1. 1 Standard Deviation of the Difference When multiple samples are analyzed by a proposed and standard methods, S d is the calculated standard deviation for the difference. S d = (  ( D i – D ) 2 / (N-1)) 1/2 S d is the standard deviation of the difference, D i is the difference between a result obtained by the standard method from that obtained by the proposed method for the same sample. D is the average of all differences.

2. 2 Example Mercury in multiple samples was determined using a standard method and a new suggested method. Six different samples were analyzed using the two procedures giving the following results in ppm Sample New Method Standard method 110.2 10.5 212.7 11.9 38.6 8.7 417.5 16.9 511.2 10.9 611.5 11.1 Find the standard deviation of the difference.

3. 3 It is wise to construct a table as below New Method Standard method D i 10.3 10.5 -0.2 12.7 11.9 +0.8 8.6 8.7 -0.1 17.5 16.9 +0.6 11.2 10.9 +0.3 11.5 11.1 +0.4 _____________________________ ________  D i = 1.8 D = 1.8/6 = 0.30

4. 4  (D i – D) 2 = { (-0.2-0.3) 2 + (+0.8-0.3) 2 + (- 0.1-0.3) 2 + (+0.6-0.3) 2 + (+0.3-0.3) 2 + (+0.4-0.3) 2 } = {0.25+0.25+0.16+0.09+0+0.01}  (D i – D) 2 = 0.76 S d = (  ( D i – D ) 2 / (N-1) ) 1/2 S d = (0.76/5) 1/2 = 0.39

5. 5 Propagation of Errors As seen earlier, each measurement has some uncertainty associated with it. During a process of calculations the uncertainty in the answer can be calculated from uncertainties in individual measurements. Calculation of error in the answer depends on whether the mathematical operation is a summation /subtraction or multiplication/division. It should be clear that in a process of calculating a final answer, as the number of mathematical operations increase, error will propagate.

6. 6 Addition and subtraction The absolute uncertainty in the answer S a can be evaluated from absolute uncertainties in individual numbers (b, c, d,.. ) as below: S a 2 = S b 2 + S c 2 + S d 2 + … Where, S a, S b, S c, and S d are absolute uncertainties in answer, b, c, and d (estimated standard deviation in answer, b, c, and d).

7. 7 Example Three samples were analyzed for iron content. The average percentage of iron in the first sample was 65.06, the second sample contained 56.13, and the third contained 62.68%. The estimated standard deviation of each of the three samples were + 0.07, + 0.01, and + 0.02%, respectively. What is the average iron content of the samples depending on these results?

8. 8 % Iron = {(65.06 + 0.07%) + (56.13 + 0.01%) + (62.68 + 0.02%)}/3 = (183.87/3) + S a % % iron = 61.29 + S a % S a 2 = (+ 0.07) 2 + ( +0.01) 2 + (+0.02) 2 = + 5.4x10 -3 S a = 7.3x10 -2 % Iron = 61.29 + 0.073% It is clear that we should retain two digits in the uncertainty as the answer is known to the nearest one hundredth (to get accurate number of significant figures). Therefore, the answer should be reported as % Iron = 61.29 + 0.07%

9. 9 Multiplication and Division The absolute uncertainty in calculations involving multiplication and division can not be estimated directly. The first step in such operations is to find the relative uncertainty in the answer from relative uncertainties in individual measurements as follows (S a 2 ) rel = (S b 2 ) rel + (S c 2 ) rel + (S d 2 ) rel + … Where, (S b ) rel = Estimated standard deviation in b (i.e. uncertainty in b)/ absolute value of b S a = Answer x (S a ) rel

10. 10 Example Chloride in a 25 mL sample was determined by titration with a 0.1167 + 0.0002 M AgNO 3 solution. If the titration required an average AgNO 3 volume of 36.78 mL and the standard deviation in the volume was 0.04 mL, find the uncertainty in the number of mmol of chloride contained in 250 mL chloride sample.

11. 11 You should remember that the standard deviation is the absolute error in volume of AgNO 3 mmol chloride = mmol AgNO 3 Since Ag + reacts with Cl - in a 1:1 mole ratio mmol AgNO 3 = molarity AgNO 3 * Volume (mL) AgNO 3 mmol AgNO 3 = (0.1167 + 0.0002) ( 36.78 + 0.04) = 4.292 + ?

12. 12 Since this is a multiplication process we use the equation for relative uncertainty (S a 2 ) rel = ( +0.0002/0.1167) 2 + ( + 0.04/36.78) 2 (S a ) rel = +2.03x10 -3 S a = 4.292 x (+2.03x10 -3 ) = 8.71x10 -3 ( This is the uncertainty in 25 mL chloride) S a in 250 mL chloride = 10 x 8.71x10 -3 = +0.0871 mmol

13. 13 If we are to report the number of mmol chloride in 250 mL sample the answer would be Answer = 42.92 + 0.0871 mmol The final answer should be 42.92 + 0.09 mmol since only two digits after the decimal points are allowed here to express actual uncertainty depending on the number of significant figures.

14. 14 Significant Figures and Propagation of Errors Unlike the argument above, we are now ready to indicate that the uncertainty (if properly calculated) defines the number of significant figures in an answer, regardless of the key number. For example, consider the mathematical value of the following calculation (73.1 + 0.2)(2.245+0.008) = 164.1 + 0.7 Here, since the absolute uncertainty in the answer is known and properly calculated as 0.7, then we are justified to include four significant figures in the answer although the key number (73.1) has only three significant figures.

15. 15 Let us look at the same calculation above but change the uncertainty of the first term to 0.7 we get the following answer (73.1 + 0.7)(2.245+0.008) = 164.1 + 1.677 It is clear that we have uncertainty associated with the integer itself, rather than the digits after the decimal point. If there is an uncertainty in an integer then the decimals are exceedingly uncertain that they can never be included. Therefore, the answer should be reported as (73.1 + 0.2)(2.245+0.008) = 164 + 2

16. 16 Look carefully at this example (101.1 + 0.9)(1.562 + 0.008) = 157.9 + 1.62 Since there is an uncertainty in an integer then the decimals are exceedingly uncertain that they can never be included. Therefore, the answer should be reported as Answer = 158 + 2 The number of significant figures is less than the key number !! Therefore, the uncertainty actually determines the number of significant figures