Let S_{A}^{2} = The sum for i=1 to 8 of (s_{i}-1.5125)^{2}
= .10875, where S_{i} is the ith sample from location A.

Let S_{B}^{2} = The sum for i=1 to 6 of (s_{i}-1.5125)^{2}
= .167673, where S_{i} is the ith sample from location B.

Let U be the statistic for testing the the hypothesis that the mean of location A does not equal the mean of location B.

U = (sqr(8+6-2)*(1.5125-1.6683)) / (sqr(1/8 + 1/6)+sqr(.10875+.167673)) = -1.693354 .

The t value, with 12 degrees of freedom, at the .95 level of significance is 1.782 .

Thus if -1.782 <= U <= 1.782 the test will pass, which it does.

Michael Shackleford, A.S.A.