1. 1 DATA ANALYSIS, ERROR ESTIMATION & TREATMENT Errors (or “uncertainty”) are the inevitable consequence of making measurements. They are divided into three categories: GROSS ERROR – e.g. Equipment malfunction, using wrong reagent. Prevents any usable measurements being obtained. SYSTEMATIC ERROR - An error with the same sign every time. Usually associated with equipment, e.g. a thermometer which always reads 1 o C higher than the actual temperature. RANDOM ERROR - Random fluctuations in the measurement as a result of the apparatus or human operator. A comprehensive set of notes (as web pages) on this topic (including sample exercises) are available in the Web Tutorials section of the Level 4 Lab Website (http://www.l4labs.soton.ac.uk/)http://www.l4labs.soton.ac.uk/

2. 2 Precision & Accuracy Two other terms which need to be understood are: PRECISION - A measure of the statistical uncertainty of the result. Closely related to random errors. ACCURACY - How close the result is to the actual value. Often associated with systematic errors.

3. 3 Evaluating Errors SYSTEMATIC ERROR Can be difficult to find - Often intuitive (based on comparison of results obtained as compared to expected results). Test for by running a calibration test of the equipment or by repeating the measurement using a different set of apparatus. RANDOM ERROR Relatively easy to identify and quantify by multiple measurements. A statistical analysis of these multiple measurements will then allow the mean ( ) and the associated error (  95 ) to be evaluated. The evaluated (random) error will be in the form of a 95% Confidence Level (  95 ), i.e. we are 95% sure (or confident) that the true value lies between -  95 and +  95.

4. 4 Estimating Errors: Digital Displays It is not always practical to make multiple observations of a given observable and in these situations, the error on a given measurement may be estimated using common sense. Estimating Errors on Digital Displays - Determination of the error depends on whether the display reading is stable or fluctuating. FLUCTUATING DISPLAY STABLE DISPLAY Estimate the range of the fluctuations, in this case between 1.47 V and 1.51 V. Value recorded is mid-point of the range, with the error being half the range: EMF = 0.5(1.47+1.51) ± 0.5(1.51-1.47) V = 1.49 ± 0.02 V Determine the minimum amount the display can change by - This is the resolution. The cited error is half the resolution. For the example above: Minimum display can change by (i.e. resolution) = 0.01 V.  Error is ± 0.01/2 V = ± 0.005 V

5. 5 Estimating Errors: Volumetric Glassware The Total Error associated with using volumetric glassware is the sum of the Manufacturer’s Error (or tolerance) and the User (or Fill) Error. The Manufacturer’s tolerance is the inherent error in the equipment, i.e. even if filled perfectly there would still be this potential error in the volume. The Fill Error is an estimation of how well the user used the equipment. Manufacturer’s Tolerance Manufacturers Tolerance used should be listed on the apparatus. In the event that it is not, a value can be taken from the table of Standard Tolerances, available from the Level 4 Laboratory Website. Note 1: If the glassware Class is also not listed on the item, it should be assumed to be Class B. Note 2: Manufacturer’s Tolerance for beakers and conical flasks should be ignored, since the Fill Error associated with these is so large (in comparison) that the Manufacturer’s Tolerance can be assumed to be negligible.

6. 6 Estimating Errors: Graduated Volumetric Glassware Fill Error - Graduated Volumetric Glassware (e.g. graduated pipettes, measuring cylinders). 1.Determine the resolution - The smallest variation you can differentiate with certainty (using the graduations). 2.The Fill Error will be half the resolution. 3.Total Error = Fill Error + Manufacturer’s Error EXAMPLE: Assume, for the 25ml Measuring Cylinder shown opposite, that I believe (when it is filled) I can differentiate levels half-way between graduations. Volume between graduations = 0.5 ml  Resolution = (1/2)*0.5 ml = 0.25 ml  Fill Error = 0.25 ml / 2 = 0.125 ml Total Error = FE + ME = 0.125 ml + 0.08 ml = 0.205 ml

