Chapter 4 Evaluating Analytical Data.pdf

CHAPTER OVERVIEW 4: Evaluating Analytical Data When we use an analytical method we make three separate evaluations of experimental error. First, before we begin the analysis we evaluate potential sources of errors to ensure they will not adversely effect our results. Second, during the analysis we...

CHAPTER OVERVIEW 4: Evaluating Analytical Data When we use an analytical method we make three separate evaluations of experimental error. First, before we begin the analysis we evaluate potential sources of errors to ensure they will not adversely effect our results. Second, during the analysis we monitor our measurements to ensure that errors remain acceptable. Finally, at the end of the analysis we evaluate the quality of the measurements and results, and compare them to our original design criteria. This chapter provides an introduction to sources of error, to evaluating errors in analytical measurements, and to the statistical analysis of data. 4.1: Characterizing Measurements and Results 4.2: Characterizing Experimental Errors 4.3: Propagation of Uncertainty 4.4: The Distribution of Measurements and Results 4.5: Statistical Analysis of Data 4.6: Statistical Methods for Normal Distributions 4.7: Detection Limits 4.8: Using Excel and R to Analyze Data 4.9: Problems 4.10: Additional Resources 4.11: Chapter Summary and Key Terms This page titled 4: Evaluating Analytical Data is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey. 1 4.1: Characterizing Measurements and Results Let’s begin by choosing a simple quantitative problem that requires a single measurement: What is the mass of a penny? You probably recognize that our statement of the problem is too broad. For example, are we interested in the mass of a United States penny or of a Canadian penny, or is the difference relevant? Because a penny’s composition and size may differ from country to country, let’s narrow our problem to pennies from the United States. There are other concerns we might consider. For example, the United States Mint produces pennies at two locations (Figure 4.1.1 ). Because it seems unlikely that a penny’s mass depends on where it is minted, we will ignore this concern. Another concern is whether the mass of a newly minted penny is different from the mass of a circulating penny. Because the answer this time is not obvious, let’s further narrow our question and ask “What is the mass of a circulating United States Penny?” Figure 4.1.1 : An uncirculated 2005 Lincoln head penny. The “D” below the date indicates that this penny was produced at the United States Mint at Denver, Colorado. Pennies produced at the Philadelphia Mint do not have a letter below the date. Source: United States Mint image. A good way to begin our analysis is to gather some preliminary data. Table 4.1.1 shows masses for seven pennies collected from my change jar. In examining this data we see that our question does not have a simple answer. That is, we can not use the mass of a single penny to draw a specific conclusion about the mass of any other penny (although we might reasonably conclude that all pennies weigh at least 3 g). We can, however, characterize this data by reporting the spread of the individual measurements around a central value. Table 4.1.1 : Masses of Seven Circulating U. S. Pennies Penny Mass (g) 1 3.080 2 3.094 3 3.107 4 3.056 5 3.112 6 3.174 7 3.198 Measures of Central Tendency One way to characterize the data in Table 4.1.1 is to assume that the masses of individual pennies are scattered randomly around a central value that is the best estimate of a penny’s expected, or “true” mass. There are two common ways to estimate central tendency: the mean and the median. Mean ¯¯¯ ¯ The mean, X , is the numerical average for a data set. We calculate the mean by dividing the sum of the individual values by the size of the data set n ¯¯¯ ¯ ∑ Xi i=1 X = n th where X is the i measurement, and n is the size of the data set. i  Example 4.1.1 What is the mean for the data in Table 4.1.1 ? 4.1.1 https://chem.libretexts.org/@go/page/401340  Solution To calculate the mean we add together the results for all measurements 3.080 + 3.094 + 3.107 + 3.056 + 3.112 + 3.174 + 3.198 = 21.821 g and divide by the number of measurements ¯¯¯ ¯ 21.821 g X = = 3.117 g 7 The mean is the most common estimate of central tendency. It is not a robust estimate, however, because a single extreme value— one much larger or much smaller than the remainder of the data—influences strongly the mean’s value [Rousseeuw, P. J. J. Chemom. 1991, 5, 1–20]. For example, if we accidently record the third penny’s mass as 31.07 g instead of 3.107 g, the mean changes from 3.117 g to 7.112 g! An estimate for a statistical parameter is robust if its value is not affected too much by an unusually large or an unusually small measurement. Median The median, X̃ , is the middle value when we order our data from the smallest to the largest value. When the data has an odd number of values, the median is the middle value. For an even number of values, the median is the average of the n/2 and the (n/2) + 1 values, where n is the size of the data set. When n = 5, the median is the third value in the ordered data set; for n = 6, the median is the average of the third and fourth members of the ordered data set.  Example 4.1.2 What is the median for the data in Table 4.1.1 ? Solution To determine the median we order the measurements from the smallest to the largest value 3.056 3.080 3.094 3.107 3.112 3.174 3.198 Because there are seven measurements, the median is the fourth value in the ordered data; thus, the median is 3.107 g. As shown by Example 4.1.1 and Example 4.1.2 , the mean and the median provide similar estimates of central tendency when all measurements are comparable in magnitude. The median, however, is a more robust estimate of central tendency because it is less sensitive to measurements with extreme values. For example, if we accidently record the third penny’s mass as 31.07 g instead of 3.107 g, the median’s value changes from 3.107 g to 3.112 g. Measures of Spread If the mean or the median provides an estimate of a penny’s expected mass, then the spread of individual measurements about the mean or median provides an estimate of the difference in mass among pennies or of the uncertainty in measuring mass with a balance. Although we often define the spread relative to a specific measure of central tendency, its magnitude is independent of the central value. Although shifting all measurements in the same direction by adding or subtracting a constant value changes the mean or median, it does not change the spread. There are three common measures of spread: the range, the standard deviation, and the variance. Problem 13 at the end of the chapter asks you to show that this is true. 4.1.2 https://chem.libretexts.org/@go/page/401340 Range The range, w, is the difference between a data set’s largest and smallest values. w = Xlargest − Xsmallest The range provides information about the total variability in the data set, but does not provide information about the distribution of individual values. The range for the data in Table 4.1.1 is w = 3.198 g − 3.056 g = 0.142 g Standard Deviation The standard deviation, s, describes the spread of individual values about their mean, and is given as −−−−−−−−−−−−− n ¯¯¯ ¯ 2 ∑ (Xi − X ) i=1 s =√ (4.1.1) n−1 where X is one of the n individual values in the data set, and X is the data set's mean value. Frequently, we report the relative i ¯¯¯ ¯ standard deviation, sr, instead of the absolute standard deviation. s sr = ¯¯¯ ¯ X The percent relative standard deviation, %sr, is s r × 100. The relative standard deviation is important because it allows for a more meaningful comparison between data sets when the individual measurements differ significantly in magnitude. Consider again the data in Table 4.1.1. If we multiply each value by 10, the absolute standard deviation will increase by 10 as well; the relative standard deviation, however, is the same.  