How to adjust this table ?

Suppose that the #x#-axis represents age in years and the #y#-axis represents the pulse rate in beats/minute. From the table, we can see that when #x# increases by a constant amount (#10#), #y# decreases by a constant amount (#9#). This is a linear relationship between #x# and #y#.

As we are given a set of data points, we only need the slope #m# of the equation to represent it in point-slope form :

#y-y_1 = m(x-x_1)#

where #(x_1, y_1)# is a point on the graph.

The slope of a linear equation is given by #m = "change in y"/"change in x"#. Given two points on the graph #(x_1, y_1), (x_2, y_2)#, this translates to #m = (y_2-y_1)/(x_2-x_1)#.

Taking the first two points on the table, we get:

#m = (166-175)/(30-20) = -9/10#

Thus, together with the point #(20, 175)#, we get our equation in point slope form:

#y - 175 = -9/10(x - 20)#

We can also solve for #y# to put it in slope-intercept form :

#y = -9/10x + 193#

Given a list of values #{x_k,y_k}, k=1,2,cdots,n#

a line can be adjusted such that the accumulated deviation error is minimum.

Let the line be given by

#y = a x + b#

then the error at point #x_k# is #e_k = y_k - (a x_k + b)#

The accumulated quadratic error will be given by

#E(a,b) = sum_(k=1)^n e_k^2 =sum_(k=1)^n( y_k - (a x_k + b))^2#

#E(a,b)# have a minimum for #a_0,b_0# such that

#(partial E)/(partial a)(a_0,b_0) = 0-> b_0 sum x_k+a_0 sumx_k^2=sum x_ky_k#
#(partial E)/(partial b)(a_0,b_0) = 0 -> n b_0 + a_0 sum x_k =sum y_k #

solving for #a_0,b_0#

#a_0 = ( (sum x_k)( sum y_k)-n sum x_ky_k)/((sum x_k)^2-nsum x_k^2)#
#b_0 = ((sum x_k)(sum x_ky_k)-(sum x_k^2)(sum y_k))/((sum x_k)^2-nsum x_k^2#

appliying it to the table

#{{20, 175}, {30, 166}, {40, 157}, {50, 148}, {60, 139}, {70, 130}}#

we obtain:

#a_0 = -0.9, b_0 = 193#

so the adjusting line is

#y = -0.9 x + 193#