Analytical Chem istry - DePauw University

126 Analytical Chemistry 2.0> dbinom(0,27,0.0111)[1] 0.73979970.740 after adjusting the significant figures. To find the probability of obtainingany outcome up to a maximum value of X, we use the pbinomfunction.pbinom(X, N, p)To find the percentage of cholesterol molecules containing 0, 1, or 2 atomsof 13 C, we enter> pbinom(2,27,0.0111)[1] 0.9967226and find that the answer is 99.7% of cholesterol molecules.Significance Te s t sFor a one-tailed F-test the command isone of the followingvar.test(X, Y, alternative = “greater”)var.test(X, Y, alternative = “less”)where “greater” is used when the alterna-2 2tive hypothesis is s > s , and “less” isX Yused when the alternative hypothesis iss< s .2 2XYR includes commands to help you complete the following significance testscovered in this chapter:• an F-test of variances• an unpaired t-test of sample means assuming equal variances• an unpaired t-test of sample means assuming unequal variances• a paired t-test for of sample means• Dixon’s Q-test for outliers• Grubb’s test for outliersLet’s use R to complete a t-test on the data in Table 4.11, which containsresults for two experiments to determine the mass of a circulating U. S.penny. To do this, enter the data from Table 4.11 into two objects.> penny1=c(3.080, 3.094, 3.107, 3.056, 3.112, 3.174, 3.198)> penny2=c(3.052, 3.141, 3.083, 3.083, 3.048)Because the data in this case are unpaired, we will use R to complete anunpaired t-test. Before we can complete a t-test we must use an F-test todetermine whether the variances for the two data sets are equal or unequal.2 2Our null hypothesis is that the variances are equal, s = s , and ourSet1 Set22 2alternative hypothesis is that the variances are not equal, s ≠ s .Set1 Set2The command for a two-tailed F-test in R, which is our choice for thisproblem, isvar.test(X, Y)where X and Y are the objects containing the data sets. Figure 4.25 showsthe output from an R session to solve this problem.R does not provide the critical value for F(0.05, 6, 4). Instead it reportsthe 95% confidence interval for F exp . Because this confidence interval of0.204 to 11.661 includes the expected value for F of 1.00, we retain thenull hypothesis and have no evidence for a difference between the variances.