Example 4.1.3 Report the standard deviation, the relative standard deviation, and the percent relative standard deviation for the data in Table 4.1.1 ? Solution To calculate the standard deviation we first calculate the difference between each measurement and the data set’s mean value (3.117), square the resulting differences, and add them together to find the numerator of Equation 4.1.1 2 2 (3.080 − 3.117 ) = (−0.037 ) = 0.001369 2 2 (3.094 − 3.117 ) = (−0.023 ) = 0.000529 2 2 (3.107 − 3.117 ) = (−0.010 ) = 0.000100 2 2 (3.056 − 3.117 ) = (−0.061 ) = 0.003721 2 2 (3.112 − 3.117 ) = (−0.005 ) = 0.000025 2 2 (3.174 − 3.117 ) = (+0.057 ) = 0.003249 2 2 (3.198 − 3.117 ) = (+0.081 ) = 0.006561 ––––––––– 0.015554 For obvious reasons, the numerator of Equation 4.1.1 is called a sum of squares. Next, we divide this sum of squares by n – 1, where n is the number of measurements, and take the square root. −−−−−−−− 0.015554 s =√ = 0.051 g 7 −1 Finally, the relative standard deviation and percent relative standard deviation are 0.051 g sr = = 0.016 3.117 g 4.1.3 https://chem.libretexts.org/@go/page/401340  %sr = (0.016) × 100 = 1.6% It is much easier to determine the standard deviation using a scientific calculator with built in statistical functions. Many scientific calculators include two keys for calculating the standard deviation. One key calculates the standard deviation for a data set of n samples drawn from a larger collection of possible samples, which corresponds to Equation 4.1.1. The other key calculates the standard deviation for all possible samples. The latter is known as the population’s standard deviation, which we will cover later in this chapter. Your calculator’s manual will help you determine the appropriate key for each. Variance Another common measure of spread is the variance, which is the square of the standard deviation. We usually report a data set’s standard deviation, rather than its variance, because the mean value and the standard deviation share the same unit. As we will see shortly, the variance is a useful measure of spread because its values are additive.  Example 4.1.4 What is the variance for the data in Table 4.1.1 ? Solution The variance is the square of the absolute standard deviation. Using the standard deviation from Example 4.1.3 gives the variance as 2 2 s = (0.051 ) = 0.0026  Exercise 4.1.1 The following data were collected as part of a quality control study for the analysis of sodium in serum; results are concentrations of Na+ in mmol/L. 140 143 141 137 132 157 143 149 118 145 Report the mean, the median, the range, the standard deviation, and the variance for this data. This data is a portion of a larger data set from Andrew, D. F.; Herzberg, A. M. Data: A Collection of Problems for the Student and Research Worker, Springer- Verlag:New York, 1985, pp. 151–155. Answer Mean: To find the mean we add together the individual measurements and divide by the number of measurements. The sum of the 10 concentrations is 1405. Dividing the sum by 10, gives the mean as 140.5, or 1.40 × 10 mmol/L. 2 Median: To find the median we arrange the 10 measurements from the smallest concentration to the largest concentration; thus 118 132 137 140 141 143 143 145 149 157 The median for a data set with 10 members is the average of the fifth and sixth values; thus, the median is (141 + 143)/2, or 142 mmol/L. Range: The range is the difference between the largest value and the smallest value; thus, the range is 157 – 118 = 39 mmol/L. Standard Deviation: To calculate the standard deviation we first calculate the absolute difference between each measurement and the mean value (140.5), square the resulting differences, and add them together. The differences are – 0.5 2.5 0.5 – 3.5 – 8.5 16.5 2.5 8.5 – 22.5 4.5 and the squared differences are 0.25 6.25 0.25 12.25 72.25 272.25 6.25 72.25 506.25 20.25 4.1.4 https://chem.libretexts.org/@go/page/401340  The total sum of squares, which is the numerator of Equation 4.1.1, is 968.50. The standard deviation is −−−−−− 968.50 s =√ = 10.37 ≈ 10.4 10 − 1 Variance: The variance is the square of the standard deviation, or 108. This page titled 4.1: Characterizing Measurements and Results is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey. 4.1: Characterizing Measurements and Results by David Harvey is licensed CC BY-NC-SA 4.0. 4.1.5 https://chem.libretexts.org/@go/page/401340 4.2: Characterizing Experimental Errors Characterizing a penny’s mass using the data in Table 4.1.1 suggests two questions. First, does our measure of central tendency agree with the penny’s expected mass? Second, why is there so much variability in the individual results? The first of these questions addresses the accuracy of our measurements and the second addresses the precision of our measurements. In this section we consider the types of experimental errors that affect accuracy and precision. Errors That Affect Accuracy Accuracy is how close a measure of central tendency is to its expected value, μ. We express accuracy either as an absolute error, e ¯¯¯ ¯ e = X −μ (4.2.1) or as a percent relative error, %e ¯¯¯ ¯ X −μ %e = × 100 (4.2.2) μ Although Equation 4.2.1 and Equation 4.2.2 use the mean as the measure of central tendency, we also can use the median. The convention for representing a statistical parameter is to use a Roman letter for a value calculated from experimental data, ¯¯¯ ¯ and a Greek letter for its corresponding expected value. For example, the experimentally determined mean is X and its underlying expected value is μ. Likewise, the experimental standard deviation is s and the underlying expected value is σ. We identify as determinate an error that affects the accuracy of an analysis. Each source of a determinate error has a specific magnitude and sign. Some sources of determinate error are positive and others are negative, and some are larger in magnitude and others are smaller in magnitude. The cumulative effect of these determinate errors is a net positive or negative error in accuracy. It is possible, although unlikely, that the positive and negative determinate errors will offset each other, producing a result with no net error in accuracy. We assign determinate errors into four categories—sampling errors, method errors, measurement errors, and personal errors—each of which we consider in this section. Sampling Errors A determinate sampling error occurs when our sampling strategy does not provide a us with a representative sample. For example, if we monitor the environmental quality of a lake by sampling from a single site near a point source of pollution, such as an outlet for industrial effluent, then our results will be misleading. To determine the mass of a U. S. penny, our strategy for selecting pennies must ensure that we do not include pennies from other countries. An awareness of potential sampling errors especially is important when we work with heterogeneous materials. Strategies for obtaining representative samples are covered in Chapter 5. Method Errors In any analysis the relationship between the signal, Stotal, and the absolute amount of analyte, nA, or the analyte’s concentration, CA, is Stotal = kA nA + Smb (4.2.3) Stotal = kA CA + Smb (4.2.4) where kA is the method’s sensitivity for the analyte and Smb is the signal from the method blank. A method error exists when our value for kA or for Smb is in error. For example, a method in which Stotal is the mass of a precipitate assumes that k is defined by a pure precipitate of known stoichiometry. If this assumption is not true, then the resulting determination of nA or CA is inaccurate. 