© 2019 Nanyang Technological University, Singapore. All Rights Reserved. Common analytical problems; Statistical treatment; Accuracy and precision; Significant figures; Gaussian distribution; Confidence interval; Student’s t test; Q test CM2011 Analytical Chemistry Topic 1: Errors and Confidence Interval © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Learning Objectives By the end of this lesson, you will be able to: 2 Explain the difference between absolute, relative and percent relative uncertainties. Discuss how errors are totaled when numbers (with errors) are combined. Apply and calculate the confidence interval. Apply the Q-test to exclude points. Explain when to use the correct number of significant figures and correct rounding. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Learning Objectives (Cont.) By the end of this lesson, you will be able to: 3 Execute the student t-test to compare different sets of data for three cases: (a) Comparing a measured result with a ‘known’ value. (b) Comparison of the means from two methods on one sample measured multiple times. (c) Paired t test for comparing individual differences from two methods but with a number of different samples. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. What is Analytical Chemistry? 4 It is the branch of chemistry that deals with the separation, identification and determination of components in a sample. Analytical approach to solving problems. 1 2 3 Identify and define the problem. Design the experimental procedure. Conduct an experiment, and gather data. 4 Analyse the experimental data. 5 Propose a solution to the problem. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. What is Analytical Chemistry? (Cont.) 5 1. Identify the problem • Determine the type of information needed (qualitative, quantitative, characterisation or fundamental) • Identify the context of the problem 2. Design the experimental procedure • Establish design criteria (accuracy, precision, scale of operation, sensitivity, selectivity, cost, speed) • Identify interferents • Select method • Establish validation criteria • Establish sampling strategy 3. • • • 5. Propose a solution • Conduct external evaluation Feedback loop 4. • • • • Analyse the experimental data Reduce or transform data Analyse statistics Verify results Interpret results Conduct an experiment Calibrate instruments and equipment Standardise reagents Gather data © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Common Analytical Problems 6 Identify what is present in a sample. How much of a constituent is present in a sample. 1. Qualitative analysis 2. Quantitative analysis An analysis in which we evaluate a sample’s chemical or physical properties. An analysis whose purpose is to improve an analytical method’s capabilities. 3. Characterisation analysis 4. Fundamental analysis CM2011 Analytical Chemistry Accuracy and Precision © 2019 Nanyang Technological University, Singapore. All Rights Reserved. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Accuracy and Precision 8 Accuracy: How close is our value to the ‘real’ one. Accurate Not accurate © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Accuracy and Precision (Cont.) Precision: How close do our values agree to each other. Reproducibility. Precise but Not accurate 9 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Errors Error – It is the difference between your answer and the ‘true’ one. Systematic or determinate – Shows consistent error. All errors are of the same size, magnitude and direction. Examples: (a) Instrumental error (b) Method error (c) Human error Random or indeterminate – These errors cannot be eliminated. They can be treated statistically. For example, instrument noise. 10 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Errors (Cont.) 11 12.35 ± 0.02 ๐๐ฟ Burette reading: Absolute uncertainty Absolute uncertainty Magnitude of measurement Relative uncertainty = Relative uncertainty = Percent relative uncertainty = 100 x Relative uncertainty % Relative uncertainty = 100 x 0.002 = 0.2% 0.02 mL/12.35 mL = 0.001619… = 0.002 (to one significant figure) CM2011 Analytical Chemistry Significant Figures and Error Propagation © 2019 Nanyang Technological University, Singapore. All Rights Reserved. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Significant Figures (How Many?) 13 100 (3sf) 1 x 102 (1sf) 0.01 (1sf) 1000 (4sf) 1 x 103 (1sf) 0.0007 (1sf) 1 x 104 (1sf) 0.000327 (3sf) 1.00 x 104 (3sf) 0.100327 (6sf) 10000 (5sf) 10000.00 (7sf) © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Significant Figures and Error Propagation Addition and subtraction When measurements are added or subtracted, the answer can contain no more decimal places than the least accurate measurement. 14 Uncertainty in addition and subtraction is given by: ๐ = √∑๐๐ 2 Here ๐๐ is the absolute uncertainty of the individual measurements. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Significant Figures and Error Propagation (Cont.) Example (1) 1.76 (±0.03) – 0.59 (±0.02) √[(0.03)2 + (0.02)2] = 0.0361 = 0.036 ≈ 0.04 (to two decimal places) = 1.17 (±0.04) (2) ๐ค = 4.52 (±0.02) cm, ๐ฅ = 2.0 (±0.2) cm, ๐ฆ = 3.0 (±0.6) cm. Find ๐ง = ๐ฅ + ๐ฆ − ๐ค and its uncertainty. √[(0.02)2 + (0.2)2 + (0.6)2] cm = 0.633 = 0.63 ≈ 0.6 cm (to one decimal place) ๐ง = 0.48 = 0.5 (±0.6) cm 15 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Significant Figures and Error Propagation (Cont.) Multiplication and division When measurements are multiplied or divided, the answer can contain no more significant figures than the least accurate measurement. 16 Uncertainty in multiplication and division is given by: ๐ = √∑(%๐๐ )2 Here %๐๐ is the percent relative uncertainty of the individual measurements. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Significant Figures and Error Propagation (Cont.) Example: 1.76(±0.03) × 1.89(±0.02) 0.59(±0.02) = 5.64 (±? %๐) (to two significant figures) Convert absolute to % relative uncertainties. 1.76(±1.7 %) × 1.89(±1.1 %) = 5.64 (±%๐) (% error to one significant figure) 0.59(±3.4 %) where %๐ = (1.7 )2 + (1.1 )2 + (3.4 )2 = 4.0 % Percentage uncertainty = 4% (to one percentage unit) Therefore, answer = 5.6 (±4%) (percentage error) Absolute uncertainty = 4.0% x 5.64 = 0.23 Therefore, alternative answer is 5.6 (±0.2) (absolute error) 17 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. % Relative Uncertainties 18 When doing calculations, keep as many digits in the calculator as possible. Report final % error to the nearest whole number. If less than 1%, then report to 1 dp. 1.7% = 2% 0.76% = 0.8% 11.3% = 11% 0.08% = 0.1% 127.1% = 127% 0.0008% = 0.001% = 0% © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Mixed Operations Example: 19 [1.76(±0.03) − 0.59(±0.02)] = 0.6190 ±? 1.89(±0.02) First, work out the difference in the numerator using absolute uncertainties. (0.03)2 + (0.02)2 = 0.036 1.76 (±0.03) − 0.59(±0.02) = 1.17(±0.036 ) Then convert into % relative uncertainties. 1.17(±0.036 ) 1.17(±3.1 %) = = 0.6190 (±3.3 %) 1.89(±0.02) 1.89(±1.1 %) because (3.1 )2 + (1.1 )2 = 3.3 The % relative uncertainty is 3.3%, so the absolute uncertainty is: 0.033 × 0.6190 = 0.0204 The final answer can be written as: 0.619(±0.020) (absolute uncertainty) or 0.619(±3%) (relative uncertainty) © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Rounding 20 ๏ง When rounding up, look at all the digits beyond the last place desired. - Round up when the digits after the desired digit are > 5. - Round down when the digits after the desired digit are < 5. For example: 120.803501 = 120.804 120.803499 = 120.803 ๏ง Rounding up and down. In the special case where the number is exactly halfway, round to the nearest even digit. 120.803500 = 120.804 120.804500 = 120.804 For example, if rounding to three significant figures. 43.55 = 43.45 = 1.425 × 10-9 = 1.42501 × 10-9 = 43.6 43.4 1.42 × 10-9 1.43 × 10-9 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Uncertainties of Other Functions Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. 21 CM2011 Analytical Chemistry Gaussian Distribution © 2019 Nanyang Technological University, Singapore. All Rights Reserved. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Gaussian Distribution This graph describes the distribution of results when an experiment is repeated several times and the errors are purely random. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. 23 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Gaussian Distribution (Cont.) 24 Equation for a Gaussian curve. ๐ฆ= 1 ๐ 2๐ 2 /2๐ 2 −(๐ฅ−๐) ๐ Here m is the population mean (or average) and s is the population standard deviation. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Gaussian Distribution (Cont.) 25 For a finite set of experimental data (N < 30). Sample mean: ๐เดค = ∑๐ ๐๐ ๐ Here N is the # of measurements. Sample (s)standard deviation: ๐ = ∑๐ ๐๐ − ๐เดค ๐−1 2 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Gaussian Distribution (Cont.) 26 s2 is called the sample variance. For an infinite set of data (๐ → ∞), ๐เดค → ๐ ๐ →๐ (๐ − 1) is called the degrees of freedom. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. Smaller the standard deviation means the more precise is the measurement. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Gaussian Distribution (Cont.) 27 68.3% of the measurements are expected to lie within one standard deviation of the mean. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. CM2011 Analytical Chemistry Confidence Interval © 2019 Nanyang Technological University, Singapore. All Rights Reserved. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Confidence Interval 29 The Confidence Interval (CI) for the mean is the range of values within which the population mean (or ‘true’ mean) m is expected to lie with a certain probability. CI for m: _ ๐=๐ฅ± ๐ก๐ ๐ where t is the student’s t. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Confidence Interval (Cont.) 30 Example: ๏ง The carbohydrate content of a glycoprotein is determined to be 12.6, 11.9, 13.0, 12.7 and 12.5 g of carbohydrate per 100 g of protein in replicate analyses. Find the 50% and 90% CI for the carbohydrate content. ๏ง First calculate the mean (=12.54) and standard deviation (= 0.40). ๏ง For the 50% confidence interval, look up 4 degrees of freedom (from N – 1). ๐ก๐ (0.741)(0. 40 ) − ๐=๐ฅ± = 12. 54 ± = 12. 54 ± 0. 13 ๐ 5 • For 90% confidence interval, m = 12. 54 ± 0. 38 . Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Rounding Sets of Data 31 Mean and standard deviation: Round to one more decimal place than your original data (if data set ≥ 10). If the data set is < 10, use the same number of decimal places in your data set (for the least accurate number). 10, 11, 14, 12, 4 Mean = 10 Std. dev = 4 10, 11, 14, 12, 4, 12, 16, 9, 8, 5 Mean = 10.1 Std. dev = 3.8 CM2011 Analytical Chemistry Student’s t Test © 2019 Nanyang Technological University, Singapore. All Rights Reserved. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Student’s t Test 33 This test will be used to determine if two sets of measurements are statistically different. We will focus on three cases: Comparing a measured result with a ‘known’ value. Comparison of the means of two methods on one sample measured multiple times. Using two methods to make single measurements on several different samples. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (a) Comparing a Measured Result with a ‘Known’ Value First compute ๐กcal ๐กcal ๐ฅาง − ๐๐๐๐ค๐ ๐ฃ๐๐๐ข๐ = ๐ ๐ If ๐กcal is greater than ๐กtable at the ๐ฅ% confidence level, the two results are considered to be significantly different at that confidence level. 34 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (a) Comparing a Measured Result with a ‘Known’ Value (Cont.) 35 Example: A new procedure for the rapid determination of the percentage of sulfur in kerosene was tested on a sample known from its method of preparation to contain 0.123% S. The results were %S = 0.112, 0.118, 0.115, and 0.119. Does the data indicate that there is a difference in the methods at the 95% confidence level? ๐กcal 0.116 − 0.123 = 0.003162 ๐ = 4.427 From the table, it can be seen that the critical value of t for 3 degrees of freedom at the 95% confidence interval is 3.18. Since 3.18 < 4.43, we conclude that there is a significant difference at the 95% confidence level. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (b) Comparison of the Means of Two Methods For two sets of data consisting of ๐1 and ๐2 measurements, with means ๐ฅ1 and ๐ฅ2 , calculate ๐กcal . ๐กcal ๐ฅ1 − ๐ฅ2 = ๐pooled ๐1 ๐2 ๐1 + ๐2 36 Here the pooled standard deviation ๐ pooled is: ๐ pooled = ๐ 12 ๐1 − 1 + ๐ 22 ๐2 − 1 ๐1 + ๐2 − 2 ๐1 + ๐2 − 2 is the degrees of freedom. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (b) Comparison of the Means of Two Methods (Cont.) ๏ง If ๐กcal is greater than ๐กtable at the ๐ฅ% confidence level, the two results are considered to be significantly different at that confidence level. ๏ง It is valid only when s is the same for both sets of measurements. One can assume that this is always the case in exercises and tests. Example: A new gravimetric method is developed for iron(III) in which the iron is precipitated in crystalline form with an organoboron “cage” compound. The accuracy of the method is checked by analysing the iron in an ore sample and comparing with the results using the standard precipitation with ammonia and weighing Fe2O3. The results, reported as %Fe for each analysis, were as follows: 37 © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (b) Comparison of the Means of Two Methods (Cont.) 38 Test method Reference method 20.10% 18.89% 20.50 19.20 18.65 19.00 19.25 19.70 19.40 19.40 19.99 Is there a significant difference between the two methods at the 90% confidence level? © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (b) Comparison of the Means of Two Methods (Cont.) ๐ฅ1 = 19.648 ๐ 1 = 0.673 ๐1 = 6 ๐1 – 1 = 5 ๐ฅ2 = 19.238 ๐ 2 = 0.324 ๐2 = 5 ๐2 – 1 = 4 ๐1 + ๐2 – 2 = 9 ๐ pooled = ๐กcal 0.6723 × 5 + 0.3224 × 4 = 0.546 9 19.648 − 19.238 = 0.546 Since 1.24 < 1.833 (๐กtable ) (for 9 degrees of freedom), we conclude that there is no significant difference in the data at the 90% confidence level. However, at the 50% CL, ๐กtable = 0.703, which would mean that at the lower level of confidence we think that there is a difference in the methods. 39 6×5 = 1.240 6+5 Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (c) Paired t Test for Comparing Individual Differences 40 This test is used to determine whether two methods give the same result for several different samples. ๐กcal ๐าง = ๐ ๐าง ๐ ๐ is the absolute value of the mean of the difference. ๐ ๐ is the standard deviation of the difference. ๐ is the number of samples. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. (c) Paired t Test for Comparing Individual Differences (Cont.) 41 Is there a difference between the methods at the 95% confidence level? ๐กcal 2.491 = 6.748 11 ๐กcal = 1.224 which is less than ๐กtable (2.228) for the 95% confidence level and 10 degrees of freedom. Since ๐กcal < ๐กtable we conclude that there is no significant difference in the data at the 95% confidence level. Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. CM2011 Analytical Chemistry Q Test for Bad Data © 2019 Nanyang Technological University, Singapore. All Rights Reserved. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Q Test for Bad Data 43 Source: D. C. Harris “Quantitative Chemical Analysis (9th Edition)", Macmillan, 2016. If ๐calculated > ๐table, discard the questionable point. For example, ๐calculated = 0.11/0.20 = 0.55 < ๐table = 0.64. Therefore, in this case do not reject the questionable point. © 2019 Nanyang Technological University, Singapore. All Rights Reserved. Summary Having gone through this lesson, you should now be able to: 44 Explain the differences between absolute, relative, and percent relative uncertainties. Understand how errors are totaled when numbers (with errors) are combined by +/- or ×/÷. Apply and calculate the confidence interval. Utilize t-tests for different scenarios. Explain when to use the correct number of significant figures and correct rounding.
