Analytical Chem istry - DePauw University

148 Analytical Chemistry 2.0mg/L. An unpaired t-test gives t exp as 0.8025. Because t exp is smaller thant(0.05, 11), which is 2.204, we have no evidence that there is a differencein the concentration of Zn 2+ between the two interfaces.Treating as Paired Data: To treat as paired data we need to calculate thedifference, d i , between the concentration of Zn 2+ at the air-water interfaceand at the sediment-water interface for each location.d i[Zn 2+2+] [Zn ]air-waterised-wateri= ( ) −( )Location 1 2 3 4 5 6d i (mg/L) 0.015 0.028 0.067 0.121 0.102 0.107The mean difference is 0.07333 mg/L with a standard deviation of 0.0441mg/L. The null hypothesis and alternative hypothesis areand the value of t exp isH : d = 0 H : d ≠ 00 A0.07333 6t exp= = 4.0730.04410Because t exp is greater than t(0.05, 5), which is 2.571, we reject the nullhypothesis and accept the alternative hypothesis that there is a significantdifference in the concentration of Zn 2+ between the air-water interfaceand the sediment-water interface.The difference in the concentration of Zn 2+ between locations is muchlarger than the difference in the concentration of Zn 2+ between the interfaces.Because out interest is in studying differences between the interfaces,the larger standard deviation when treating the data as unpaired increasesthe probability of incorrectly retaining the null hypothesis, a type 2 error.Click here to return to the chapter.Practice Exercise 4.12You will find small differences between the values given here for t exp andF exp , and for those values shown with the worked solutions in the chapter.These differences arise because Excel does not round off the results ofintermediate calculations.The two snapshots of Excel spreadsheets shown in Figure 4.29 providesolutions to these two examples.Click here to return to the chapter.