4.2.1 https://chem.libretexts.org/@go/page/401341 We can minimize a determinate error in kA by calibrating the method. A method error due to an interferent in the reagents is minimized by using a proper method blank. Measurement Errors The manufacturers of analytical instruments and equipment, such as glassware and balances, usually provide a statement of the item’s maximum measurement error, or tolerance. For example, a 10-mL volumetric pipet (Figure 4.2.1 ) has a tolerance of ±0.02 mL, which means the pipet delivers an actual volume within the range 9.98–10.02 mL at a temperature of 20 oC. Although we express this tolerance as a range, the error is determinate; that is, the pipet’s expected volume, μ , is a fixed value within this stated range. Figure 4.2.1 : Close-up of a 10-mL volumetric pipet showing that it has a tolerance of ±0.02 mL at 20 oC. Volumetric glassware is categorized into classes based on its relative accuracy. Class A glassware is manufactured to comply with tolerances specified by an agency, such as the National Institute of Standards and Technology or the American Society for Testing and Materials. The tolerance level for Class A glassware is small enough that normally we can use it without calibration. The tolerance levels for Class B glassware usually are twice that for Class A glassware. Other types of volumetric glassware, such as beakers and graduated cylinders, are not used to measure volume accurately. Table 4.2.1 provides a summary of typical measurement errors for Class A volumetric glassware. Tolerances for digital pipets and for balances are provided in Table 4.2.2 and Table 4.2.3. Table 4.2.1 : Measurement Errors for Type A Volumetric Glassware Transfer Pipets Volumetric Flasks Burets Capacity (mL) Tolerance (mL) Capacity (mL) Tolerance (mL) Capacity (mL) Tolerance (mL) 1 ±0.006 5 ±0.02 10 ±0.02 2 ±0.006 10 ±0.02 25 ±0.03 5 ±0.01 25 ±0.03 50 ±0.05 10 ±0.02 50 ±0.05 20 ±0.03 100 ±0.08 25 ±0.03 250 ±0.12 50 ±0.05 500 ±0.20 100 ±0.08 1000 ±0.30 2000 Table 4.2.2 : Measurement Errors for Digital Pipets Pipet Range Volume (mL or μL) Percent Measurement Error 10–100 μL 10 ±3.0% 50 ±1.0% 100 ±0.8% 100–1000 μL 100 ±3.0% 500 ±1.0% 1000 ±0.6% 4.2.2 https://chem.libretexts.org/@go/page/401341  Pipet Range Volume (mL or μL) Percent Measurement Error 1–10 mL 1 ±3.0% 5 ±0.8% 10 ±0.6% The tolerance values for the volumetric glassware in Table 4.2.1 are from the ASTM E288, E542, and E694 standards. The measurement errors for the digital pipets in Table 4.2.2 are from www.eppendorf.com. Table 4.2.3 : Measurement Errors for Selected Balances Balance Capacity (g) Measurement Error Precisa 160M 160 ±1 mg A & D ER 120M 120 ±0.1 mg Metler H54 160 ±0.01 mg We can minimize a determinate measurement error by calibrating our equipment. Balances are calibrated using a reference weight whose mass we can trace back to the SI standard kilogram. Volumetric glassware and digital pipets are calibrated by determining the mass of water delivered or contained and using the density of water to calculate the actual volume. It is never safe to assume that a calibration does not change during an analysis or over time. One study, for example, found that repeatedly exposing volumetric glassware to higher temperatures during machine washing and oven drying, led to small, but significant changes in the glassware’s calibration [Castanheira, I.; Batista, E.; Valente, A.; Dias, G.; Mora, M.; Pinto, L.; Costa, H. S. Food Control 2006, 17, 719–726]. Many instruments drift out of calibration over time and may require frequent recalibration during an analysis. Personal Errors Finally, analytical work is always subject to personal error, examples of which include the ability to see a change in the color of an indicator that signals the endpoint of a titration, biases, such as consistently overestimating or underestimating the value on an instrument’s readout scale, failing to calibrate instrumentation, and misinterpreting procedural directions. You can minimize personal errors by taking proper care. Identifying Determinate Errors Determinate errors often are difficult to detect. Without knowing the expected value for an analysis, the usual situation in any analysis that matters, we often have nothing to which we can compare our experimental result. Nevertheless, there are strategies we can use to detect determinate errors. The magnitude of a constant determinate error is the same for all samples and is more significant when we analyze smaller samples. Analyzing samples of different sizes, therefore, allows us to detect a constant determinate error. For example, consider a quantitative analysis in which we separate the analyte from its matrix and determine its mass. Let’s assume the sample is 50.0% w/w analyte. As we see in Table 4.2.4 , the expected amount of analyte in a 0.100 g sample is 0.050 g. If the analysis has a positive constant determinate error of 0.010 g, then analyzing the sample gives 0.060 g of analyte, or an apparent concentration of 60.0% w/w. As we increase the size of the sample the experimental results become closer to the expected result. An upward or downward trend in a graph of the analyte’s experimental concentration versus the sample’s mass (Figure 4.2.2 ) is evidence of a constant determinate error. Table 4.2.4 : Effect of a Constant Determinate Error on the Analysis of a Sample That is 50.0% w/w Analyte Experimental Expected Mass Experimental Mass of Sample (g) Constant Error (g) Concentration of Analyte of Analyte (g) Mass of Analyte (g) (% w/w) 0.100 0.050 0.010 0.060 60.0 0.200 0.100 0.010 0.110 55.0 0.400 0.200 0.010 0.210 52.5 4.2.3 https://chem.libretexts.org/@go/page/401341  Experimental Expected Mass Experimental Mass of Sample (g) Constant Error (g) Concentration of Analyte of Analyte (g) Mass of Analyte (g) (% w/w) 0.800 0.400 0.010 0.410 51.2 1.600 0.800 0.010 0.810 50.6 Figure 4.2.2 : Effect of a constant positive determinate error of +0.01 g and a constant negative determinate error of –0.01 g on the determination of an analyte in samples of varying size. The analyte’s expected concentration of 50% w/w is shown by the dashed line. A proportional determinate error, in which the error’s magnitude depends on the amount of sample, is more difficult to detect because the result of the analysis is independent of the amount of sample. Table 4.2.5 outlines an example that shows the effect of a positive proportional error of 1.0% on the analysis of a sample that is 50.0% w/w in analyte. Regardless of the sample’s size, each analysis gives the same result of 50.5% w/w analyte. Table 4.2.5 : Effect of a Proportional Determinate Error on the Analysis of a Sample That is 50.0% w/w Analyte Experimental Expected Mass Proportional Experimental Mass of Sample (g) Concentration of Analyte of Analyte (g) Error (%) Mass of Analyte (g) (% w/w) 0.100 0.050 1.00 0.0505 50.5 0.200 0.100 1.00 0.101 50.5 0.400 0.200 1.00 0.202 50.5 0.800 0.400 1.00 0.404 50.5 1.600 0.800 1.00 0.808 50.5 One approach for detecting a proportional determinate error is to analyze a standard that contains a known amount of analyte in a matrix similar to our samples. Standards are available from a variety of sources, such as the National Institute of Standards and Technology (where they are called Standard Reference Materials) or the American Society for Testing and Materials. Table 4.2.6 , for