D. Harvey - Modern Analytical Chemistry (794078), страница 28
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This is known as a type 2 error, and itsprobability of occurrence is β. Unfortunately, in most cases β cannot be easily calculated or estimated.1400-CH04 9/8/99 3:54 PM Page 8585Chapter 4 Evaluating Analytical DataThe probability of a type 1 error is inversely related to the probability of a type2 error.
Minimizing a type 1 error by decreasing α, for example, increases the likelihood of a type 2 error. The value of α chosen for a particular significance test,therefore, represents a compromise between these two types of error. Most of theexamples in this text use a 95% confidence level, or α = 0.05, since this is the mostfrequently used confidence level for the majority of analytical work. It is notunusual, however, for more stringent (e.g. α = 0.01) or for more lenient(e.g. α = 0.10) confidence levels to be used.4F Statistical Methods for Normal DistributionsX–The most commonly encountered probability distribution is the normal, or Gaussian, distribution. A normal distribution is characterized by a true mean, µ, and vari–ance, σ2, which are estimated using X and s2.
Since the area between any two limitsof a normal distribution is well defined, the construction and evaluation of significance tests are straightforward.4F.1 Comparing X– to µt exp =µ−X × nsntexpsntexpsX+ntexpsn(b)t (α,ν)sX–X–texpsnnt (α,ν)sX+nX+t (α,ν)sX–Rearranging equation 4.14X+(a)X–One approach for validating a new analytical method is to analyze a standardsample containing a known amount of analyte, µ. The method’s accuracy is judged–by determining the average amount of analyte in several samples, X, and using–a significance test to compare it with µ.
The null hypothesis is that X and µ arethe same and that any difference between the two values can be explained by in–determinate errors affecting the determination of X . The alternative hypothesis is–that the difference between X and µ is too large to be explained by indeterminateerror.The equation for the test (experimental) statistic, texp, is derived from the confidence interval for µt exp sµ = X±4.14ntexpsntexpsnt (α,ν)sX+n(c)4.15gives the value of texp when µ is at either the right or left edge of the sample’s apparent confidence interval (Figure 4.11a). The value of texp is compared with acritical value, t(α ,ν), which is determined by the chosen significance level, α , thedegrees of freedom for the sample, ν, and whether the significance test is onetailed or two-tailed.
Values for t(α ,ν) are found in Appendix 1B. The criticalvalue t(α ,ν) defines the confidence interval that can be explained by indeterminate errors. If texp is greater than t(α ,ν), then the confidence interval for the datais wider than that expected from indeterminate errors (Figure 4.11b). In this case,the null hypothesis is rejected and the alternative hypothesis is accepted. If texp isless than or equal to t(α ,ν), then the confidence interval for the data could be attributed to indeterminate error, and the null hypothesis is retained at the statedsignificance level (Figure 4.11c).–A typical application of this significance test, which is known as a t-test of X toµ, is outlined in the following example.Figure 4.11Relationship between confidence intervalsand results of a significance test.
(a) Theshaded area under the normal distributioncurves shows the apparent confidenceintervals for the sample based on texp. Thesolid bars in (b) and (c) show the actualconfidence intervals that can be explained byindeterminate error using the critical value of(α,ν). In part (b) the null hypothesis isrejected and the alternative hypothesis isaccepted. In part (c) the null hypothesis isretained.t-testStatistical test for comparing two meanvalues to see if their difference is toolarge to be explained by indeterminateerror.1400-CH04 9/8/99 3:54 PM Page 8686Modern Analytical ChemistryEXAMPLE 4.16Before determining the amount of Na2CO3 in an unknown sample, a studentdecides to check her procedure by analyzing a sample known to contain98.76% w/w Na2CO3. Five replicate determinations of the %w/w Na2CO3 in thestandard were made with the following results98.71%98.59%98.62%98.44%98.58%Is the mean for these five trials significantly different from the accepted value atthe 95% confidence level (α = 0.05)?SOLUTIONThe mean and standard deviation for the five trials are–X = 98.59s = 0.0973–Since there is no reason to believe that X must be either larger or smaller thanµ, the use of a two-tailed significance test is appropriate.
