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Introduction to Analytical Chemistry CHAPTER 19 STATISTICAL AIDS TO HYPOTHESIS TESTING AND GROSS ERRORS 19A Statistical Aids To Hypothesis Testing Scientists and engineers frequently must judge whether a numerical difference is a manifestation of the random errors inevitable in all measurements. Tests of this kind make use of a null hypothesis, which assumes that the numerical quantities being compared are, in fact, the same. 19-2 Copyright © 2011 Cengage Learning 19A-1 Comparing an Experimental Mean with the True Value A common way of testing for bias in an analytical method is to use the method to analyze a sample whose composition is accurately known in Figure 19-1. Method A has no bias, so the population mean μA is the true value xt . Method B has a systematic error, or bias, that is given by (19-1) 19-3 Copyright © 2011 Cengage Learning 19A-1 Comparing an Experimental Mean with the True Value It is likely that the experimental mean will differ from the accepted value xt as shown in the figure; the judgment must then be made whether this difference is the consequence of random error or, alternatively, a systematic error. 19-4 Copyright © 2011 Cengage Learning 19A-1 Comparing an Experimental Mean with the True Value In treating this type of problem statistically, the difference is compared with the difference that could be caused by random error. If the observed difference is less than that computed for a chosen probability level, the null hypothesis that are the same cannot be rejected; that is, no significant systematic error has been demonstrated. 19-5 Copyright © 2011 Cengage Learning 19A-1 Comparing an Experimental Mean with the True Value The critical value for rejecting the null hypothesis is (19-2) where N is the number of replicate measurements used in the test. If a good estimate of σis available, Equation 19-2 can be modified by replacing t with z and s with σ. 19-6 Copyright © 2011 Cengage Learning Example 19-1 A new procedure for the rapid determination of sulfur in kerosenes was tested on a sample known from its method of preparation to contain 0.123% S (xt). The results were % S = 0.112, 0.118, 0.115, and 0.119. Do the data indicate that there is bias in the method? 19-7 Copyright © 2011 Cengage Learning Example 19-1 19-8 Copyright © 2011 Cengage Learning Example 19-1 From Table 3-6 (Chapter 3), we find that at the 95% confidence level, t has a value of 3.18 for three degrees of freedom. The values of t from the tables are often called critical values and symbolized tcrit . The test value is calculated from 19-9 Copyright © 2011 Cengage Learning Example 19-1 If we reject the null hypothesis at the confidence level chosen. The absolute value of t is used. This type of test is often called a two-tailed test. In our case Since 4.375 ＞ 3.18, the critical value of t at the 95% confidence level, we conclude that a difference this large is significant and reject the null hypothesis. 19-10 Copyright © 2011 Cengage Learning 19A-2 Comparing Two Experimental Means Since we know from Equation 6-5 that the standard deviation of the mean is and likewise for 19-11 Copyright © 2011 Cengage Learning 19A-2 Comparing Two Experimental Means Thus, the variance sd2 of the difference (d = x1 – x2) between the means is given by 19-12 Copyright © 2011 Cengage Learning 19A-2 Comparing Two Experimental Means If we then assume that the spooled standard deviation spooled is a good estimate of both sm1 and sm2 , then 19-13 Copyright © 2011 Cengage Learning 19A-2 Comparing Two Experimental Means Substituting this equation into Equation 19-2 (and also for xt), we find that (19-3) test value of t is given by (19-4) 19-14 Copyright © 2011 Cengage Learning Example 19-2 Two barrels of wine were analyzed for their alcohol content to determine whether they were from different sources. On the basis of six analyses, the average content of the first barrel was established to be 12.61% ethanol. Four analyses of the second barrel gave a mean of 12.53% alcohol. The ten analyses yielded a pooled value of s = 0.070%. Do the data indicate a difference between the wines? Here we employ Equation 19-4 to calculate the test statistic t. 19-15 Copyright © 2011 Cengage Learning Example 19-2 The critical value of t at the 95% confidence level for 10 – 2 = 8 degrees of freedom is 2.31. Since 1.771 ＜ 2.31, we accept the null hypothesis at the 95% confidence level and conclude that there is no difference in the alcohol content of the wines. 19-16 Copyright © 2011 Cengage Learning 19A-2 Comparing Two Experimental Means In Example 19-2, no significant difference between the alcohol content of the two wines was indicated at the 95% confidence level. Note that this statement is equivalent to saying that is equal to with a certain probability, but the tests do not prove that the wines come from the same source. 19-17 Copyright © 2011 Cengage Learning 19A-2 Comparing Two Experimental Means In contrast, the finding of one significant difference in any test would clearly show that the two wines are different. Thus, the establishment of a significant difference by a single test is much more revealing than the establishment of an absence of difference. 19-18 Copyright © 2011 Cengage Learning 19B Detecting Gross Errors A data point that differs excessively from the mean in a data set is termed an outlier. 19-19 Copyright © 2011 Cengage Learning 19B-1 Using the Q Test the absolute value of the difference between the questionable result xq and its nearest neighbor xn is divided by the spread w of the entire set to give the quantity Qexp : (19-5) This ratio is then compared with rejection values Qcrit found in Table 19-1. If Qexp is greater than Qcrit , the questionable result can be rejected with the indicated degree of confidence. 19-20 Copyright © 2011 Cengage Learning Table 19-1 19-21 Copyright © 2011 Cengage Learning Example 19-3 The analysis of a calcite sample yielded CaO percentages of 55.95, 56.00, 56.04, 56.08, and 56.23. The last value appears anomalous; should it be retained or rejected? The difference between 56.23 and 56.08 is 0.15%. The spread (56.23－55.95) is 0.28%. Thus, For five measurements, Qcrit at the 90% confidence level is 0.64. Because 0.54＜0.64, we must retain the outlier at the 90% confidence level. 19-22 Copyright © 2011 Cengage Learning 19B-2 A Word of Caution about Rejecting Outliers Statistical tests like the Q test, assume that the distribution of the population data is normal, or Gaussian. Unfortunately, this condition cannot be proved or disproved for samples that have many fewer than 50 results. Consequently, statistical rules, should be used with extreme caution when applied to samples containing only a few data. 19-23 Copyright © 2011 Cengage Learning THE END 19-24 Copyright © 2011 Cengage Learning