example, lists certified values for several analytes in a standard sample of Gingko biloba leaves. Another approach is to compare our analysis to an analysis carried out using an independent analytical method that is known to give accurate results. If the two methods give significantly different results, then a determinate error is the likely cause. Table 4.2.6 : Certified Concentrations for SRM 3246: Gingko bilbo (Leaves) Class of Analyte Analyte Mass Fraction (mg/g or ng/g) Flavonoids/Ginkgolide B (mass fraction in Qurecetin 2.69 ± 0.31 mg/g) Kaempferol 3.02 ± 0.41 Isorhamnetin 0.517 ± 0.0.99 Total Aglycones 6.22 ± 0.77 Selected Terpenes (mass fraction in mg/g) Ginkgolide A 0.57 ± 0.28 4.2.4 https://chem.libretexts.org/@go/page/401341  Class of Analyte Analyte Mass Fraction (mg/g or ng/g) Ginkgolide B 0.470 ± 0.090 Ginkgolide C 0.59 ± 0.22 Ginkgolide J 0.18 ± 0.10 Bilobalide 1.52 ± 0.40 Total Terpene Lactones 3.3 ± 1.1 Selected Toxic Elements (mass fraction in Cadmium 20.8 ± 1.0 ng/g) Lead 995 ± 30 Mercury 23.08 ± 0.17 The primary purpose of this Standard Reference Material is to validate analytical methods for determining flavonoids, terpene lactones, and toxic elements in Ginkgo biloba or other materials with a similar matrix. Values are from the official Certificate of Analysis available at www.nist.gov. Constant and proportional determinate errors have distinctly different sources, which we can define in terms of the relationship between the signal and the moles or concentration of analyte (Equation 4.2.3 and Equation 4.2.4). An invalid method blank, Smb, is a constant determinate error as it adds or subtracts the same value to the signal. A poorly calibrated method, which yields an invalid sensitivity for the analyte, kA, results in a proportional determinate error. Errors that Affect Precision As we saw in Section 4.1, precision is a measure of the spread of individual measurements or results about a central value, which we express as a range, a standard deviation, or a variance. Here we draw a distinction between two types of precision: repeatability and reproducibility. Repeatability is the precision when a single analyst completes an analysis in a single session using the same solutions, equipment, and instrumentation. Reproducibility, on the other hand, is the precision under any other set of conditions, including between analysts or between laboratory sessions for a single analyst. Since reproducibility includes additional sources of variability, the reproducibility of an analysis cannot be better than its repeatability. The ratio of the standard deviation associated with reproducibility to the standard deviation associated with repeatability is called the Horowitz ratio. For a wide variety of analytes in foods, for example, the median Horowtiz ratio is 2.0 with larger values for fatty acids and for trace elements; see Thompson, M.; Wood, R. “The ‘Horowitz Ratio’–A Study of the Ratio Between Reproducibility and Repeatability in the Analysis of Foodstuffs,” Anal. Methods, 2015, 7, 375–379. Errors that affect precision are indeterminate and are characterized by random variations in their magnitude and their direction. Because they are random, positive and negative indeterminate errors tend to cancel, provided that we make a sufficient number of measurements. In such situations the mean and the median largely are unaffected by the precision of the analysis. Sources of Indeterminate Error We can assign indeterminate errors to several sources, including collecting samples, manipulating samples during the analysis, and making measurements. When we collect a sample, for instance, only a small portion of the available material is taken, which increases the chance that small-scale inhomogeneities in the sample will affect repeatability. Individual pennies, for example, may show variations in mass from several sources, including the manufacturing process and the loss of small amounts of metal or the addition of dirt during circulation. These variations are sources of indeterminate sampling errors. During an analysis there are many opportunities to introduce indeterminate method errors. If our method for determining the mass of a penny includes directions for cleaning them of dirt, then we must be careful to treat each penny in the same way. Cleaning some pennies more vigorously than others might introduce an indeterminate method error. 4.2.5 https://chem.libretexts.org/@go/page/401341 Finally, all measuring devices are subject to indeterminate measurement errors due to limitations in our ability to read its scale. For example, a buret with scale divisions every 0.1 mL has an inherent indeterminate error of ±0.01–0.03 mL when we estimate the volume to the hundredth of a milliliter (Figure 4.2.3 ). Figure 4.2.3 : Close-up of a buret showing the difficulty in estimating volume. With scale divisions every 0.1 mL it is difficult to read the actual volume to better than ±0.01–0.03 mL. Evaluating Indeterminate Error Indeterminate errors associated with our analytical equipment or instrumentation generally are easy to estimate if we measure the standard deviation for several replicate measurements, or if we monitor the signal’s fluctuations over time in the absence of analyte (Figure 4.2.4 ) and calculate the standard deviation. Other sources of indeterminate error, such as treating samples inconsistently, are more difficult to estimate. Figure 4.2.4 : Background noise in an instrument showing the random fluctuations in the signal. To evaluate the effect of an indeterminate measurement error on our analysis of the mass of a circulating United States penny, we might make several determinations of the mass for a single penny (Table 4.2.7 ). The standard deviation for our original experiment (see Table 4.1.1) is 0.051 g, and it is 0.0024 g for the data in Table 4.2.7. The significantly better precision when we determine the mass of a single penny suggests that the precision of our analysis is not limited by the balance. A more likely source of indeterminate error is a variability in the masses of individual pennies. Table 4.2.7 : Replicate Determinations of the Mass of a Single Circulating U. S. Penny Replicate Mass (g) Replicate Mass (g) 1 3.025 6 3.023 2 3.024 7 3.022 3 3.028 8 3.021 4 3.027 9 3.026 5 3.028 10 3.024 In Section 4.5 we will discuss a statistical method—the F-test—that you can use to show that this difference is significant. Error and Uncertainty Analytical chemists make a distinction between error and uncertainty [Ellison, S.; Wegscheider, W.; Williams, A. Anal. Chem. 1997, 69, 607A–613A]. Error is the difference between a single measurement or result and its expected value. In other words, error 4.2.6 https://chem.libretexts.org/@go/page/401341 is a measure of bias. As discussed earlier, we divide errors into determinate and indeterminate sources. Although we can find and correct a source of determinate error, the indeterminate portion of the error remains. Uncertainty expresses the range of possible values for a measurement or result. Note that this definition of uncertainty is not the same as our definition of precision. We calculate precision from our experimental data and use it to estimate the magnitude of indeterminate errors. Uncertainty accounts for all errors—both determinate and indeterminate—that reasonably might affect a measurement or a result. Although we always try to correct determinate errors before we begin an analysis, the correction itself is subject to uncertainty. Here is an example to help illustrate the difference between precision and uncertainty. Suppose you purchase a 10-mL Class A pipet from a laboratory supply company and use it