CORRECTION OF DATA FOR ERRORS 319 



Solution: 



Calculate the standard deviation of the five tests for each trans- 

 mitter. Find the average value of these 1000 values and divide it by 

 .8407. This follows from the fact that the average 5 of the observed 

 standard deviation for a series of samples of size n dreiwn from a 

 normal distribution with standard deviation a is 



where the symbol [^ is equivalent to T {X-\-l) 

 Thus for n = 5 we get ^^ s = .8407o-. 



Fig. 7 also presents the values of the ratio — for reference and with 



(T 



sufficient accuracy for solving problems similar to the example cited. 

 Greater accuracy than that afforded by the curves can be secured by 

 direct substitution in the equations for s and 5 or by referring to the 

 original tables. 



"We will recall with interest how closely the observed average, J = .798(r, of the 

 1000 values of 5 corresponding to the 1000 samples of four herein presented checked 

 the theoretical average of .80 lo-. 



