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by: Carmela Kilback


Carmela Kilback
GPA 3.92


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This 19 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 461 at University of Washington taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/192607/chem-461-university-of-washington in Chemistry at University of Washington.


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Date Created: 09/09/15
a 4 T i 39 MENT or ANALYTICAL DATA WW 2 5 G when Imagine that you are determining by analysis some variable quantity x In two sets of quadruplicate measurements on two different materials A and B you obtain the data shown in Figure 21 The question is does material A really differ from material B An accurate but useless answer is Maybe It is possible to arses quantitatively the con dence with which you might imlead answer Yes or No It is pathetic to guess at the answer to such questions incredible wastes of time and money can result For example the materials A and B might be two batches of a manufactured product If they truly differ one must be discarded at great expense Alternatively the materials A and B might be products of a chemical reaction carried out under two slightly different sets of conditions If the difference in conditions has actually affected the composition of the product a new theory is proven or required Anyone who undertakes chemical measurements can either understand the statistical methods which allow useful answers to practical questions or can accept the role of an analytical instrument a mindless performer of laboratory procedures Consider a second example the quantity x has been determined by measurement ofa b c and d xab quot av 26 22724 Number of observations Number of observations Figure 2 1 Histograms showing results for the determination of the variable x in two materials A and B How do uncertainties in a b c and d affect the result How can the uncertainty in x be computed from the known uncertainties in the measured quantities The questions posed above are very practical The techniques which provide useful answers are easily mastered and this chapter offers a brief treatment of some basic statistical techniques A knowledge of calculus is helpful but by no means required for the understanding of this material and we beg the reader unfamiliar with calculus not to faint at the rst differential but instead to get a translation from the instructor NATURE OF QUANTITATIVE MEASUREMENTS Signi cant Figures In the examples below the number of signi cant gures in each quantity is given in parentheses 1 12270g 5 6 43062 ml 5 2 123g 3 7 43100 ml 3 3 10g 1 8 40000 ml 1 4 000524M 3 9 10000 5 5 0005 M 1 10 100 3 11 0010 2 The number of signi cant gures is always equal to or greater than the number of nonzero digits When zeroes are used only to locate the decimal point they do not count as signi cant gures Thus in l 9 10 and II the trailing zeroes are not required to locate the decimal point but instead indicate that the measurement has been made with the indicated accuracy Similarly in 6 the interior zero is signi cant In 3 7 and 8 the trailing zeroes may be signi cant or may be used only to locate the decimal point Scienti c notation furnishes the most convenient way of overcoming this ambiguity although an alternative used by some workers is the marking of signi cant trailing zeroes for example 466 indicates that all three gures are signi cant In scienti c notation the same number would be written 400 X 102 If there were only one signi cant gure the number would be written 4 x 102 To overcome uncertainties associated with various roundoff conventions some workers include the rst insigni cant digit in numerical data The insigni cant gure TREATMENT OF ANALYTICAL DATA w as trequently written as a subscript For example in the number 4052 the 2 is iii isggni ca nt Caution number bases are sometimes noted in the same W39l th 6243 might represent an octal rather than a decimal number i y us Absolute uncertainty and relative uncertainty Uncertainty in nIC39ISll d values may be considered from two distinct viewpoints Absolute uncertainty is exprcsxsztl directly in units of the measurement A weight expressed as 102 g is presuniabl I valid Within a tenth ofa gram so the absolute uncertainty is one tenth of Similarly a volume measurement written as 4626 ml indicates an absolfitimm certainty of one hundredth of a milliliter Absolute uncertainties are expressed in f1quot same units as the quantity being measured grams liters and so on m Relative uncertainty is expressed in terms of the magnitude of the quantit bei measured The weight 102 g is valid to within one tenth ofa gram and t1 cntii g quantity represents 102 tenths ofa gram so the relative uncertainty is about one ar in parts The volume written as 4626 ml is correct to within one hundredchf milliliter in 4626 hundredths ofa milliliter so the relative uncertainty is one art in 4626 parts or about 02 part in a thousand It is customary but by no mcnnspncc I sary to express relative uncertainties as parts per hundred per cent as arts cs Itahousand orl as parts per million Relative uncertainties do not have difnensig Su sireirtlipurgigertainty is Simply a ratio between two numbers having the I To distinguish further between absolute and relative uncertainty consider th weighing oftwo different objects on an analytical balance one object weighin 0 002l g and the other 05432 g As written the absolute uncertainty of each numb is on ten thousandth of a gram yet the relative uncertainties differ widely one art it 20 for the rst weight and one part in approximately 5000 for the other valu Signi cant gures in mathematical operations In addition and subtraction the absolute uncertainty in the result must be equal to the largest absolute uncertaint am th components Consider three examples y mg g 100051 05362 19724 42598 00014 00003 42595 025 l 19778 3 079 Infill rst two cases all the component numbers have the same absolute uncertainty an etermination ofthe correct number ofsigni cant gures in the result is a simple s the sicond case two components with ve signi cant gures yield a result with only ope the thg d example the absolute uncertainties are not equal and the number 0 stgni cant gures in the result is determined b the 39 