without any additional calibration. The pipet’s tolerance of ±0.02 mL is its uncertainty because your best estimate of its expected volume is 10.00 mL ± 0.02 mL. This uncertainty primarily is determinate. If you use the pipet to dispense several replicate samples of a solution and determine the volume of each sample, the resulting standard deviation is the pipet’s precision. Table 4.2.8 shows results for ten such trials, with a mean of 9.992 mL and a standard deviation of ±0.006 mL. This standard deviation is the precision with which we expect to deliver a solution using a Class A 10-mL pipet. In this case the pipet’s published uncertainty of ±0.02 mL is worse than its experimentally determined precision of ±0.006 ml. Interestingly, the data in Table 4.2.8 allows us to calibrate this specific pipet’s delivery volume as 9.992 mL. If we use this volume as a better estimate of the pipet’s expected volume, then its uncertainty is ±0.006 mL. As expected, calibrating the pipet allows us to decrease its uncertainty [Kadis, R. Talanta 2004, 64, 167–173]. Table 4.2.8 : Experimental Results for Volume Dispensed by a 10–mL Class A Transfer Pipet Replicate Volume (ml) Replicate Volume (mL) 1 10.002 6 9.983 2 9.993 7 9.991 3 9.984 8 9.990 4 9.996 9 9.988 5 9.989 10 9.999 This page titled 4.2: Characterizing Experimental Errors is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey. 4.2: Characterizing Experimental Errors by David Harvey is licensed CC BY-NC-SA 4.0. 4.2.7 https://chem.libretexts.org/@go/page/401341 4.3: Propagation of Uncertainty Suppose we dispense 20 mL of a reagent using the Class A 10-mL pipet whose calibration information is given in Table 4.2.8. If the volume and uncertainty for one use of the pipet is 9.992 ± 0.006 mL, what is the volume and uncertainty if we use the pipet twice? As a first guess, we might simply add together the volume and the maximum uncertainty for each delivery; thus (9.992 mL + 9.992 mL) ± (0.006 mL + 0.006 mL) = 19.984 ± 0.012 mL It is easy to appreciate that combining uncertainties in this way overestimates the total uncertainty. Adding the uncertainty for the first delivery to that of the second delivery assumes that with each use the indeterminate error is in the same direction and is as large as possible. At the other extreme, we might assume that the uncertainty for one delivery is positive and the other is negative. If we subtract the maximum uncertainties for each delivery, (9.992 mL + 9.992 mL) ± (0.006 mL – 0.006 mL) = 19.984 ± 0.000 mL we clearly underestimate the total uncertainty. So what is the total uncertainty? From the discussion above, we reasonably expect that the total uncertainty is greater than ±0.000 mL and that it is less than ±0.012 mL. To estimate the uncertainty we use a mathematical technique known as the propagation of uncertainty. Our treatment of the propagation of uncertainty is based on a few simple rules. A Few Symbols A propagation of uncertainty allows us to estimate the uncertainty in a result from the uncertainties in the measurements used to calculate that result. For the equations in this section we represent the result with the symbol R, and we represent the measurements with the symbols A, B, and C. The corresponding uncertainties are uR, uA, uB, and uC. We can define the uncertainties for A, B, and C using standard deviations, ranges, or tolerances (or any other measure of uncertainty), as long as we use the same form for all measurements. The requirement that we express each uncertainty in the same way is a critically important point. Suppose you have a range for one measurement, such as a pipet’s tolerance, and standard deviations for the other measurements. All is not lost. There are ways to convert a range to an estimate of the standard deviation. See Appendix 2 for more details. Uncertainty When Adding or Subtracting When we add or subtract measurements we propagate their absolute uncertainties. For example, if the result is given by the equation R = A+B−C the the absolute uncertainty in R is −−−−−−−−−−− 2 2 2 uR = √ u +u +u (4.3.1) A B C  Example 4.3.1 If we dispense 20 mL using a 10-mL Class A pipet, what is the total volume dispensed and what is the uncertainty in this volume? First, complete the calculation using the manufacturer’s tolerance of 10.00 mL±0.02 mL, and then using the calibration data from Table 4.2.8. Solution To calculate the total volume we add the volumes for each use of the pipet. When using the manufacturer’s values, the total volume is V = 10.00 mL + 10.00 mL = 20.00 mL and when using the calibration data, the total volume is 4.3.1 https://chem.libretexts.org/@go/page/401342  V = 9.992 mL + 9.992 mL = 19.984 mL Using the pipet’s tolerance as an estimate of its uncertainty gives the uncertainty in the total volume as 2 2 uR = (0.02 ) + (0.02 ) = 0.028 mL = 0.028 mL and using the standard deviation for the data in Table 4.2.8 gives an uncertainty of 2 2 uR = (0.006 ) + (0.006 ) = 0.0085 mL Rounding the volumes to four significant figures gives 20.00 mL ± 0.03 mL when we use the tolerance values, and 19.98 ± 0.01 mL when we use the calibration data. Uncertainty When Multiplying or Dividing When we multiple or divide measurements we propagate their relative uncertainties. For example, if the result is given by the equation A×B R = C then the relative uncertainty in R is −−−−−−−−−−−−−−−−−−−−− − 2 2 2 uR uA uB uC = √( ) +( ) +( ) (4.3.2) R A B C  Example 4.3.2 The quantity of charge, Q, in coulombs that passes through an electrical circuit is Q = i ×t where i is the current in amperes and t is the time in seconds. When a current of 0.15 A ± 0.01 A passes through the circuit for 120 s ± 1 s, what is the total charge and its uncertainty? Solution The total charge is Q = (0.15 A) × (120 s) = 18 C Since charge is the product of current and time, the relative uncertainty in the charge is −−−−−−−−−−−−−−−−− 2 2 uR 0.01 1 = √( ) +( ) = 0.0672 R 0.15 120 and the charge’s absolute uncertainty is uR = R × 0.0672 = (18 C) × (0.0672) = 1.2 C Thus, we report the total charge as 18 C ± 1 C. Uncertainty for Mixed Operations Many chemical calculations involve a combination of adding and subtracting, and of multiply and dividing. As shown in the following example, we can calculate the uncertainty by separately treating each operation using Equation 4.3.1 and Equation 4.3.2 as needed.  Example 4.3.3 For a concentration technique, the relationship between the signal and the an analyte’s concentration is Stotal = kA CA + Smb 4.3.2 https://chem.libretexts.org/@go/page/401342  What is the analyte’s concentration, CA, and its uncertainty if Stotal is 24.37 ± 0.02, Smb is 0.96 ± 0.02, and kA is 0.186 ± 0.003 ppm ? −1 Solution Rearranging the equation and solving for CA Stotal − Smb 24.37 − 0.96 23.41 CA = = = = 125.9 ppm −1 −1 kA 0.186 ppm 0.186 ppm gives the analyte’s concentration as 126 ppm. To estimate the uncertainty in CA, we first use Equation 4.3.1 to determine the uncertainty for the numerator. −−−−−−−−−−−−− 2 2 uR = √ (0.02 ) + (0.02 ) = 0.028 The numerator, therefore, is 23.41 ± 0.028. To complete the calculation we use Equation 4.3.2 to estimate the relative uncertainty in CA. −−−−−−−−−−−−−−−−−−− 2 2 uR 0.028 0.003 = √( ) +( ) = 0.0162 R 23.41 0.186 The absolute uncertainty in the analyte’s concentration is uR = (125.9 ppm) × (0.0162) = 2.0 ppm Thus, we report the analyte’s concentration as 126 ppm ± 2 ppm.  