Analytical Chem istry - DePauw University

108 Analytical Chemistry 2.0tsAµ A= X A±nA4.17tsBµ B= X B±nB4.18where n A and n B are the sample sizes for A and B. Our null hypothesis,H : µ = µ0 A B, is that and any difference between m A and m B is the resultof indeterminate errors affecting the analyses. The alternative hypothesis,H : µ ≠ µA A B, is that the difference between m A and m B means is too largeto be explained by indeterminate error.To derive an equation for t exp , we assume that m A equals m B , and combineequations 4.17 and 4.18.XAt s t sexp Aexp± = X ±BnnABBProblem 9 asks you to use a propagationof uncertainty to show that equation 4.19is correct.Solving for XA− X and using a propagation of uncertainty, givesBs sAX − X = t × +A B expn n2 2BAB4.19Finally, we solve for t expSo how do you determine if it is okay topool the variances? Use an F-test.texp =XA− XB2 2s s4.20A B+n nAand compare it to a critical value, t(a,n), where a is the probability of atype 1 error, and n is the degrees of freedom.Thus far our development of this t-test is similar to that for comparingX to m, and yet we do not have enough information to evaluate the t-test.Do you see the problem? With two independent sets of data it is unclearhow many degrees of freedom we have.Suppose that the variances s A2 and s B2 provide estimates of the same s 2 .In this case we can replace s A2 and s B2 with a pooled variance, (s pool ) 2 , thatprovides a better estimate for the variance. Thus, equation 4.20 becomestexp =spoolwhere s pool , the pooled standard deviation, isXAB− X X − XBA B nnA B= ×1 1 s n + npoolA B4.21+n nAB