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Least Squares, Medians, and the Indy 500

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Figure 138 shows the situation in general We have taken the baseline along the x-axis, with the leftmost vertex of the triangle at the point (0, 0) This simpli es the calculations greatly and does not limit the generality of our arguments

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(x2,y2)

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m3 -- -- -- -- -- -> (x2/2,y2/2) ((x1+x2)/2,y2/2)

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m1-- ->

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m2-- -- > (x1,0)

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(0,0)

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First the point (x, y) is the point ((x1 + x2 )/3, (y2 /3)), so the point (x, y) is 1/3 of the distance from the baseline toward the vertex, P2 Now we must show that the medians meet at that point The equations of the three medians shown are as follows: m1: m2: m3: y2 x x1 + x 2 x 1 y2 y2 y= + x x1 2x2 x2 2x1 y= y= x 1 y2 2y2 + x 2x2 x1 2x2 x1

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It is easy to verify that the point ((x1 + x2 )/3, (y2 /3)) lies on each of the lines and hence is the point of intersection of these median lines

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These facts suggest two ways to determine the equation of the median median line They are as follows: 1 First, determine the slope of the line joining P1 and P3 This is the slope of the median median line Second, determine the point (x, y) Finally, nd the median median line using the slope and a point on the line 2 Determine the slope of the line joining P1 and P3 Then determine the equations of two of the medians and solve them simultaneously to determine the point of intersection The median median line can be found using the slope and a point on the line To these two methods, suggested by the facts above, we add a third method 3 Determine the intercept of the line joining P1 and P3 and the intercept of the line through P2 with the slope of the line through P1 and P3 The intercept of the median median line is the average of twice the rst intercept plus the second intercept (and its slope is the slope of the line joining P1 and P3 ) Method 1 is by far the easiest of the three methods although all are valid Methods 2 and 3 are probably useful only if one wants to practice nding equations of lines and doing some algebra! Method 2 is simply doing the proof above with actual data We will not prove that method 3 is valid but the proof is fairly easy

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When Are the Lines Identical

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It turns out that if x2 = x, then the least squares line and the median median line are identical To show this, consider the diagram in Figure 139 where we have taken the base of the triangle along the x-axis and the vertex of the triangle at (x1 /2, y2 ) since 0 + x1 + 3 x1 2 = x1 2

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Least Squares, Medians, and the Indy 500

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Now = Here

n n i=1 (xi x)(yi y) n 2 i=1 (xi x)

(xi x)(yi y) = 0

x1 2 x1 2

y2 x1 x1 + 3 2 2 y2 3 =0

y2

y2 3

+ x1

so = 0 Also, = y x = y = y2 /3 So the least squares line is y = y2 /3 But this is also the median median line since the median median line passes through (x, y) and has slope 0 It is not frequent that x2 = x, but if these values are close, we expect the least squares line and the median median line to also be close We now proceed to an example using the Indianapolis 500-mile race winning speeds The reason for using these speeds as an example is that the data are divided naturally into three parts due to the fact that the race was not held during World War I (1917 and 1918) and World War II (1942 1946) We admit that the three periods of data are far from equal in size The data have been given above by year; we now show the median points for each of the three time periods (Table 137 and Figure 1310)

Table 137 Period 1911 1916 1919 1941 1947 2008 Median (years) 19135 1930 19775 Median (speed) 8060 10045 14995