Analytical Chem istry - DePauw University

110 Analytical Chemistry 2.03. 117 −3.081 7t exp=× × 5.0.0459 7+ 5= 134The critical value for t(0.05,10), from Appendix 4, is 2.23. Because t exp isless than t(0.05,10) we retain the null hypothesis. For a = 0.05 there is noevidence that the two sets of pennies are significantly different.Example 4.20One method for determining the %w/w Na 2 CO 3 in soda ash is an acid–base titration. When two analysts analyze the same sample of soda ash theyobtain the results shown here.Determine whether the difference in the mean values is significant ata=0.05.So l u t i o nAnalyst AAnalyst B86.82 81.0187.04 86.1586.93 81.7387.01 83.1986.20 80.2787.00 83.94We begin by summarizing the mean and standard deviation for each analyst.X = 86. 83% X = 82. 71%ABs = 032 . s = 2.16ATo determine whether we can use a pooled standard deviation, we firstcomplete an F-test of the following null and alternative hypotheses.2 2H s = s H : s ≠ s:0 A BCalculating F exp , we obtain a value ofB2 2A A B2( 216 . )F exp= = 45.62( 032 . )Because F exp is larger than the critical value of 7.15 for F(0.05,5,5) fromAppendix 5, we reject the null hypothesis and accept the alternative hypothesisthat there is a significant difference between the variances. As aresult, we cannot calculate a pooled standard deviation.