39 39 absol t third number added y u c uncertainty m the In multiplication and division the relative uncertainty in the result must be equal to the largest relative uncertainty among the components For example in the operation 0 12 X 0628234 the correct product is 12 Expressing the result as 11614 would be un justi ed because the relative uncertainty in the rst factor is lone part in twelve Precision and Accuracy The terms precision and accuracy are not synonymous and are de ned here by contrasting examples A balance which on repeated trials gives the weight of an TREATMENT OF ANALYTICAL DATA 8 object as 1307 1308 1305 and 1307 g is more arerise than one which gives the weights 1302 1316 1305 and 1310 g I rrririmi relates to the degree of scatter in a set ofdata the scatter in these examples clearly being greater in the second set A balance which gives the weight of a known 10000g stiinrliird 15 10001 g is more accurate than one which gives the weight of the same standard as 10008 g Accuracy relates to the difference between a measured quantity and its true value Errors Systematic errors Systematic errors are frequently related to improper design or adjustment of experimental apparatus such errors reduce accuracy by systemati cally skewing or offsetting the observed data A careful study of measurements ofsome standard can reveal the nature ofthe error and accuracy can be regained by adjusting the apparatus or applying some correction For example if a balance showed the weight of a tengram standard to be 10080g and of a onegram standard to be 1008 g we could conclude that the balance was misadjusted so that it weighed 080 per cent high It would be possible to apply a correction to all the data but it would be far better to repair the balance Alternatively a balance might show the weight ofa onegram standard as 1050 g and the weight of a tengram standard as 10050 g In this case it could be concluded that the error was not proportional to the weight but was instead a constant 0050g offset Again a correction could be applied or preferably the balance could be repaired These examples also illustrate the value of systematic observations during instrument troubleshooting because the nature of the errors observed would naturally lead to certain types of adjustments The constant 0050 g offset for example must be due to a misplaced zero point on the mass scale the 080 per cent relative error is probably due to improper positioning of the balance arm 1 Random errors Random errors result from insulficicntly controlled variations in measurement conditions Many different effects acting together lead to small variations in the observed value Repeated observations will scatter randomly around the true value and it is thus clear that the size and frequency ofthcsc random errors will determine the precision ofa given measurement For example in the case oflhe analytical balance sources of random error include fluctuations in room temperature and humidity small variations in the placement ofweights variations in the position ing of the knife edge on its bearing and the subjectivity of the operator who reads the final weight from some uniformly graduated scale Careful consideration of the conditions of measurement can substantially reduce random errors but unlike systematic errors they cannot be eliminated nor can some formula be derived to correct an observation for their influence As we shall learn in the following sections random errors are amenable to statistical treatment and repeated measurements of the same variable can have the effect of reducing their importance BASIC STATISTICAL CONCEPTS Frequency Distributions Random errors and the normal distribution Consider the quantitative determination of for example iron in ferric oxide Figure 2 2 shows a sequence of three graphs illustrating the effect ofrandom errors In the first case only four quite TREATMENT OF ANALYTICAL DATA Number of observations 4 A 0 Number of n 32 observations A B 1 zazaaa39 n 70 N m v obgersaeiiooiis VVV VV A quot39 J C Frequency of observations D 532 534 536 538 540 542 oFe Figure 2 2 A B and C are histograms showing the distribu tion of results for a series of iron analyses D shows the normal distribution which would be obtained if an in nite number of observations could be made widely scattered observations were obtained More observations were obtained in the second and third cases and it is clear that the distribution of results is tending toward the smooth curve drawn at the bottom of the gure The smooth curve represents the universe of all possible determinations of iron in iron oxide and each of the smaller sets of observations is a sample drawn from that universe The height of the universe distribution at any given value of per cent iron is a measure of the frequency with which observations of that value will be obtained during sampling The relationship is known generally as a frequency distribution The shape of the frequency distribution is given by the normal law of error and the curve is known as the normal distribution Many other names are also applied but indicate exactly the same curve Some of the most commonly used synonyms are normal curve of error Gaussian distribution and probability distribution 9ther frequency distributions In Figure 2 3 two other fundamentally different frequency distributions are shown in order to further emphasize the concept If dice are thrown or cards are cut and the dice are not loaded and the deck is not stacked it is equally likely that any oneof the six sides of a die will come up or that TREATMENT OF ANALYTICAL DATA any one of the thirteen possible playingcard denominations will be cut The fre quency distribution representing39 these situations is flat If pictures are taken at ran dom intervals of a swinging pendulum the observed pendulum positions will be distributed as shown where 9 is the angle between the pendulum and its center of swings This frequency distribution arises because the pendulum is most likely to be observed where its velocity is low at the ends of its swmg and IS least likely to be caught by the camera at its center of swing where its velocityis at a maximum The se distributions apply to specialized situations The normal distribution applies in near y all cases where random errors occur a and quot of Y I For