TREATMENT OF EXPERIMENTAL DATA 21 

 The mathematical problem is to minimize 



E(2/ — 2/exp) 2 = E(« + bx — 2/exp) 2 - 



We take the derivative with respect to a and b in turn and set the deriva- 

 tive equal to zero, obtaining 



= E2(a + bx — i/exp), 



= L2(a + bx — y exp ) (x). 



Factoring out the 2's, and multiplying as indicated in the second equation, 

 we obtain 



E<* + bEz — HVexp = = na + 6(2» — El/exp, 



Lax + bY,x 2 — E^exp = = a(E-*0 + 6(2> 2 ) — E^exp- 



We can solve these two equations for a and b to give us the values of 

 a and b which minimize the deviations from the theoretical line: 



(Es 2 )(E2/exp) — (EzyexpXEs) 



n(E* 2 ) - (Ex) 2 



b _ n(£,xy exp ) — (E-g)(E^/exp) 

 "(E* 2 ) - (E*) 2 



To show the use of these expressions, and also to demonstrate that 

 the expressions are much more fearful-looking than actually harmful, 

 we illustrate by obtaining the straight line used in the previous section 

 on chi squared. Because we will need the sums, we tabulate the data 

 again, along with the operations needed for this application. 



Ex = 1+2 + 3 + 4 + 5= 15 



E2/exp = 19 



E^ 2 = 55 



E^exp = 70 



We substitute the various numbers in the expressions for a and b, 

 obtaining 



(55)(19) - (70)(15) ^5 

 (5) (55) - (15) 2 50 



(5) (70) - (15) (19) 65 

 5(55) - (15) 2 " 50 " ' 



The least-squares straight line for these data then is 



y = -0.1 + 1.3a;. 



