Analytical Chem istry - DePauw University

Chapter 4 Evaluating Analytical Data113Is there a significant difference between the methods at a = 0.05?So l u t i o nAcquiring samples over an extended period of time introduces a substantialtime-dependent change in the concentration of monensin. Because thevariation in concentration between samples is so large, we use paired t-testwith the following null and alternative hypotheses.H : d = 0 H : d ≠ 00 ADefining the difference between the methods asd X X= ( ) −( )i elect i micro iwe can calculate the following differences for the samples.Sample 1 2 3 4 5 6 7 8 9 10 11d i 2.8 1.4 -3.0 6.0 -6.6 -0.5 9.7 12.7 -1.6 4.0 -0.2The mean and standard deviation for the differences are, respectively, 2.25and 5.63. The value of t exp is225 . 11t exp= = 133 .563 .which is smaller than the critical value of 2.23 for t(0.05,10) from Appendix4. We retain the null hypothesis and find no evidence for a significantdifference in the methods at a = 0.05.Practice Exercise 4.11Suppose you are studying the distribution of zinc in alake and want to know if there is a significant differencebetween the concentration of Zn 2+ at the sedimentwaterinterface and its concentration at the air-waterinterface. You collect samples from six locations—nearthe lake’s center, near its drainage outlet, etc.—obtainingthe results (in mg/L) shown in the table. Using thedata in the table shown to the right, determine if thereis a significant difference between the concentration ofZn 2+ at the two interfaces at a = 0.05.Complete this analysis treating the data as (a) unpaired, and (b) paired. Briefly comment on your results.Click here to review your answers to this exercise.One important requirement for a paired t-test is that determinate andindeterminate errors affecting the analysis must be independent of the ana-LocationAir-Water Sediment-WaterInterface Interface1 0.430 0.4152 0.266 0.2383 0.457 0.3904 0.531 0.4105 0.707 0.6056 0.716 0.609