any set of data our aim is to summarize it quickly and usefully Everyone is familiar with summarizing a set of measurements by computing an averagetbu t this is strictly a measurement of central tendency and tells nothing about the dutrzbutzan of the measurements In order to summarize a set of data we requtre in addition some measurement of spread or dispersion of the individual data pomts V Central tendency The most common measurement of central tendency is the mean or average Usually we denote the mean value ofsome variable by placmg a bar over its symbol thus 2 n where 2 is the mean ofn observations ofx We must distinguish between x the mean Figure 23 Frequency distributions A for A the cutting of cards or rolling of a die and B observations of a pendulum at random time intervals See text for discussion 8 1 039 n L u X B gt o c a 3 0 9 LL o 6 TREATMENT OF ANALYTICAL DATA l i quot and the I I preceding data I crorc we even saw it given i ofsome finite sample and u the true value or universe mean Notice that could be calculated 39 4 a points In general I 2 alculated from dala 39 hm x u I degrccs or frcchIn 1 number of const inls n m t 2 2 3 Other useful measures of central tendency are the median or middle value ofsome 9quot sample and the mode or most frequently observed value of sonic sample In partic ular the appearance of more than one mode in a distribution requires careful study The existence of a bimodal distribution might indicate that a sample is in fact drawn from two different universes For example a bimodal distribution in examina 391n eclinmle of the universe variance The relationship lar to that between A and u The sample variance is l between 52 and 02 5 sum tion grades is frequently observed in Classes which mix undergraduate and graduate llm 52 quot2 students quotno Measures of dispersion The simplest measure of dispersion is the range Z x A2 2 x 02 of values found in some sample It is however a poor indicator of the shape of the lim F M 1 distribution because it depends only on the highest and lowest values By far the best quotMD 1 1 and most useful measure of dispersion is given by the variance which is the mean square deviation of all observations where the deviation is the di erence between an observation and the mean for the standard deviation The expressions apid computation of the sample variance and permit much more rapid calcu Alternati39ve expressions given above are inconvenient for the r The following relations are entirely equivalent T 2 1 tion a Z x 1 2 1 a 2 x2 22 n 2 Z 2 T quot 2 2 2b Me 32 if n l l The variance has the dimenSions of x2 and its value does not convey any feeling for the n amount of scatter in x Thus the standard deviation or rootmeansquare deviation is often used It is assigned the symbol 6 and is also de ned by equation 2 1 Notice that the formula can be used only when the universe mean u is known Thus equation 2 1 relates only to the universe variance and standard deviation of and 7 These parameters must be carefully distinguished from the sample variance and c standard deviation for the following results i Apollo ll negraincd material ange the data in a small table as Example 2 Compute the sampl obtained for the analysis of carbon in lunar 501 I30 162 160 122 ppm It is convenient to arr shown below standard deviation 32 and 5 which are discussed below x 2 Degrees of freedom An expression for the calculation of the variance and standard deviation of the sample as opposed to those of the universe is 130 16900 152 26244 25 600 x 39 2 160 i V if E 1x 22a 122 14884 r M Z x 574 2 x2 83628 quot 4 where Vz is the variance of x This expression differs from that given for 0 in two ways First 5 has replaced it and second the denominator is n I rather than n ation 2 2c we obtain The effect of the second change which decreases the denominator is to increase the Substituting these results in equ value for 53 This is appropriate because we have stacked the deck by substituting a 2 5742 for u a change which has the effect of minimizing the numerator This comes about 2 x2 g1 83628 because we have calculated 3 from the individual x values thus 1 represents the 2 71 i center of our sample but not necessarily that of the universe More fundamentally 5 1 1 3 I J I J n 1 represents the number of degrees of quot or r calculations which are possible within the sample after 2 has been calculated Consider a sample with a data points First we calculate k and then we begin to calculate deviations from the mean x When we reach the nth data point the x z 205 ppm comparison is no longer an independent one because the x value for that nth point 83628 82369 4197 ppm2 3 z illerencc between two large numbers Ilius 39 Int the numerator is the d Nome t alculation Sliderule accuracy is quot 39dinthcc A better representation of the numerator would be given by 2 x u but it is conventional many signi cant figuics must be eat me uate in unambiguous cases to omit the subscripts and interval notations for 2 usually maqu TREATMENT OF ANALYTICAL DATA TREATMENT OF ANALYTICAL DATA Standard deviation calculated from multiple samples The standard deviation is generally a function of the method of analysis and is not dependent on the speci c material analyzed Tonsider the use ofa particular technique to analyze ten different materials each in triplicate Thirty analyses have been performed and much experience about the precision ofthc method has been obtained Ifa standard deviation is calculated for each individual sample ten different standard deviations are obtained probably some are high some are low and each is based on only two degrees offreedom In this situation anyone would be tempted whether he knew much about statistics or not to calculate an average standard deviation in order to obtain a better estimate of 0393 It happens that for the case mentioned with equal numbers of observations in each sample a simple averaging procedure would give a perfectly correct answer In general when means 1 2 3 have been calculated from I11 n2 n3 observations 522Xi 112ZX2i222xa532quot39 z quoti 1 2 lquota lquot39 z z 22 z Ly 1 2 3 2 4b quotl Iquot2l 713 139 52 sum of deviations2 2 3 W 2 421 The last expression has been given before and is repeated here to stress its general applicability The reader who takes time to understand the equivalence