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By the end of the lesson you should be able to answer this regression exam question

**Carbon percentages against melting point**

y = -4.0476x + 36.214

40 35 30 25 20 15 10 5 Rather than draw a line a best fit by eye you can calculate and plot accurately a line that minimises the distance between the data plotted and the line.

0 0 1 2 3 4 5

**Carbon percentage**

6 7 8 9 The vertical distance between the points and the line of best fit are called

**residuals**

(red lines) This is why the regression line is sometimes called the

**least squares regression line**

y = a + bx a is the value of the melting point when carbon is 0% b is the rate at which the melting point reduces as the carbon percentage increases

The y axis shows the dependent (or response) variable. These variables are determined by the x values 40 35 30 25 20 15

**Carbon percentages against melting point**

y = -4.0476x + 36.214

10 5 0 0 1 2 3 4 5

**Carbon percentage**

6 7 8 9 The x axis shows the independent (or explanatory) variable. It is set independently of the other variable

Sxx = Σx² - (Σx)² n Sxy = Σxy - ΣxΣy n b = Sxy Sxx a = y - bx Regression line equation y = a + bx

Example The results from an experiment in which different masses were placed on a spring and the resulting length of the spring measured, are shown below Mass, x (kg) Length, y (cm) 20 48 40 55.1

60 56.3

80 61.2

100 68 Σx = 300, Σx²=22000, x = 60, Σxy = 18238, Σy² = 16879.14, Σy = 288.6, y = 57.72

a) Calculate Sxx and Sxy b) Calculate the regression line of y on x c) Calculate the length of the spring when a mass of 50kg is added d) Calculate the length of the spring when a mass of 140kg is added. Give a reason why this may or may not be a reliable answer.

Mass, x (kg) Length, y (cm) 20 48 40 55.1

Σx = 300, Σx²=22000, x = 60, Σxy = 18238, Σy² = 16879.14, Σy = 288.6, y = 57.72

a) Calculate Sxx and Sxy Sxx = 22000 - 300 ² = 4000 5 Sxy = 18238 – 300x288.6 = 922 5 b) Calculate the regression line of y on x b = Sxy = 922 = 0.2305

Sxx 4000 60 56.3

a = y – bx a = 57.72 – 0.2305 x 60 = 43.89

y = 43.89 + 0.2305x

80 61.2

100 68

c) Calculate the length of the spring when a mass of 50kg is added d) Calculate the length of the spring when a mass of 140kg is added. Give a reason why this is or is not a reliable answer y = 43.89 + 0.2305x

c) Y = 43.39 + 0.2305x50 = 54.915cm

d) Y = 43.39 + 0.2305 x 140 = 75.66cm

This may not a reliable answer as it has been calculated using extrapolation. It could be unreliable140kg is outside the range of data given and used to calculate the regression line.