Analytical Chem istry - DePauw University

106 Analytical Chemistry 2.0So l u t i o nThe variance for the sample of 10 tablets is 4.3. The null hypothesis andalternative hypotheses areH02 2: s =σ H s≠σA : 2 2The value for F exp isFexpσ 2 25= = = 58 .2s 43 .The critical value for F(0.05,∞,9) from Appendix 5 is 3.333. Since F expis greater than F(0.05,∞,9), we reject the null hypothesis and accept thealternative hypothesis that there is a significant difference between thesample’s variance and the expected variance. One explanation for the differencemight be that the aspirin tablets were not selected randomly.4F.3 Comparing Two Sample VariancesWe can extend the F-test to compare the variances for two samples, A andB, by rewriting equation 4.16 asFss2= Aexp 2Bdefining A and B so that the value of F exp is greater than or equal to 1.Example 4.18Table 4.11 shows results for two experiments to determine the mass ofa circulating U.S. penny. Determine whether there is a difference in theprecisions of these analyses at a = 0.05.So l u t i o nThe variances for the two experiments are 0.00259 for the first experiment(A) and 0.00138 for the second experiment (B). The null and alternativehypotheses areand the value of F exp is2 2H s = s H : s ≠ s:0 A BFexp2 2A A B2sA0.00259= = = 188 .2s 0.00138BFrom Appendix 5, the critical value for F(0.05,6,4) is 9.197. BecauseF exp < F(0.05,6,4), we retain the null hypothesis. There is no evidence ata = 0.05 to suggest that the difference in precisions is significant.