of all three expressions above will be well rewarded In addition to allowing a better under standing of the standard deviation these expressions demonstrate that many statisti cal expressions which appear complex when written out in abstract form are in essence very simple An 39quot of the I I quot 39 of this 39 L I coupled with data coding see below is provided by Example 2 2 Coding The labor required in the calculation of a standard deviation can be considerably reduced by subtracting some constant from each observation in order to obtain smaller numbers In other calculations we often do this mentally For example anyone would nd the average of 61 63 65 and 67 by adding 1 3 5 7 dividing by four and adding the result to 60 This type of coding can also be way Also it is sometimes convenient to multiply or divide by some constant factor This too is permissible but the reverse operation must be performed after the calculations are complete in order to obtain a correct 3 An example is given below analyze four different samples Estimate the standard deviation of the analytical procedure Data Sample 1 1125 11301131 1131 Cl Sample 2 1426 1427 1430 1432 Cl Sample 3 1872 1865 18607001 Sample 4 1650 1645 1642 Cl 39 The coded data can be conveniently tabulated as follows1 Sample 1 x1 o Cl 1125100 Sample 2 x2 C1 1425100 Sample 3 x3 CI 1860100 Sample 4 x4 C1 1640100 TREATMENT OF ANALYTICAL DATA Given this coding we have i s 100rl l 4 Sample 1 Sample 2 Sample 3 S39imp e x 2 xI x42 quot quotS quot7 f 124 m 5 25 2 4 5 2339 52 2 6 3G 5 25 0 O 6 3G 7 49 I7 169 17 129 F 5 7 15 79 Applying equation 2 41 we have 5 4 l41 3 l3 1 gr 72 79 56 169 96 129 96 3322 25237333lamp4 10 5 0039 pet ccnt 5 100 For a similar problem see the first part of Example 2 13 Probabilities Derived from the Normal Distribution 39 39 39 the 00d for something Imagine Pumng it crUdelyl tiidl b iiihtliil 11121s beef applied hundreds of times than we had soni anivyre ard s as essentially equivalent to Us Couldwe not then and thhat w1blilitylthat 1 indiilidual measurement was within some given distance Sttfue t39liiisris easily possible but the method reqtiircssome introduciipnrorm or he 0 u39Mathematical expression for the normal distribution to normal distribution curve is given by LII V1 T 0 exp O f dx 2 5 t d x dx IS the probabili y fthe universe between A an 39here dN N the fraction 0 t m tcrms of x iim r willtalte a value between x and x dx An expresswnCSC Furthermore add 0 is not as conveniently evaluated as one With a Single varia Iquot a Ihc notauo ex 1nd cates that the argument IS the Jewel t0 Wlncll E 111115 be used IlIUS n p n 2 n flat 101131 eponents occur ills notation eliminates IlluCh COHlUSIOII e p Vi he i TREATMENT OF ANALYTICAL DATA I the Udn 39 itfg W39 V lr W 339l S u 39on 0 W L C H Hquot SI U 0 S i applgca 1 I x y till I ll It at l lli39ll ll 0 ill i i I i OWCVCF tllC Illa8 does not change WilatCVC the values ofx l and 7 I I and this constancy is 39 i seen in the fact that the e 39 r Single reduced variable u xp ession can be cast in terms ofa et 14 x 2 6 Differentiating equation 2 6 with respect to x we obtain a x 0391 du 24 Substituting equations 2 6 and 2 7 in equation 2 5 we obtain dN l 112 l u2 E ex 27r I p 2 0392 du exp 2 du 2 8 The 25c 022in3e elct expresses all deviations in terms of the standard deviation xam e ntheseries ofiron anal 39 yses noted abov F Fe and aquot 020 Calculate u for x 5358 7 Fe 6 lgure 2 2 T 5378 Substitution in equation 2 6 yields 0 x i 5358 5378 020 10 a 020 020 11 The value x 55t34 Fe would be termed one standard deviation out that one standard deViation one u unit away from the mean Is I Example 2 4 For the nal examination in a certain chemistry course 75 pomts and o z IO pomts Calculate u for x 100 points i u Substituting the appropriate values in equation 2 6 we obtain ux 4100 75 25 25 a2 10 io 39 On this examination a 39 perfect score is 25 standard deviations ou v I t see what the probability of this occurrence is we Shall soon univgflaltionship getween probability and area Every observation in any e 165 somew ere under the normal distribution 39 curve for that univers Stating this fact mathematically we can write the integral directly below 6 uco f a NN 10 u ao 0 TREATMENT or ANALYTICAL DATA The small sketch at the left illustrates the area given by the integral Similarly we could observe that half of the observations will lie on each side of the mean no ua I dNN 05 f dNN u m 0 o o u u 39 The small sketches illustrate the areas given by the two integrals Because the distri bution is symmetrical the probability is 05 that x will lie somewhere on the positive side of it and 05 that x will lie somewhere on the negative side of u What is the probability that x will take some value such that u 2 20 That is what is the chance that x will be two or more standard deviations out on the pdsitive side This prob ability is represented by the area under the normal distribution curve in the range 20 g u s 00 and can be determined by means of either procedure A or procedure B below A probability that x 2 it 20 f deN u20 u20 B probability that x 2 i 27 05 f dNN u0 The areas representing the integrals in A and B are shown in Figure 2 4 Either procedure is correct depending on which form oftabulation of the normal distribution is available Although most mathematical handbooks list only one or the other we present both forms in Table 21 For a tabulation of form A the probability that x will lie two or more standard deviations above the mean is given directly by the inte gral representing the area under the curve from u 20 to u 00 This area or B Figure 2 4 Normal error curves showing the areas which are given in the two alternative tabulation lbrms for the normal distribution See text for discussion TREATMENT OF ANALYTICAL DATA 7 I value of the integralais listed at u 20 in tables of form A and is given in Table 2 as 00227 prOViding directly the information that there is a 227 per cent chance t at an observation Will fall two or more standard deviations away from the mean When a table of form B is consulted the area or value of the integral tabulated at it 20 rep eslqpts the probability that x will lie between u and u 2039 This area is given in a e 2 1 as 04773 and the robabilit th 39 39 z quot 39 05000 04773 or 00227 p y at x hes outrzde