Exercise 4.3.1 To prepare a standard solution of Cu2+ you obtain a piece of copper from a spool of wire. The spool’s initial weight is 74.2991 g and its final weight is 73.3216 g. You place the sample of wire in a 500-mL volumetric flask, dissolve it in 10 mL of HNO3, and dilute to volume. Next, you pipet a 1 mL portion to a 250-mL volumetric flask and dilute to volume. What is the final concentration of Cu2+ in mg/L, and its uncertainty? Assume that the uncertainty in the balance is ±0.1 mg and that you are using Class A glassware. Answer The first step is to determine the concentration of Cu2+ in the final solution. The mass of copper is 74.2991 g − 73.3216 g = 0.9775 g Cu The 10 mL of HNO3 used to dissolve the copper does not factor into our calculation. The concentration of Cu2+ is 0.9775 g Cu 1.000 mL 1000 mg 2 + × × = 7.820 mg Cu /L 0.5000 L 250.0 mL g Having found the concentration of Cu2+, we continue with the propagation of uncertainty. The absolute uncertainty in the mass of Cu wire is −−−−−−−−−−−−−−−−− 2 2 ug Cu = √ (0.0001 ) + (0.0001 ) = 0.00014 g The relative uncertainty in the concentration of Cu2+ is −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 2 2 2 2 umg/L 0.00014 0.20 0.006 0.12 =√ ( ) +( ) +( ) +( ) = 0.00603 7.820 mg/L 0.9775 500.0 1.000 250.0 Solving for umg/L gives the uncertainty as 0.0472. The concentration and uncertainty for Cu2+ is 7.820 mg/L ± 0.047 mg/L. 4.3.3 https://chem.libretexts.org/@go/page/401342 Uncertainty for Other Mathematical Functions Many other mathematical operations are common in analytical chemistry, including the use of powers, roots, and logarithms. Table 4.3.1 provides equations for propagating uncertainty for some of these function where A and B are independent measurements and where k is a constant whose value has no uncertainty. Table 4.3.1 : Propagation of Uncertainty for Selected Mathematical Functions Function uR Function uR uA R = kA uR = kuA R = ln(A) uR = A −−−−−−− uA 2 2 R = A +B uR = √u +u R = log(A) uR = 0.4343 × A B A −−−−−−− 2 2 A uR R = A −B uR = √u +u R = e = uA A B R −−−−−−−−−−−− 2 uR uA uB 2 A uR R = A ×B = √( ) +( ) R = 10 = 2.303 × uA R A B R −−−−−−−−−−−− 2 A uR uA uB 2 uR uA k R = = √( ) +( ) R = A = k× B R A B R A  Example 4.3.4 If the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, what is the [H+] and its uncertainty? Solution The concentration of H+ is + −pH −3.72 −4 [H ] = 10 = 10 = 1.91 × 10 M or 1.9 × 10−4 M to two significant figures. From Table 4.3.1 the relative uncertainty in [H+] is uR = 2.303 × uA = 2.303 × 0.03 = 0.069 R The uncertainty in the concentration, therefore, is −4 −5 (1.91 × 10 M) × (0.069) = 1.3 × 10 M We report the [H+] as 1.9(±0.1) × 10 −4 M, which is equivalent to 1.9 × 10 −4 M ± 0.1 × 10 −4 M.  Exercise 4.3.2 A solution of copper ions is blue because it absorbs yellow and orange light. Absorbance, A, is defined as P A = − log T = − log( ) Po where, T is the transmittance, Po is the power of radiation as emitted from the light source and P is its power after it passes through the solution. What is the absorbance if Po is 3.80 × 10 and P is 1.50 × 10 ? If the uncertainty in measuring Po and P 2 2 is 15, what is the uncertainty in the absorbance? Answer The first step is to calculate the absorbance, which is 2 P 1.50 × 10 A = − log T = − log = − log = 0.4037 ≈ 0.404 2 Po 3.80 × 10 Having found the absorbance, we continue with the propagation of uncertainty. First, we find the uncertainty for the ratio P/Po, which is the transmittance, T. 4.3.4 https://chem.libretexts.org/@go/page/401342  −−−−−−−−−−−−−−−−−−−−−−−−−− 2 2 uT 15 15 = √( ) +( ) = 0.1075 2 2 T 3.80 × 10 1.50 × 10 Finally, from Table 4.3.1 the uncertainty in the absorbance is uT −2 uA = 0.4343 × = (0.4343) × (0.1075) = 4.669 × 10 T The absorbance and uncertainty is 0.40 ± 0.05 absorbance units. Is Calculating Uncertainty Actually Useful? Given the effort it takes to calculate uncertainty, it is worth asking whether such calculations are useful. The short answer is, yes. Let’s consider three examples of how we can use a propagation of uncertainty to help guide the development of an analytical method. One reason to complete a propagation of uncertainty is that we can compare our estimate of the uncertainty to that obtained experimentally. For example, to determine the mass of a penny we measure its mass twice—once to tare the balance at 0.000 g and once to measure the penny’s mass. If the uncertainty in each measurement of mass is ±0.001 g, then we estimate the total uncertainty in the penny’s mass as −−−−−−−−−−−−−−− 2 2 uR = √ (0.001 ) + (0.001 ) = 0.0014 g If we measure a single penny’s mass several times and obtain a standard deviation of ±0.050 g, then we have evidence that the measurement process is out of control. Knowing this, we can identify and correct the problem. We also can use a propagation of uncertainty to help us decide how to improve an analytical method’s uncertainty. In Example 4.3.3 , for instance, we calculated an analyte’s concentration as 126 ppm ± 2 ppm, which is a percent uncertainty of 1.6%. Suppose we want to decrease the percent uncertainty to no more than 0.8%. How might we accomplish this? Looking back at the calculation, we see that the concentration’s relative uncertainty is determined by the relative uncertainty in the measured signal (corrected for the reagent blank) 0.028 = 0.0012 or 0.12% 23.41 and the relative uncertainty in the method’s sensitivity, kA, −1 0.003 ppm = 0.016 or 1.6% −1 0.186 ppm Of these two terms, the uncertainty in the method’s sensitivity dominates the overall uncertainty. Improving the signal’s uncertainty will not improve the overall uncertainty of the analysis. To achieve an overall uncertainty of 0.8% we must improve the uncertainty in kA to ±0.0015 ppm–1.  Exercise 4.3.3 Verify that an uncertainty of ±0.0015 ppm–1 for kA is the correct result. Answer An uncertainty of 0.8% is a relative uncertainty in the concentration of 0.008; thus, letting u be the uncertainty in kA −−−−−−−−−−−−−−−−−− 2 2 0.028 u 0.008 = √ ( ) +( ) 23.41 0.186 Squaring both sides of the equation gives 2 2 0.028 u −5 6.4 × 10 =( ) +( ) 23.41 0.186 4.3.5 https://chem.libretexts.org/@go/page/401342  Solving for the uncertainty in kA gives its value as 1.47 × 10 −3 or ±0.0015 ppm–1. Finally, we can use a propagation of uncertainty to determine which of several procedures provides the smallest uncertainty. When we dilute a stock solution usually there are several combinations of volumetric glassware that will give the same final concentration. For instance, we can dilute a stock solution by a factor of 10 using a 10-mL pipet and a 100-mL volumetric flask, or using a 25-mL pipet and a 250-mL volumetric flask. We also can accomplish the same dilution in two steps using a 50-mL pipet and 100-mL volumetric flask for the first dilution, and a 10-mL pipet and a 50-mL volumetric flask for the second dilution. The overall uncertainty in the final concentration—and, therefore, the best option for the dilution—depends on the uncertainty of the volumetric pipets and volumetric flasks. As shown in the following example, we can use the tolerance values for volumetric glassware to determine the optimum dilution strategy [Lam, R. B.; Isenhour, T. L. Anal. Chem. 1980, 52, 1158–1161].  