7. 7 Estimating Errors: Single Fill Volumetric Glassware Fill Error - Single Fill Volumetric Glassware (e.g. bulb pipettes, volumetric flasks). 1.Estimate (when you used it least accurately) how far the actual filled level could have been from the fill mark (  x). 2.Estimate the internal diameter of the apparatus at the fill mark (d). 3.These two values can be used to calculate the volume of a cylinder =  (d/2) 2  x. This is the Fill Error. EXAMPLE: Assume, for the (very badly filled) 100 ml Volumetric Flask shown opposite, that I estimate  x = 0.3 cm and d = 1.0 cm. Fill Error =  (d/2) 2  x = (3.141)(1.0cm/2) 2 (0.3cm) = 0.24 ml Total Error = FE + ME = 0.24 ml + 0.20ml = 0.44 ml

8. 8 Propagation of Errors The final result of an experiment usually depends on the combination of a number of measured properties, each of which has some associated error. The combined error may be calculated by using either mathematical relationships, or worst case scenarios. For “Worst Case Scenarios” the answer is first calculated using the values without taking into account their errors. The “Worst Cases” are then calculated, i.e. recalculate the answer having added or subtracted the errors to the values in such a way as to give the maximum or minimum possible values for the answer. The difference between the actual and maximum answer value is then determined, as is the difference between the minimum and actual answer value. The bigger of these two difference would then be cited as the  Error.

9. 9 Error Treatment: Worst Case Scenarios EXAMPLE 1: If y = a / b, where a = 1.2 ± 0.3 kg and b = 7.0 ± 0.6 m 3 y = 1.2 / 7.0 = 0.171 kg/m 3 Worst case values of y are: y max = (1.2 + 0.3) / (7.0 - 0.6) = 0.234 kg/m 3 y min = (1.2 - 0.3) / (7.0 + 0.6) = 0.118 kg/m 3 y max - y = 0.063 kg/m 3 ; y - y min = 0.053 kg/m 3  y = (0.17 ± 0.06) kg/m 3 EXAMPLE 2: Solution A is made by taking 5.00±0.03 ml of a 1.50±0.01 M Stock and making up to 25.0±0.2 ml. C A = C Stock V Stock /V Total = (1.5M)(5.00ml)/(25ml) = 0.3 M C A,Max = C Stock,Max V Stock,Max /V Total,Min = (1.51M)(5.03ml)/(24.8ml) = 0.3063M C A,Min = C Stock,Min V Stock,Min /V Total,Max = (1.49M)(4.97ml)/(25.2ml) = 0.2939M C A,Max - C A = 0.0063 M; C A -C A,Min = 0.0061 M C A = 0.3000 ± 0.0063 M or C A = 0.300 ± 0.006 M

10. 10 Citing Final Values Final values must be cited to an appropriate number of significant figures and decimal places: Round the error to one or two significant figures and cite the value to the same number of decimal places as the rounded error. EXAMPLE 1: A value of 1.234 x 10 -4 M with an error of ± 5.678 x 10 -5 M should be cited as (1.2 ± 0.6) x 10 -4 M or (1.23 ±0.57) x 10 -4 M EXAMPLE 2: A value of 98765 J/mol with an error of ± 4321 J/mol should be cited as 99 ± 4 kJ/mol or 98.8 ± 4.3 kJ/mol

11. 11 Statistical Tools - Regression A means of determining the line of best fit with error estimates for the gradient and intercept is via a Linear Regression (aka Least Square Fitting). Although a Linear Regression (and the calculation of the mean with its associated 95% Confidence Level) can be performed by hand, it is far easier to let a spreadsheet handle the calculation. The Excel spreadsheet package can be used to perform the above statistical analysis and extract from the results the relevant parameters along with the associated 95% Confidence Levels (or Limits).