this range is Notice that because it is symmetrical onl 39 39 39 39 39 y half the distribution 15 given In able 2 l this is stressed by tabulating the absolute value of u Practical calculations ased on equation 2 6 give u some sign which must be ignored when consulting the table Conversely when a certain In is found to correspond to some speci ed area it is necessary to think about whether the desired x is less than or greater than u and to aSSIgn a plus or minus sign to it accordingly An example follows Example 2 5 Your company makes steelbelted radial ply tires Extensive com mi a average me life of pt 50000 miles and a standard deviation on the is ri ution o 0 served tire lives of a 4300 miles For I l 39 what mileage can you guarantee I on y Per cent redemptions We require X to be chosen so that the shaded area in the sketch on the next page 15001 or 1 percent of the total We begin by consulting a table of the normal dis tribution Ifit is ofform A we look for area 00100 Ifit is of form B we look for 4 504918 04938 04953 04965 3904974 39 04987 O O O O O O vXltr I N In to V 50000 0 o 9quot 47 to 58600 62900 V area 04900 In either case we find u 233 X u dz Then for X lt 1 X It 2336 40000 miles 39 Example 2 6 For the examination mentioned in Example 2 4 what fraction of the students can be expected to obtain a pcrfcct score We have already found u to be 25 Consulting a table of form A we nd the area to be 00002 indicating that 062 per cent of the students will get a perfect score The area found in a table of form B is 04938 Example 2 7 What is the probability that any single data point lies within two standard deviations of the mean This probability is given by the integral u20 u20 uw J dNN 2 dNN 205 f dNN u 20 i0 u20 The second integral is listed in a table of form B from which we nd that the area is 04773 Thus the probability that a single observation lies within two standard deviations of the mean is 204773 or 9546 per cent We could obtain the same result by using a table of form A in which for u 20 the area is 00227 Notice that ugqaniet tsthisHaulageivgjbazaa iacr cen t con dcnt that u is within 2039 of any individual x Thus if we have for example x 250 and dz 01 we can be 95 per cent con dent that x 20 g u s x 20 or 248 g u g 252 ThWo en expressed for an experimental result in this wayvgr l250 con dence limits areinvaluable iii the interpraa tion of39e39xpe39rimlental data bul we39vvill th stress the eon39 i39e39pt39 it this point because the relations we have thus far derived require knowledge of the true 1 This is unlikely In practical situations we usually deal with 5 and a useful technique for the computation of con dence limits based on s will be given in a later section of this chapter PROPAGATION OF ERRORS The General Case Consider some result 10 obtained as a function of three experimentally deter mined independent variables xy and z The variances ofxy and z are known How TREATMENT OF ANALYTICAL DATA can they be combined to determine the variance ol tu An approximatc39cxprcssion is given by 3w 2 aw2 8w 2 V V V V quot39 Bx 2 6 quot396z z 2 9 map 394 n Speci c Cases Addition and subtraction Then Consider the case to fx z x 4 z and the general relationship given in equation 2 9 takes the form Vw 12Va 12Vz Va Vv V Rewriting equation 2 10a in terms of standard deviations we obtain xi I 52 s 2401 rw J5 32 if 2 10c Notice that 3 is not 5 rv 5 but is instead considerably smaller This occurs because it is unlikely that x y and 2 will simultaneously take values far above their means For example at the same time x happens to39be two standard deviations high y and z are likely to be much nearer their mean values and possibly even negative with respect to their means This opportunity for random errors in one variable to offset random errors in another variable is responsible for the form of the expression for Vw When coef cients are involved the calculation is only slightly more complex waxbycz 3w 3w 3w a 5 8x a z Vw anz szv 62Vz 2 11 Note that if the signs ofany ofthe coef cients happened to be negative ie ifwe were dealing with subtraction there would be no difference in the result because the derivatives are always squared Example 2 8 Suppose that the weight ofa sample we is determined by noting rst the weight ofa sample bottle plus sample w b and then the weight of the sample bottle alone wb w wab wb I Assume that the standard deviation of single is known to be r 1 mg Calculate 50 In accord with equation 2 10a we can write weight observations on the balance used Vw unn Vwb TREATMENT OF ANALYTICAL DATA v Since 10 H and llquot are both singlewright lS l tlllll lwm quot0 I l 39 quot7 l t 39 339 7 azxm mgz lhus I m l 1 l 2 mg and ru g I Y from Nam11 24 An analvsis is made by means ol a Single nu asurcmt n 4 which 31 where l is a background correction factor must be subtracted Ve can write I Y 3 where 1 is the analytical result Suppose that the variable If has been qulte prectscly measured in 0 ll x ht rcas the quantity X is known more poorly xx 0l Calculate t I Using equation 2 l we obtain VI39 121 l 321quot l39 10quot2 l 9l39 109 1w 11 1 x l0quot1 Standard deviation of thgjncan An interesting application of these ex W m pressions allows calculation of the standard deviation ofsome mean x derived from n observations xn711x1x2x339xn 2 fxls X2 x3 39 39 39 2 l 2 I 2 l 239 1 2423 V 7 V 7 V Va Vquot Because all the x observations are from the same universe r 2 l3 Vp L13VTVI 2 1210 2 12c As the number of observations in a sample increases the standard devnatlonbofttllig mean decreases In this sense four measurements are twrce as good as 11116 I u SC measurements are required if the precision is to be doubled again 1r n cnceof onn quotquot His 1 o r mumlii iedgfiofi In lixainple 2 1 above the standard deviation 2 pl atsar ifrl oflunar carbon measurements was calculated to be 205 ppm Calculate t re s anfour deviation of the mean 2 5744 l44 ppm The number of observations It Is Applying equation 2 12c we obtain TREATMENT OF ANALYTICAL DATA 2 393 s Functions involving multiplication and division C the expressmn onsider as an example w fxy 2 39 2 First we can write 63 m g aw awe go aw xxu 3x 52 x a 52 j z 2 Following the general expression 2 9 we obtain w 2 2 2 Vw Hm 2 le ilV x y z 2 Notice that values for x pendent of these values bec affect w