Example 4.3.5 : Which of the following methods for preparing a 0.0010 M solution from a 1.0 M stock solution provides the smallest overall uncertainty? (a) A one-step dilution that uses a 1-mL pipet and a 1000-mL volumetric flask. (b) A two-step dilution that uses a 20-mL pipet and a 1000-mL volumetric flask for the first dilution, and a 25-mL pipet and a 500-mL volumetric flask for the second dilution. Solution The dilution calculations for case (a) and case (b) are 1.000 mL case (a): 1.0 M × = 0.0010 M 1000.0 mL 20.00 mL 25.00 mL case (b): 1.0 M × × = 0.0010 M 1000.0 mL 500.0mL Using tolerance values from Table 4.2.1, the relative uncertainty for case (a) is −−−−−−−−−−−−−−−−−−−− 2 2 0.006 0.3 uR = √ ( ) +( ) = 0.006 1.000 1000.0 and for case (b) the relative uncertainty is −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− 2 2 2 2 0.03 0.3 0.03 0.2 uR = √ ( ) +( ) +( ) +( ) = 0.002 20.00 1000 25.00 500.0 Since the relative uncertainty for case (b) is less than that for case (a), the two-step dilution provides the smallest overall uncertainty. Of course we must balance the smaller uncertainty for case (b) against the increased opportunity for introducing a determinate error when making two dilutions instead of just one dilution, as in case (a). This page titled 4.3: Propagation of Uncertainty is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey. 4.3: Propagation of Uncertainty by David Harvey is licensed CC BY-NC-SA 4.0. 4.3.6 https://chem.libretexts.org/@go/page/401342 4.4: The Distribution of Measurements and Results Earlier we reported results for a determination of the mass of a circulating United States penny, obtaining a mean of 3.117 g and a standard deviation of 0.051 g. Table 4.4.1 shows results for a second, independent determination of a penny’s mass, as well as the data from the first experiment. Although the means and standard deviations for the two experiments are similar, they are not identical. The difference between the two experiments raises some interesting questions. Are the results for one experiment better than the results for the other experiment? Do the two experiments provide equivalent estimates for the mean and the standard deviation? What is our best estimate of a penny’s expected mass? To answer these questions we need to understand how we might predict the properties of all pennies using the results from an analysis of a small sample of pennies. We begin by making a distinction between populations and samples. Table 4.4.1 : Results for Two Determinations of the Mass of a Circulating U. S. Penny First Experiment Second Experiment Penny Mass (g) Penny Mass (g) 1 3.080 1 3.052 2 3.094 2 3.141 3 3.107 3 3.083 4 3.056 4 3.083 5 3.112 5 3.048 6 3.174 7 3.198 ¯¯¯ X ¯ 3.117 3.081 s 0.051 0.037 Populations and Samples A population is the set of all objects in the system we are investigating. For the data in Table 4.4.1 , the population is all United States pennies in circulation. This population is so large that we cannot analyze every member of the population. Instead, we select and analyze a limited subset, or sample of the population. The data in Table 4.4.1 , for example, shows the results for two such samples drawn from the larger population of all circulating United States pennies. Probability Distributions for Populations Table 4.4.1 provides the means and the standard deviations for two samples of circulating United States pennies. What do these samples tell us about the population of pennies? What is the largest possible mass for a penny? What is the smallest possible mass? Are all masses equally probable, or are some masses more common? To answer these questions we need to know how the masses of individual pennies are distributed about the population’s average mass. We represent the distribution of a population by plotting the probability or frequency of obtaining a specific result as a function of the possible results. Such plots are called probability distributions. There are many possible probability distributions; in fact, the probability distribution can take any shape depending on the nature of the population. Fortunately many chemical systems display one of several common probability distributions. Two of these distributions, the binomial distribution and the normal distribution, are discussed in this section. The Binomial Distribution The binomial distribution describes a population in which the result is the number of times a particular event occurs during a fixed number of trials. Mathematically, the binomial distribution is defined as N! X N −X P (X, N ) = ×p × (1 − p ) X!(N − X)! 4.4.1 https://chem.libretexts.org/@go/page/401343 where P(X , N) is the probability that an event occurs X times during N trials, and p is the event’s probability for a single trial. If you flip a coin five times, P(2,5) is the probability the coin will turn up “heads” exactly twice. The term N! reads as N-factorial and is the product N × (N – 1) × (N – 2) × ⋯ × 1. For example, 4! is 4 × 3 × 2 × 1 = 24. Your calculator probably has a key for calculating factorials. A binomial distribution has well-defined measures of central tendency and spread. The expected mean value is μ = Np and the expected spread is given by the variance 2 σ = N p(1 − p) or the standard deviation. −−−−−−−− σ = √ N p(1 − p) The binomial distribution describes a population whose members have only specific, discrete values. When you roll a die, for example, the possible values are 1, 2, 3, 4, 5, or 6. A roll of 3.45 is not possible. As shown in Worked Example 4.4.1 , one example of a chemical system that obeys the binomial distribution is the probability of finding a particular isotope in a molecule.  Example 4.4.1 12 13 Carbon has two stable, non-radioactive isotopes, C and C, with relative isotopic abundances of, respectively, 98.89% and 1.11%. (a) What are the mean and the standard deviation for the number of 13C atoms in a molecule of cholesterol (C27H44O)? (b) What is the probability that a molecule of cholesterol has no atoms of 13C? Solution The probability of finding an atom of 13C in a molecule of cholesterol follows a binomial distribution, where X is the number of 13C atoms, N is the number of carbon atoms in a molecule of cholesterol, and p is the probability that an atom of carbon in 13C. For (a), the mean number of 13C atoms in a molecule of cholesterol is μ = N p = 27 × 0.0111 = 0.300 with a standard deviation of −−−−−−−− −−−−−−−−−−−−−−−−−−−−− σ = √ N p(1 − p) = √ 27 × 0.0111 × (1 − 0.0111) = 0.544 For (b), the probability of finding a molecule of cholesterol without an atom of 13C is 27! 0 27−0 P (0, 27) = × (0.0111 ) × (1 − 0.0111 ) = 0.740 0! (27 − 0)! There is a 74.0% probability that a molecule of cholesterol will not have an atom of 13C, a result consistent with the observation that the mean number of 13C atoms per molecule of cholesterol, 0.300, is less than one. A portion of the binomial distribution for atoms of 13C in cholesterol is shown in Figure 4.4.1. Note in particular that there is little probability of finding more than two atoms of 13C in any molecule of cholesterol. 4.4.2 https://chem.libretexts.org/@go/page/401343  Figure 4.4.1 : Portion of the binomial distribution for the number of naturally occurring 13C atoms in a molecule of cholesterol. Only 3.6% of cholesterol molecules contain more than one atom of 13C, and only 0.33% contain more than two atoms of 13C. The Normal Distribution A binomial distribution describes a population whose members have only certain discrete values. This is the case with the number of 13 C atoms in cholesterol. A molecule of cholesterol, for example, can have two 13C atoms, but it can not have 2.5 atoms of 13C. A population is continuous if its members may take on any value. The efficiency of extracting cholesterol from a sample, for example, can take on any value between 0% (no cholesterol is extracted) and 100% (all cholesterol is extracted). The most common continuous distribution is the Gaussian, or normal distribution, the equation for which is 2 ( X −μ) 1 − 2 f (X) = −−−−e 2σ √2πσ 2 where μ is the expected mean for a population with n members n ∑ Xi i=1 μ = n and σ is the population’s variance. 