D. Harvey - Modern Analytical Chemistry (794078), страница 56
Текст из файла (страница 56)
Once Ks is known, themass of sample needed to achieve a desired relative standard deviation for samplingcan be calculated.EXAMPLE7.6The following data were obtained in a preliminary determination of theamount of inorganic ash in a breakfast cereal.Mass of Cereal(g)Percentage Ash(w/w)0.99560.99811.00360.99941.00671.341.291.321.261.28Determine Ks and the amount of sample needed to give a relative standarddeviation for sampling of ±2.0%. Predict the percent relative standarddeviation and the absolute standard deviation if samples of 5 g are collected.SOLUTIONTo determine Ks we need to know the average mass of the cereal samples andthe relative standard deviation for the %(w/w) ash.
The average mass of the fivecereal samples is 1.0007 g. The average %(w/w) ash and the absolute standarddeviation are, respectively, 1.298% and 0.03194. The percent relative standarddeviation, therefore, isR=ss0.03194× 100 =× 100 = 2.46%X1.298ThusKs = mR2 = (1.0007 g)(2.46)2 = 6.06 gThe amount of sample needed to give a relative standard deviation of ±2%,therefore, isK6.06 g= 1.5 gm = 2s =(2.0)2RIf we use 5.00-g samples, then the expected percent relative standard deviation isR=Ks=m6.06 g= 1.10%5.00 gand the expected absolute standard deviation issR = s × 100Xss =RX(1.10)(1.298)== 0.01431001001891400-CH07 9/8/99 4:03 PM Page 190190Modern Analytical ChemistryWhen the target population is segregated, or stratified, equation 7.5 provides a poorestimate of the amount of sample needed to achieve a desired relative standard deviation for sampling.
A more appropriate relationship, which can be applied to bothsegregated and nonsegregated samples, has been proposed.7s s2 =AB+mns ns7.6where ns is the number of samples to be analyzed, m is the mass of each sample, A isa homogeneity constant accounting for the random distribution of analyte in thetarget population, and B is a segregation constant accounting for the nonrandomdistribution of analyte in the target population. Equation 7.6 shows that samplingvariance due to the random distribution of analyte can be minimized by increasingeither the mass of each sample or the total number of samples. Sampling errors dueto the nonrandom distribution of the analyte, however, can only be minimized byincreasing the total number of samples.
Values for the homogeneity constant andheterogeneity constant are determined using two sets of samples that differ significantly in mass.EXAMPLE 7.7To develop a sampling plan for the determination of PCBs in lake sediments,the following two experiments are conducted. First, 15 samples, each with amass of 1.00 g, are analyzed, giving a sampling variance of 0.0183. In a secondexperiment, ten samples, each with a mass of 10.0 g, are analyzed, giving asampling variance of 0.0069.
If samples weighing 5.00 g are to be collected, howmany are needed to give a sampling variance of 0.0100? If five samples are to becollected, how much should each sample weigh?SOLUTIONSubstituting known values for the two experiments into equation 7.6 gives thefollowing pair of simultaneous equations0.0183 =AB+(1.00)(15) 150.0069 =AB+(10.0)(10) 10Solving for A and B gives values of 0.228 and 0.0462, respectively. The numberof 5.00-g samples is determined by solving0.0100 =0.2280.0462+(5.00)nnfor n, giving n = 9.2 ≈ 9 samples. When using five samples, the mass of each isgiven by the equation0.0100 =for which m is 60.0 g.0.228 0.0462+m(5)51400-CH07 9/8/99 4:03 PM Page 191Chapter 7 Obtaining and Preparing Samples for Analysis7B.4 How Many Samples to CollectIn the previous section we considered the amount of sample needed to minimizethe sampling variance. Another important consideration is the number of samplesrequired to achieve a desired maximum sampling error.
If samples drawn from thetarget population are normally distributed, then the following equation describesthe confidence interval for the sampling errortsµ = X± snswhere ns is the number of samples and ss is the sampling standard deviation. Rear–ranging and substituting e for the quantity (µ – X), gives the number of samples asns =s2st 2 s s2e27.7e2where and are both expressed as absolute uncertainties or as relative uncertainties. Finding a solution to equation 7.7 is complicated by the fact that the value of tdepends on ns. As shown in Example 7.8, equation 7.7 is solved iteratively.EXAMPLE 7.8In Example 7.6 we found that an analysis for the inorganic ash content of abreakfast cereal required a sample of 1.5 g to establish a relative standarddeviation for sampling of ±2.0%.