Example 2 11 Suppose that the overall f 39 i ormatxon constant of th 2 IS determlned by making measurements of the ratio MXZ EM ercoamndlfil and 2 must be inserted The magnitude of Vw is not inde ause they determine the extent to which each variable can MX X 39 2 w 4 M X12 x12 For solutions in which 7 is 10 with a standard deviation of05 and X is 0 l M with a standard deviation ofO 01 M 39 what IS the standard devi 39 formation constant 32 anon or the cal Following the general expression 2 9 we obtain Va giftr f xt cula ted overall Evaluating the partial derivatives we nd ampE2 2 32 Gr X2 7 am X3 2l Substitutionin the general expression yields 12 211 2 lm From information given above g rX2 lO0l2 I 10339 inserti result along With the values for other terms gives on or this l 3 V 53 25 x 102 4 x 10 5 21 x 102 TREATMENT OF ANALYTICAL DATA 22 quotu THE tDISTRIBUTION AND ITS APPLICATIONS Although we have been careful to de ne the difference between 3 and 6 we have performed a number of calculations in which it was assumed that ax was known If 1000 or even 100 observations have been made this is e ectively true The 3 calculated from such a large sample is bound to be a very good estimate of 6 However in the typical situation where a sample contains only two three or four observations the calculated 5 is a very uncertain estimate of 1 When trying to calcu late con dence limits for example this uncertainty in 3 must be taken into account and the tdistribution provides the only way of doing so Con dence Limits on the Means of Small Samples Con dence limits using oz In Example 27 above we calculated 95 per cent con dence limits based on oz In order to stress this important concept let us repeat the procedure here in a slightly different and more formal way 1 We begin by asking What limits must be placed on u in order to include 95 per cent of the area under the normal distribution 2 See Figure 2 5 The shaded area totals 2st The unshadcd area is therefore 10 20 3 In order to include 95 per cent of the universe we want to choose 11 such that 10 2a 095 therefore or 0025 4 Searching a table of form A for area 0025 or one of form B for area 04750 we would nd that u 196 this limit is more accurate than the approxi mate one of u 20 given in Example 2 7 5 Thus for any single observation drawn from the universe we can be 95 per cent con dent that u is within 1960 of the observed x and we can write the 95 per cent con dence limits on x as x l 110025o x t 1900 There are only 5 chances in 100 or I chance in 20 that II will be outside the 95 per cent con dence limits 6 In 39gencral we can write the l00l 20 per cent con dence limits as x t 111039 2 14 7 An entirely equivalent expression holds for confidence limits on the mean With reference to equation 212c this relation can be formulated as uua 71 E j 1102 f j J 245 Figure 2 5 A view of the normal dis tribution used to clarify the concept of con dence limits See discussion in text TREATMENT OF ANALYTICAL DATA 23 Con dence limits using 1 When 5 replaces 01 a true value is replaced by an estimate which may be in error In order to take the uncertainty in 139 into account this substitution of 339 for c1z requires an increase in the coef cient ucl in equations l4 and 2 15 in order to widen the con dence limits Ifn is large the new coef cient can be about the same value as 14 because a large 71 provides a relatively accurate estimate of 0 When n is small and 3 is a relatively poor estimate of 0 the coef cient must be considerably larger than u The new coef cient is law a func tion not only of at but also city the number ofdegrees offreedom For a sample from which one mean value has been calculated 9 n l The general expressions written above for the 1001 2d per cent con dence limits can be rewritten as d 246 an 217 A tabulation of the tdistribution is given in Table 2 2 It is instructive to examine how tom changes as p varies Note rst that for p 00 1002 196 the same value that is obtained from the normal distribution This occurs of course because 1 0 when p 90 In general 1 um As 0 decreases l increases although t is still within l0 per cent ofu for p 13 Further decreases in 71 result in sharp ingeasesj nj In the extreme case where q 1 that is where r is determined from r a pair of observations the con dence limits correctly calculated using 002500 ex39ceed those which are sometimes incorrectly calculated using uo025s by a factor if 12706l960 or 65 Example 2 12 Calculate the 90 95 and 99 per cent con dence limits on the mean carbon content in Apollo 11 lunar nes The required data are in Examples 2 1 and 2 10 2 144 ppm xi 102 ppm 71 4 An expression for the con dence limits is given by equation 2 17 In this case p n l 3 For the 90 per cent con dence limits at 005 for the 95 per cent con dence limits at 0025 and for the 99 per cent con dence limits at 0005 The desired results are as follows Con dence Level t Con dence Limits E 90 2353 i2353102 i24 ppm 95 3182 i3182102 332 ppm 99 5841 i5841102 3560 ppm Unless otherwise noted the 95 per cent con dence limits are used in practical work I Signi cance Testing Con dence limits on the difference betuJeen two means In Figure 2 1 at the beginning of this chapter two small data samples are shown and the question of whether they truly di er is posed From each sample a mean can be calculated Call these means i and 5239 and note that each is an estimate of a universe mean u and 12 The original question can be rephrased does 21 uz It is entirely TREATMENT OF ANALYTICAL DATA 2 1 4443 i Table 22 The tDlstribution Wot W tp a 005 0005 00005 biota l1quot 1 6314 12706 63657 63662 2 2920 4303 9925 31598 fit 3 2353 31825 5841 12924 6L 4 2 132 2776 4604 8610 MAM 5 2r v is 5 2015 2571 4032 6869 M0 W 6 1943 2447 3707 5959 7 1895 2365 3499 5408 quot t d 8 1860 2306 3355 5041 GigiInd 9 1833 2262 3250 4781 thN AUWD 10 1812 2228 3169 4587 39 9 11 1796 2201 3106 4437 12 1782 2179 3055 4318 W m 4 51 13 1771 2160 3012 4221 M 14 1761 2145 2977 4140 W W 15 1753 2131 2947 4073 a NA 20 1725 2086 2845 3850 wt 30 1697 2042 2750 3646 60 1671 2000 2660 3460 00 1645 2576 3291 lam18 1 288825 35 possible that f and 2 could diller that is Aquot 2 A sf 0simply because of random scatter in the data This difference might occur even 1f the two samples were drawn from the same universe that is u 122 Note that Ax1s