D. Harvey - Modern Analytical Chemistry (794078), страница 29
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An example ofthis application of the F-test is shown in the following example.EXAMPLE 4.18Tables 4.1 and 4.8 show results for two separate experiments to determine themass of a circulating U.S. penny. Determine whether there is a difference in theprecisions of these analyses at α = 0.05.SOLUTIONLetting A represent the results in Table 4.1 and B represent the results in Table4.8, we find that the variances are sA2 = 0.00259 and sB2 = 0.00138.
A two-tailedsignificance test is used since there is no reason to suspect that the results forone analysis will be more precise than that of the other. The null andalternative hypotheses areH0:sA2 = sB2HA:sA2 ≠ sB2and the test statistic isFexp =s A20.00259== 1.8820.00138sBThe critical value for F(0.05, 6, 4) is 9.197.
Since Fexp is less than F(0.05, 6, 4),the null hypothesis is retained. There is no evidence at the chosen significancelevel to suggest that the difference in precisions is significant.4F.4 Comparing Two Sample Meansunpaired dataTwo sets of data consisting of resultsobtained using several samples drawnfrom a single source.paired dataTwo sets of data consisting of resultsobtained using several samples drawnfrom different sources.The result of an analysis is influenced by three factors: the method, the sample, andthe analyst. The influence of these factors can be studied by conducting a pair of experiments in which only one factor is changed.
For example, two methods can becompared by having the same analyst apply both methods to the same sample andexamining the resulting means. In a similar fashion, it is possible to compare twoanalysts or two samples.Significance testing for comparing two mean values is divided into two categories depending on the source of the data. Data are said to be unpaired when eachmean is derived from the analysis of several samples drawn from the same source.Paired data are encountered when analyzing a series of samples drawn from different sources.––Unpaired Data Consider two samples, A and B, for which mean values, XA and XB,and standard deviations, sA and sB, have been measured. Confidence intervals for µAand µB can be written for both samples1400-CH04 9/8/99 3:55 PM Page 89Chapter 4 Evaluating Analytical DataµA = XA ±ts AµB = XB ±tsB4.17nA4.18nBwhere nA and nB are the number of replicate trials conducted on samples A and B.
A––comparison of the mean values is based on the null hypothesis that XA and XB areidentical, and an alternative hypothesis that the means are significantly different.A test statistic is derived by letting µA equal µB, and combining equations 4.17and 4.18 to givetstsX A ± A = XB ± BnAnB– –Solving for XA – XB and using a propagation of uncertainty, givesX A − XB = t ×s A2s2+ Bn A nBFinally, solving for t, which we replace with texp, leaves us witht exp =X A − XB4.19(s 2A / nA ) + (s B2 / nB )The value of texp is compared with a critical value, t(α, ν), as determined by the chosen significance level, α, the degrees of freedom for the sample, ν, and whether thesignificance test is one-tailed or two-tailed.It is unclear, however, how many degrees of freedom are associated with t(α, ν)since there are two sets of independent measurements. If the variances sA2 and sB2 estimate the same σ2, then the two standard deviations can be factored out of equation4.19 and replaced by a pooled standard deviation, spool, which provides a better estimate for the precision of the analysis.
Thus, equation 4.19 becomest exp =X A − XB4.20s pool (1/ nA ) + (1/ nB )with the pooled standard deviation given asspool =(n A − 1)s A2 + (nB − 1)sB2n A + nB − 24.21As indicated by the denominator of equation 4.21, the degrees of freedom for thepooled standard deviation is nA + nB – 2.If sA and sB are significantly different, however, then texp must be calculatedusing equation 4.19.
In this case, the degrees of freedom is calculated using the following imposing equation.ν=[(s A2 /nA ) + (s B2 /nB )]222[(s A /nA ) /(nA + 1)] + [(s B2 /nB )2 /(nB+ 1)]–24.22Since the degrees of freedom must be an integer, the value of ν obtained usingequation 4.22 is rounded to the nearest integer.Regardless of whether equation 4.19 or 4.20 is used to calculate texp, the null hypothesis is rejected if texp is greater than t(α, ν), and retained if texp is less than orequal to t(α, ν).891400-CH04 9/8/99 3:55 PM Page 9090Modern Analytical ChemistryEXAMPLE 4.19Tables 4.1 and 4.8 show results for two separate experiments to determine themass of a circulating U.S. penny.
