D. Harvey - Modern Analytical Chemistry (794078), страница 26
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The chief limitation to thecentral limit theorem is that the sources of indeterminate error must be independent and of similar magnitude so that no one source of error dominates the finaldistribution.Estimating µ and σ2 Our comparison of the histogram for the data in Table 4.12–to a normal distribution assumes that the sample’s mean, X, and variance, s2, areappropriate estimators of the population’s mean, µ, and variance, σ2.
Why did we–select X and s2, as opposed to other possible measures of central tendency and–spread? The explanation is simple; X and s2 are considered unbiased estimators of µand σ2.7,8 If we could analyze every possible sample of equal size for a given population (e.g., every possible sample of five pennies), calculating their respective meansand variances, the average mean and the average variance would equal µ and σ2. Al–though X and s2 for any single sample probably will not be the same as µ or σ2, theyprovide a reasonable estimate for these values.Figure 4.8Histogram for data in Table 4.12. A normal–distribution curve for the data, based on X2and s , is superimposed on the histogram.central limit theoremThe distribution of measurementssubject to indeterminate errors is often anormal distribution.1400-CH04 9/8/99 3:54 PM Page 8080Modern Analytical ChemistryDegrees of Freedom Unlike the population’s variance, the variance of a sample includes the term n – 1 in the denominator, where n is the size of the sample∑i =1 ( Xins2 =degrees of freedomThe number of independent values onwhich a result is based (ν).− X )2n −14.12Defining the sample’s variance with a denominator of n, as in the case of the population’s variance leads to a biased estimation of σ2.
The denominators of the variance equations 4.8 and 4.12 are commonly called the degrees of freedom for thepopulation and the sample, respectively. In the case of a population, the degrees offreedom is always equal to the total number of members, n, in the population. For–the sample’s variance, however, substituting X for µ removes a degree of freedomfrom the calculation. That is, if there are n members in the sample, the value of the–nth member can always be deduced from the remaining n – 1 members and X.
Forexample, if we have a sample with five members, and we know that four of themembers are 1, 2, 3, and 4, and that the mean is 3, then the fifth member of thesample must be–(X × n) – X1 – X2 – X3 – X4 = (3 × 5) – 1 – 2 – 3 – 4 = 54D.5 Confidence Intervals for SamplesEarlier we introduced the confidence interval as a way to report the most probablevalue for a population’s mean, µ, when the population’s standard deviation, σ, isknown. Since s2 is an unbiased estimator of σ2, it should be possible to constructconfidence intervals for samples by replacing σ in equations 4.10 and 4.11 with s.Two complications arise, however.
The first is that we cannot define s2 for a singlemember of a population. Consequently, equation 4.10 cannot be extended to situations in which s2 is used as an estimator of σ2. In other words, when σ is unknown,we cannot construct a confidence interval for µ by sampling only a single memberof the population.The second complication is that the values of z shown in Table 4.11 are derivedfor a normal distribution curve that is a function of σ2, not s2. Although s2 is an unbiased estimator of σ2, the value of s2 for any randomly selected sample may differsignificantly from σ2.
To account for the uncertainty in estimating σ2, the term z inequation 4.11 is replaced with the variable t, where t is defined such that t ≥ z at allconfidence levels. Thus, equation 4.11 becomesµ = X ±tsn4.13Values for t at the 95% confidence level are shown in Table 4.14. Note that t becomes smaller as the number of the samples (or degrees of freedom) increase, approaching z as n approaches infinity. Additional values of t for other confidence levels can be found in Appendix 1B.EXAMPLE 4.15What is the 95% confidence interval for the data in Table 4.1?SOLUTIONThe mean and standard deviation for this sample are, respectively, 3.117 g and0.051 g.
Since the sample consists of seven measurements, there are six degrees1400-CH04 9/8/99 3:54 PM Page 81Chapter 4 Evaluating Analytical Dataof freedom. The value of t from Table 4.14, is 2.45. Substituting into equation4.13 givesµ = X ±ts= 3.117 ±(2.45)(0.051)n= 3.117 ± 0.047 g7Thus, there is a 95% probability that the population’s mean is between 3.070and 3.164 g.Table 4.14Values of t for the 95%Confidence IntervalDegrees ofFreedomt1234567891012141618203050∞12.714.303.182.782.572.452.362.312.262.232.182.142.122.102.092.042.011.964D.6 A Cautionary StatementThere is a temptation when analyzing data to plug numbers into an equation, carryout the calculation, and report the result.
This is never a good idea, and you shoulddevelop the habit of constantly reviewing and evaluating your data. For example, ifanalyzing five samples gives an analyte’s mean concentration as 0.67 ppm with astandard deviation of 0.64 ppm, then the 95% confidence interval isµ = 0.67 ±(2.78)(0.64)= 0.64 ± 0.80 ppm5This confidence interval states that the analyte’s true concentration lies within therange of –0.16 ppm to 1.44 ppm. Including a negative concentration within the confidence interval should lead you to reevaluate your data or conclusions. On furtherinvestigation your data may show that the standard deviation is larger than expected,811400-CH04 9/8/99 3:54 PM Page 8282Modern Analytical Chemistrymaking the confidence interval too broad, or you may conclude that the analyte’sconcentration is too small to detect accurately.*A second example is also informative.
When samples are obtained from a normally distributed population, their values must be random. If results for severalsamples show a regular pattern or trend, then the samples cannot be normally distributed. This may reflect the fact that the underlying population is not normallydistributed, or it may indicate the presence of a time-dependent determinate error.For example, if we randomly select 20 pennies and find that the mass of each pennyexceeds that of the preceding penny, we might suspect that the balance on whichthe pennies are being weighed is drifting out of calibration.4E Statistical Analysis of DataValues(a)In the previous section we noted that the result of an analysis is best expressed as aconfidence interval.
For example, a 95% confidence interval for the mean of five results gives the range in which we expect to find the mean for 95% of all samples ofequal size, drawn from the same population. Alternatively, and in the absence of determinate errors, the 95% confidence interval indicates the range of values in whichwe expect to find the population’s true mean.The probabilistic nature of a confidence interval provides an opportunity to askand answer questions comparing a sample’s mean or variance to either the acceptedvalues for its population or similar values obtained for other samples.
For example,confidence intervals can be used to answer questions such as “Does a newly developed method for the analysis of cholesterol in blood give results that are significantly different from those obtained when using a standard method?” or “Is there asignificant variation in the chemical composition of rainwater collected at differentsites downwind from a coalburning utility plant?” In this section we introduce ageneral approach to the statistical analysis of data.
Specific statistical methods ofanalysis are covered in Section 4F.4E.1 Significance TestingValues(b)Values(c)Figure 4.9Three examples of possible relationshipsbetween the probability distributions for twopopulations. (a) Completely separatedistributions; (b) Distributions with a greatdeal of overlap; (c) Distributions with someoverlap.Let’s consider the following problem. Two sets of blood samples have been collectedfrom a patient receiving medication to lower her concentration of blood glucose.One set of samples was drawn immediately before the medication was administered; the second set was taken several hours later.
The samples are analyzed andtheir respective means and variances reported. How do we decide if the medicationwas successful in lowering the patient’s concentration of blood glucose?One way to answer this question is to construct probability distribution curvesfor each sample and to compare the curves with each other.
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