an estimate of Ala and rephrase the question again does All 0 We can answer this questlon by determining whether the con dence limits for A37 are larger than Ax itself Ifthey are there is a good chance that AA is zero and that Alu O smce Af con dence limit s Alu g A2 con dence limit In general con dence limits for anything jglmwanumhm 2 18 where anything has 2 degrees of freedom In this case we are intct39cstcd in the con dence limits for Atquot which is de ned by equation 2 l9 A 7 52 2 l9 Our rst task is to de ne 5 which llCI C takes the role ofsanwmng Considering equa tion 2 19 and equation 2401 we obtain VAX Vil If2 2 20a In accord with equation 2 12b the individual variances are given by 2 21a 2 2lb TREATMENT OF ANALYTICAL DATA 25 m 1 where 711 and n are the numbers of observations in each sample When results from the analyses of two materials are being compared it is quite generally true that VaEl Va VI because this variance is a function of the method of analysis only Then substituting equations 2 21a and 2 2lb in 2 203 we obtain V V i i 39 VA 7 7 V 5 2 20b Recasting equation 2 20b in terms of standard deviations we get 1 1 1 2 5 51E 2 20C Following equation 2 18 the con dence limits for A are given by 12 con dence limits for Ai CL itchy itmvxreq 2 22 where p n n2 2 two constants 1 and 2 having been calculated from the data in this case In such cases at is conventionally set equal to 0025 Then if the con dence limits are smaller than Af the chances are 19 in 20 that Alul gt 0 Au is said to be signi cant and it is judged that there is a real difference between the samples 1 75 2 Example 213 Evidence for the identity of an organic compound can be obtained by measuring the time required for passage of the compound through a chromatographic column and comparing this retention time to the retention time ofa known standard Equal retention times suggest but do not prove that the un known and known compounds are identical Suppose that an unknown compound is passed through a chromatographic column three times and that retention times of 1020 1035 and 1025 min are obtained In addition suppose that standard n octane has been run eight times and that the observed retention times are 1024 1028 1031 1032 1034 1036 1036 and 1037 min Might the unknown be noctane 1 Determination qf 1 In each case code the data such that x tn 1020100 unknown standard noctane in x1 x12 tn x2 x22 1020 0 O 1024 4 16 1035 15 225 1028 8 64 1025 5 25 1031 I 11 121 I 1027 E 55 0 1032 12 144 711 3 1034 14 196 1036 16 256 1036 16 256 1037 17 289 I 03 98 1342 712 8 TREATMENT OF ANALYTICAL DATA Z Using equation 2 1b we obtain r 20 K 982 n 2gt0 3 11312 T 25 1 quot 3 18 1 9 s 100 V2t 7 54 w 5quot 0054 min 2 Determination Jcmi a mua li171i1rfnr At If we utilize equation 2 22 the result is as follows 1 l 2 LL ilz ws lgt qr n1 1 n2 2 9 for 95 CL 0t 0025 CL 122620054 quot2 j226200540677 008 min 3 Comparison oft12 C11t ltllt d mn dence limits will Atquot At RI in 1027 1032 005 min Since l 005 lt 008 there is a good chance that A 0 Accordingly the un known could be noctane Alternatively we could note that the indicated con dence interval is 013 g All g 003 min and tlia39t this range indicates the possibility of At 0 The ttcst In the preceding section we have judged the Signi cance of A by computing its con dence limits The essence of this approach can be summar ized as follows If the confidence limits for A2 are greater than A then the dillerence between the true means might be zero 2 23a Ift wt 2 then possibly Alu 0 2 23b The arithmetic of signi cance testing is made simpler if this approach is slightly changed A value for 1 no subscripts is computed from the relation 1 M i l 2 24 1 i SAi xmnl Compare equations 2 23b and 2 24 Note that ift lt law the con dence limits will be greater than A and therefore Apt 0 If 21 11 u2 This method of calculating a test value for l and comparing it to some critical value listed in a table is known as the Nest Example 2 14 Two limestone samples are analyzed for their magnesium con tent The results obtained are Sample 1 122 125 126 Mg Sample 2 131 134 135 Mg Do the samples diller signi cantly TREATMENT OF ANALYTICAL DATA Z7 4 1 Calculation ofsm the standard deviation of the analysis Code the data Sample l x1 Z Mg l20100 Sample 2 x2 70 Mg l30100 Sample 1 Sample 2 Mg 1 x12 Mg x2 922 122 2 4 131 1 T 125 5 25 134 4 16 126 135 5 25 Mgl24 13 65 M g133 B 312 Using equation 2 4b we obtain 52 1005 2 Bi WU B Uo n rm 0021 per cent 2 Calculation oft according to equation 2 24 AW 133 124 L LW 1 1 In quotI l39 n2 0021 009 t 52 0017 9 3 Comparison of t to oritical values lated from the data and the number n2 233 24 At the 95 per cent con dence lev el tX 0025 and the critical value f I I t sviSund 1p Table 2 2 In th1s case to39mquot 2776 Since 2776 is less than 52 tiii p6 no a cu 1Lahtjed abqve the difference between the samples is signi cant at the 95 r cen con ence evel That is there is less than one 39 I cha t d1fference this large could occur between two samples from mfg Li iiiiiii s hat a 99 H1gher levels of possxble signi cance can be tested by decreasing at For th 4 6amperltc n2tgcop dfrfle level at 0005 Table 2 2 shows t0 005 4 4 604 Sine tieierencbt H dencc level e e ween samples 15 stgm cant at the 99 per cent eon Table 2 2 shows to on 8 610 39 I 054 Since 8610 gt 529 the observed di not s1gni cant at the 999 per cent con dence level That is differences this gilezl iisiii OCCUI between two samples from tllC same universe 111016 il equ ll 1 C tl an no time out Detection limits In this case two constants have been calcu of degrees of freedom is given by to n1 Particularly in the a 39 naly51s of trace constituent 39 39 m s the mini inpm detIectable amount or concentration of some analyzed component is of reat ni 1xslttly tdifslraskedf What is the mlmmum quantity which produces a resultgsig 1 erent rom zero 3 When an anal t39 l 39 39 39 39 date I I yica technique IS ut1hzed near its 521meltion hmits it IS always observed that not only are the results on legitimate p es scattered by the effects of random error but also the results on blank TREATMENT OF ANALYTICAL DATA 29 samples are scattered about zero by the effects ofrandom error The determination of detection limits thus requires consideration of the minimum significant lilfercnce between two uncertain