Determine whether there is a difference in themeans of these analyses at α = 0.05.SOLUTIONTo begin with, we must determine whether the variances for the two analysesare significantly different. This is done using an F-test as outlined in Example4.18. Since no significant difference was found, a pooled standard deviationwith 10 degrees of freedom is calculateds pool ==(nA − 1)s A2 + (nB − 1)s B2nA + nB − 2(7 − 1)(0.00259) + (5 − 1)(0.00138)7+5−2= 0.0459where the subscript A indicates the data in Table 4.1, and the subscript Bindicates the data in Table 4.8. The comparison of the means for the twoanalyses is based on the null hypothesis––H0: XA = XBand a two-tailed alternative hypothesis––XA ≠ XBHA:Since the standard deviations can be pooled, the test statistic is calculated usingequation 4.20t exp =X A − XBs pool (1/ nA + 1/ nB )=3.117 − 3.0810.0459 (1/ 7 + 1/ 5)= 1.34The critical value for t(0.05, 10), from Appendix 1B, is 2.23.
Since texp is less thant(0.05, 10) the null hypothesis is retained, and there is no evidence that the twosets of pennies are significantly different at the chosen significance level.EXAMPLE 4.20The %w/w Na2CO3 in soda ash can be determined by an acid–base titration.The results obtained by two analysts are shown here.
Determine whether thedifference in their mean values is significant at α = 0.05.Analyst AAnalyst B86.8287.0486.9387.0186.2087.0081.0186.1581.7383.1980.2783.941400-CH04 9/8/99 3:55 PM Page 91Chapter 4 Evaluating Analytical DataSOLUTIONWe begin by summarizing the mean and standard deviation for the datareported by each analyst. These values are–XA = 86.83%s = 0.32–AXB = 82.71%sB = 2.16A two-tailed F-test of the following null and alternative hypothesesH0:sA2 = sB2HA:sA2 ≠ sB2is used to determine whether a pooled standard deviation can be calculated.The test statistic isFexp =sB2(2.16)2== 45.62sA(0.32)2Since Fexp is larger than the critical value of 7.15 for F(0.05, 5, 5), the nullhypothesis is rejected and the alternative hypothesis that the variances aresignificantly different is accepted.
As a result, a pooled standard deviationcannot be calculated.The mean values obtained by the two analysts are compared using a twotailed t-test. The null and alternative hypotheses are– –– –H0: XA = XBHA: XA ≠ XBSince a pooled standard deviation could not be calculated, the test statistic, texp,is calculated using equation 4.19t exp =X A − XB(s 2A / nA ) + (s B2 / nB )=86.83 − 82.71[(0.32)2 /6] + [(2.16)2 /6]= 4.62and the degrees of freedom are calculated using equation 4.22ν=[(0.322 /6) + (2.162 /6)]2− 2 = 5.3 ≈ 5{(0.322 /6)2 /(6 + 1)} + {(2.162 /6)2 /(6 + 1)}The critical value for t(0.05, 5) is 2.57. Since the calculated value of texp isgreater than t(0.05, 5) we reject the null hypothesis and accept the alternativehypothesis that the mean values for %w/w Na2 CO 3 reported by the twoanalysts are significantly different at the chosen significance level.Paired Data In some situations the variation within the data sets being comparedis more significant than the difference between the means of the two data sets.
Thisis commonly encountered in clinical and environmental studies, where the databeing compared usually consist of a set of samples drawn from several populations.For example, a study designed to investigate two procedures for monitoring theconcentration of glucose in blood might involve blood samples drawn from ten patients. If the variation in the blood glucose levels among the patients is significantlylarger than the anticipated variation between the methods, then an analysis in whichthe data are treated as unpaired will fail to find a significant difference between the911400-CH04 9/8/99 3:55 PM Page 9292Modern Analytical Chemistrymethods.
In general, paired data sets are used whenever the variation being investigated is smaller than other potential sources of variation.In a study involving paired data the difference, di, between the paired values for–each sample is calculated. The average difference, d, and standard deviation of the–differences, sd, are then calculated. The null hypothesis is that d is 0, and that thereis no difference in the results for the two data sets. The alternative hypothesis is that–the results for the two sets of data are significantly different, and, therefore, d is notequal to 0.–The test statistic, texp, is derived from a confidence interval around d0=d ±tsdnwhere n is the number of paired samples. Replacing t with texp and rearranging givest exp =paired t-testStatistical test for comparing paired datato determine if their difference is toolarge to be explained by indeterminateerror.d nsdThe value of texp is then compared with a critical value, t(α, ν), which is determinedby the chosen significance level, α, the degrees of freedom for the sample, ν, andwhether the significance test is one-tailed or two-tailed.
For paired data, the degreesof freedom is n – 1. If texp is greater than t(α, ν), then the null hypothesis is rejectedand the alternative hypothesis is accepted. If texp is less than or equal to t(α, ν), thenthe null hypothesis is retained, and a significant difference has not been demonstrated at the stated significance level. This is known as the paired t-test.EXAMPLE 4.21Marecek and colleagues developed a new electrochemical method for the rapidquantitative analysis of the antibiotic monensin in the fermentation vats usedduring its production.9 The standard method for the analysis, which is basedon a test for microbiological activity, is both difficult and time-consuming.
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