302 
BULLETIN OF THE BUREAU OF FISHERIES 
N 2 are the total number of items in the respective frequencies. S' is the summation 
of these values for each class. ^ 
The probability, P, that chance alone would cause the same or a greater diver- 
gence between two random samples of the same population is obtained by the formula: 
p = e -M X 2( i + A- + 
"d 
,y— 3 
2 2X4 2X4X6 
+ 
+ 
2 X 4 X 6 X • 
X (y 
-a) 
where y equals the number of classes, and e equals the base of the Napierian system 
of logarithms. 
From the calculation we obtain P as 0.621, meaning that the age frequencies 
of 6 out of every 10 samples of the same population would differ as much as these 
two frequencies. 
Application of the formula to the size frequencies gives a value for P of 0.0898, 
quite different from that of the age readings. This may be due to erroneous age 
Z 4 6 8 10 IZ 14 16 
AGE IN YEARS 
Figure 40.— Age histograms of Shuyak Strait and lower Kachemak Bay for 1926 
readings or to chance, but an examination of the length frequencies (Table 31) 
causes one to believe that it is largely due to growth as the two frequencies are very 
similar, the chief difference being that the modal length in Kachemak Bay is half 
a centimeter greater than in Shuyak Strait. As mentioned above the Kachemak 
Bay samples were taken six weeks later than those of Shuyak Strait. The period 
thus covered, from July 15 to about August 27, is one in which the herring make a 
large part of the season’s growth. These facts could account for the half centi- 
meter difference in modal length, which would cause P to be very low, whereas 
without the growth in the period intervening between the collection of the samples 
P would undoubtedly be much higher. 