2 n 2 ∑ (Xi − μ) 2 i=1 σ = (4.4.1) n Examples of three normal distributions, each with an expected mean of 0 and with variances of 25, 100, or 400, respectively, are shown in Figure 4.4.2. Two features of these normal distribution curves deserve attention. First, note that each normal distribution has a single maximum that corresponds to μ , and that the distribution is symmetrical about this value. Second, increasing the population’s variance increases the distribution’s spread and decreases its height; the area under the curve, however, is the same for all three distributions. Figure 4.4.2 : Normal distribution curves for: (a) μ = 0; σ = 25 (b) μ = 0; σ = 100 (c) μ = 0; σ = 400. 2 2 2 The area under a normal distribution curve is an important and useful property as it is equal to the probability of finding a member of the population within a particular range of values. In Figure 4.4.2 , for example, 99.99% of the population shown in curve (a) have values of X between –20 and +20. For curve (c), 68.26% of the population’s members have values of X between –20 and +20. Because a normal distribution depends solely on μ and σ , the probability of finding a member of the population between any two 2 limits is the same for all normally distributed populations. Figure 4.4.3 , for example, shows that 68.26% of the members of a normal distribution have a value within the range μ ± 1σ , and that 95.44% of population’s members have values within the range μ ± 2σ. 4.4.3 https://chem.libretexts.org/@go/page/401343 Only 0.27% members of a population have values that exceed the expected mean by more than ± 3σ. Additional ranges and probabilities are gathered together in the probability table included in Appendix 3. As shown in Example 4.4.2 , if we know the mean and the standard deviation for a normally distributed population, then we can determine the percentage of the population between any defined limits. Figure 4.4.3 : Normal distribution curve showing the area under the curve for several different ranges of values of X.  Example 4.4.2 The amount of aspirin in the analgesic tablets from a particular manufacturer is known to follow a normal distribution with μ = 250 mg and σ = 5. In a random sample of tablets from the production line, what percentage are expected to contain between 243 and 262 mg of aspirin? Solution We do not determine directly the percentage of tablets between 243 mg and 262 mg of aspirin. Instead, we first find the percentage of tablets with less than 243 mg of aspirin and the percentage of tablets having more than 262 mg of aspirin. Subtracting these results from 100%, gives the percentage of tablets that contain between 243 mg and 262 mg of aspirin. To find the percentage of tablets with less than 243 mg of aspirin or more than 262 mg of aspirin we calculate the deviation, z, of each limit from μ in terms of the population’s standard deviation, σ X −μ z = σ where X is the limit in question. The deviation for the lower limit is 243 − 250 zlower = = −1.4 5 and the deviation for the upper limit is 262 − 250 zupper = = +2.4 5 Using the table in Appendix 3, we find that the percentage of tablets with less than 243 mg of aspirin is 8.08%, and that the percentage of tablets with more than 262 mg of aspirin is 0.82%. Therefore, the percentage of tablets containing between 243 and 262 mg of aspirin is 100.00% − 8.08% − 0.82% = 91.10% Figure 4.4.4 shows the distribution of aspiring in the tablets, with the area in blue showing the percentage of tablets containing between 243 mg and 262 mg of aspirin. 4.4.4 https://chem.libretexts.org/@go/page/401343  Figure 4.4.4 : Normal distribution for the population of aspirin tablets in Example 4.4.2. The population’s mean and standard deviation are 250 mg and 5 mg, respectively. The shaded area shows the percentage of tablets containing between 243 mg and 262 mg of aspirin.  Exercise 4.4.1 What percentage of aspirin tablets will contain between 240 mg and 245 mg of aspirin if the population’s mean is 250 mg and the population’s standard deviation is 5 mg. Answer To find the percentage of tablets that contain less than 245 mg of aspirin we first calculate the deviation, z, 245 − 250 z = = −1.00 5 and then look up the corresponding probability in Appendix 3, obtaining a value of 15.87%. To find the percentage of tablets that contain less than 240 mg of aspirin we find that 240 − 250 z = = −2.00 5 which corresponds to 2.28%. The percentage of tablets containing between 240 and 245 mg of aspiring is 15.87% – 2.28% = 13.59%. Confidence Intervals for Populations If we select at random a single member from a population, what is its most likely value? This is an important question, and, in one form or another, it is at the heart of any analysis in which we wish to extrapolate from a sample to the sample’s parent population. One of the most important features of a population’s probability distribution is that it provides a way to answer this question. Figure 4.4.3 shows that for a normal distribution, 68.26% of the population’s members have values within the range μ ± 1σ. Stating this another way, there is a 68.26% probability that the result for a single sample drawn from a normally distributed population is in the interval μ ± 1σ. In general, if we select a single sample we expect its value, Xi is in the range Xi = μ ± zσ (4.4.2) where the value of z is how confident we are in assigning this range. Values reported in this fashion are called confidence intervals. Equation 4.4.2, for example, is the confidence interval for a single member of a population. Table 4.4.2 gives the confidence intervals for several values of z. For reasons discussed later in the chapter, a 95% confidence level is a common choice in analytical chemistry. When z = 1, we call this the 68.26% confidence interval. Table 4.4.2 : Confidence Intervals for a Normal Distribution z Confidence Interval 0.50 38.30 1.00 68.26 4.4.5 https://chem.libretexts.org/@go/page/401343  z Confidence Interval 1.50 86.64 1.96 95.00 2.00 95.44 2.50 98.76 3.00 99.73 3.50 99.95  Example 4.4.3 What is the 95% confidence interval for the amount of aspirin in a single analgesic tablet drawn from a population for which μ is 250 mg and for which σ is 5? Solution Using Table 4.4.2 , we find that z is 1.96 for a 95% confidence interval. Substituting this into Equation 4.4.2 gives the confidence interval for a single tablet as Xi = μ ± 1.96σ = 250 mg ± (1.96 × 5) = 250 mg ± 10mg A confidence interval of 250 mg ± 10 mg means that 95% of the tablets in the population contain between 240 and 260 mg of aspirin. Alternatively, we can rewrite Equation 4.4.2 so that it gives the confidence interval is for μ based on the population’s standard deviation and the value of a single member drawn from the population. μ = Xi ± zσ (4.4.3)  Example 4.4.4 The population standard deviation for the amount of aspirin in a batch of analgesic tablets is known to be 7 mg of aspirin. If you randomly select and analyze a single tablet and find that it contains 245 mg of aspirin, what is the 95% confidence interval for the population’s mean? Solution The 95% confidence interval for the population mean is given as μ = Xi ± zσ = 245 mg ± (1.96 × 7) mg = 245 mg ± 14 mg Therefore, based on this one sample, we estimate that there is 95% probability that the population’s mean, μ , lies within the range of 231 mg to 259 mg of aspirin. Note the qualification that the prediction for μ is based on one sample; a different sample likely will give a different 95% confidence interval. Our result here, therefore, is an estimate for μ based on this one sample. It is unusual to predict the population’s expected mean from the analysis of a single sample; instead, we collect n samples drawn from a population of known σ, and report the mean, X. The standard deviation of the mean, σ , which also is known as the ¯ ¯ X