How many samples are needed to obtain arelative sampling error of no more than 0.80% at the 95% confidence level?SOLUTIONBecause the value of t depends on ns, and the value of ns is not yet known, webegin by letting ns = ∞ and use the associated value of t. From Appendix 1B,the value for t is 1.96. Substituting known values into equation 7.7ns =(1.96)2 (2.0)2= 24(0.80)2Letting ns = 24, the value of t from Appendix 1B is 2.075. Recalculating ns givesns =(2.075)2 (2.0)2= 26.9 ≈ 27(0.80)2When ns = 27, the value of t is 2.066.
Recalculating ns, we findns =(2.066)2 (2.0)2= 26.7 ≈ 27(0.80)2Since two successive calculations give the same value for n s , an iterativesolution has been found. Thus, 27 samples are needed to achieve the desiredsampling error.Equation 7.7 only provides an estimate for the smallest number of samples expected to produce the desired sampling error. The actual sampling error may besubstantially higher if the standard deviation for the samples that are collected issignificantly greater than the standard deviation due to sampling used to calculate ns.1911400-CH07 9/8/99 4:03 PM Page 192192Modern Analytical ChemistryThis is not an uncommon problem.
For a target population with a relative samplingvariance of 50 and a desired relative sampling error of ±5%, equation 7.7 predictsthat ten samples are sufficient. In a simulation in which 1000 samples of size 10were collected, however, only 57% of the samples resulted in sampling errors of lessthan ±5%.8 By increasing the number of samples to 17 it was possible to ensure thatthe desired sampling error was achieved 95% of the time.7B.5 Minimizing the Overall VarianceA final consideration in developing a sampling plan is to minimize the overall variance for the analysis. Equation 7.2 shows that the overall variance is a function ofthe variance due to the method and the variance due to sampling. As we have seen,we can improve the variance due to sampling by collecting more samples of propersize.
Increasing the number of times we analyze each sample improves the variance2 , then the method’s variancedue to the method. If ss2 is significantly greater than smcan be ignored and equation 7.7 used to estimate the number of samples to analyze.Analyzing any sample more than once will not improve the overall variance, sincethe variance due to the method is insignificant.2 is significantly greater than s2, then we only need to collect and analyze aIf smssingle sample. The number of replicate analyses, nr, needed to minimize the errordue to the method is given by an equation similar to equation 7.7nr =2t 2sme2Unfortunately, the simple situations just described are often the exception. Inmany cases, both the sampling variance and method variance are significant, andboth multiple samples and replicate analyses of each sample are required.
The overall error in this circumstance is given by s2s2 e = t s + m ns ns nr 1/27.8Equation 7.8 does not have a unique solution because different combinations of nsand nr give the same overall error. The choice of how many samples to collect andhow many times each sample should be analyzed is determined by other concerns,such as the cost of collecting and analyzing samples, and the amount of availablesample.EXAMPLE 7.9A certain analytical method has a relative sampling variance of 0.40% and arelative method variance of 0.070%. Evaluate the relative error (α = 0.05) if(a) you collect five samples, analyzing each twice; and, (b) you collect twosamples, analyzing each five times.SOLUTIONBoth sampling strategies require a total of ten determinations.
Using Appendix 1B,we find that the value of t is 2.26. Substituting into equation 7.8, we find that therelative error for the first sampling strategy is 0.40 0.070 e = 2.26+(5)(2) 51/2= 0.67%1400-CH07 9/8/99 4:03 PM Page 193Chapter 7 Obtaining and Preparing Samples for Analysisand that for the second sampling strategy is 0.40 0.070 e = 2.26+(2)(5) 21/2= 1.0%As expected, since the relative method variance is better than the relativesampling variance, a sampling strategy that favors the collection of moresamples and few replicate analyses gives the better relative error.7C Implementing the Sampling PlanAfter a sampling plan has been developed, it is put into action. Implementing asampling plan normally involves three steps: physically removing the sample fromits target population, preserving the sample, and preparing the sample for analysis.Except for in situ sampling, the analysis of a sample occurs after removing it fromthe target population.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
