add (9, 15) to both lists
   • \(y=1.65x+13.5\), \(r=0.600\)

3. (c) Describe the effect of the new point.

   • The gradient changes from 3.08 to 1.65
   • |r| falls from 0.997 to 0.600
   • the new point lies well below the trend, so it flattens the line and weakens the correlation

A study of 40 adults finds the regression line \(y=78-1.7x\) for resting heart rate y (beats per minute) on weekly exercise x (hours), with x from 1 to 9 and r = −0.46.

1. (a) Interpret the intercept. Explain why it should be treated with care.

   • The intercept 78 is the value of y when x = 0
   • The model gives 78 beats per minute for someone who does no exercise
   • x = 0 is outside the range 1 to 9: an extrapolation

2. (b) Predict y when x = 6 and when x = 20. Comment on each prediction.

   • \(y=78-1.7\times6=67.8\): inside the range, but r = −0.46 is weak, so the estimate is rough
   • \(y=78-1.7\times20=44\): outside the range 1 to 9, so unreliable