The null andalternative hypotheses are––H0: X = µHA: X ≠ µThe test statistic ist exp =µ−X × ns=98.76 − 98.59 × 50.0973= 3.91The critical value for t(0.05,4), as found in Appendix 1B, is 2.78. Since texp isgreater than t(0.05, 4), we must reject the null hypothesis and accept thealternative hypothesis. At the 95% confidence level the difference between–X and µ is significant and cannot be explained by indeterminate sources oferror.
There is evidence, therefore, that the results are affected by a determinatesource of error.If evidence for a determinate error is found, as in Example 4.16, its sourceshould be identified and corrected before analyzing additional samples. Failing toreject the null hypothesis, however, does not imply that the method is accurate, butonly indicates that there is insufficient evidence to prove the method inaccurate atthe stated confidence level.–The utility of the t-test for X and µ is improved by optimizing the conditions–used in determining X. Examining equation 4.15 shows that increasing the number of replicate determinations, n, or improving the precision of the analysis enhances the utility of this significance test. A t-test can only give useful results,however, if the standard deviation for the analysis is reasonable.
If the standarddeviation is substantially larger than the expected standard deviation, σ, the con–fidence interval around X will be so large that a significant difference between–X and µ may be difficult to prove. On the other hand, if the standard deviation is–significantly smaller than expected, the confidence interval around X will be too–small, and a significant difference between X and µ may be found when none exists. A significance test that can be used to evaluate the standard deviation is thesubject of the next section.1400-CH04 9/8/99 3:54 PM Page 87Chapter 4 Evaluating Analytical Data4F.2 Comparing s2 to σ2When a particular type of sample is analyzed on a regular basis, it may be possibleto determine the expected, or true variance, σ2, for the analysis. This often is thecase in clinical labs where hundreds of blood samples are analyzed each day.
Replicate analyses of any single sample, however, results in a sample variance, s2. A statistical comparison of s2 to σ2 provides useful information about whether the analysisis in a state of “statistical control.” The null hypothesis is that s2 and σ2 are identical,and the alternative hypothesis is that they are not identical.The test statistic for evaluating the null hypothesis is called an F-test, and isgiven as eithers2σ2Fexp = 2 or Fexp = 2σs4.16(s2 > σ2)(σ2 > s2)depending on whether s2 is larger or smaller than σ2. Note that Fexp is defined suchthat its value is always greater than or equal to 1.If the null hypothesis is true, then Fexp should equal 1.
Due to indeterminate errors, however, the value for Fexp usually is greater than 1. A critical value, F(α, νnum,νden), gives the largest value of F that can be explained by indeterminate error. It ischosen for a specified significance level, α, and the degrees of freedom for the variances in the numerator, νnum, and denominator, νden.
The degrees of freedom for s2is n – 1, where n is the number of replicates used in determining the sample’s variance. Critical values of F for α = 0.05 are listed in Appendix 1C for both one-tailedand two-tailed significance tests.EXAMPLE 4.17A manufacturer’s process for analyzing aspirin tablets has a known variance of25. A sample of ten aspirin tablets is selected and analyzed for the amount ofaspirin, yielding the following results254249252252249249250247251252Determine whether there is any evidence that the measurement process is notunder statistical control at α = 0.05.SOLUTIONThe variance for the sample of ten tablets is 4.3. A two-tailed significance test isused since the measurement process is considered out of statistical control ifthe sample’s variance is either too good or too poor.
The null hypothesis andalternative hypotheses areH0:s2 = σ2The test statistic isFexp =HA:s2 ≠ σ2σ225== 5.84.3s2The critical value for F(0.05, ∞, 9) from Appendix 1C is 3.33. Since F is greaterthan F(0.05,∞, 9), we reject the null hypothesis and accept the alternativehypothesis that the analysis is not under statistical control. One explanation forthe unreasonably small variance could be that the aspirin tablets were notselected randomly.F-testStatistical test for comparing twovariances to see if their difference is toolarge to be explained by indeterminateerror.871400-CH04 9/8/99 3:55 PM Page 8888Modern Analytical Chemistry4F.3 Comparing Two Sample VariancesThe F-test can be extended to the comparison of variances for two samples, A andB, by rewriting equation 4.16 asFexp =s A2sB2where A and B are defined such that sA2 is greater than or equal to sB2.
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