numbers the sample result and the blank result If the average blank result is denoted as A7 and the average sample result as 5 the difference between them will be Afsmb and its confidence limits are given by CL irulmm ii 13 3 2 25 quots quotb where 1 and 5 are the standard deviations of the sample and blank respectively and n and 12 are the number of observations on the sample and blank respectively Near the limit of detection 5 will be approximately equal to J which can be con veniently measured by repeated observation of blank results The preceding ex pression then takes the form 1 l CL ilk me L 2 26 715 11 The minimum detectable A s can be denoted by Ailim to determine its value we need only specify that it exceed the con dence limits set by equation 2 26 l l Axum gt lawIzM 2 27 s n b where p is the number of degrees of freedom used in the determination oft Example 2 15 Ten blank experiments using a new technique for the deter mination of traces of sulfur dioxide in air gave39values of 2 3 12 l 1 0 5 10 ll and 8 ppb parts per billion of sulfur dioxide in air a Deter mine the minimum concentration of 02 detectable at the 99 per cent confidence level by means of this technique ifa single sample measurement is compared with a single blank measurement b What is the detection limit at the 99 per cent con dence level for the situation where triplicate observations are made of the blank and of the sample 1 Determination ofrrb Calculation of the standard deviation in the usual way see Example 2 1 yields the result r1quot 52 ppb 2 Determination ofdeteclion limits In part a we have ns ab l and ct 0005 From the preceding calculation we obtained 5 52 ppb and q 9 Insertion of these values into equation 2 27 gives Afilm gt 32552VJl 24 ppb Thus any concentration of sulfur dioxide greater than 24 ppb will differ signi cantly at the 99 per cent con dence level from zero In part b the only change is that n II 3 thus the detection limit is computed to be Aim gt 32532i t 14 ppl REGRESSION It often happens that an analytical instrument is calibrated with a series of standard samples A graph of the instrument output versus known sample input is constructed by drawing a line through the calibration points A typical example is shown in Figure 2 6 The simple graphic solution of such problems in instrument TREATMENT OF ANALYTICAL DATA 29 09 08 r 03 Instrument Readout absorbance 02 o a l g 0 7 2 3 4 5 6 7 5 Feb M x 105 2111 5 6 A typical instrument calibration graph in this case it is a plot of ironII con With 533ili3an aqueottas solution containing a complexing agent which gives an intense color versus e Instrument readout absorbanc 39 l l e at a certain wav I statistical derivation of the calibration line is discussed in the text 6 ength The calibration is unsatisfactory in several respects First the graph must be ver hr 391 more than two signi cant gures are to be determined Second the placemeyht ofgfll line 18 an emotional process The experimenter is very likelynot to draw the li 16 which best ts the points and an improper t can cause large errors It is in ii cases better to derive a mathematical relationship which expresses the instiume t response function Comparison of the derived coef cients on a daytoda bali allovgs the experimenter to monitor instrument performance and the dribiasii 3u3tical procedure offers the only way to achieve line placement without The problem of deriving such relationships comes under the general headin of regress1071 analysis in statistics Here we shall be concerned with only the simgl t case the tting of straightline relationships by the method of least squares p es TREATMENT OF ANALYTICAL DATA 30 am a Wu The Distribution of Calibration Data The chemical input for example the iron11 solutions of known concentration used in establishing the graph of Figure 2 6 is regarded as a variable which is free of error 039 0 Everything which follows con cerning the leastsquares determination of calibration lines assumes that the standard chemical inputs are known with perfect accuracy or at least that errors on this axis are insigni cant when compared with errors on the instrument output axis If this is not true a difTerent approach to data analysis is required The observed instrument response points Each instrument output reading is a single observation drawn from the universc of all possible instrument outputs for a given chemical input This situation is depicted in Figure 2 7 For the input standard with value 03 for example the small normal distribution curve which is sketched on the graph represents the universe of all possible instrument outputs for a chemical input of 0300 Point A observed during calibration is near the center of this distribution well within the 3120 limits marked by the shading In fact we were quite lucky to obtain a calibration point so close to the universe mean Even so we do not want the calibration line to go through that point but instead through the exact center qftlze distribution which is marked by the arrow Similar comments apply to the distributions indicated for standard inputs of 0600 and 0900 The known chemical inputs 80 60 observed output y 40 20 0 05 10 input standards X Figure 2 7 Another view of an instrument calibration graph The small plots of dNN versus u represent the normal distri bution of outputs for each of three particular inputs There fore they refer only to the dNN and axes and do not have any signi cance on the 39 axiS Sec text for further discussion TREATMENT OF ANALYTICAL DATA 3 m Principle of Operation we m ompm Olsgtpbtaip estimates ofthe center ofcach distribution by recording several accurac fro ipns or each input but this is not necessary to obtain fullest possible y m tie method ofleast squares which determines the locus of universe by H m l g 2 t 8 1 ns CC I 0 lllCallS Ill I 12 H d V121 O 2 thW H tllC ailblat on data p 115 Ind the y a bx 2 28 Th quotmt suml39of squares of the dCVla OnS is minimized by adjusting a and b 1nd isla ziiearfimctwnal relationsn between x and this has the effect of puitin ih chresston me through the best estimate ofthe true mean values Thus oints Hg Ci asAvsielllas pomt A aid in the determination of uo a and so on W I i an at iematieally we derive solution 39 l I s for a and b 1 ex ressin 39 2 which we denote by Q as a function ofa and b y p g 2 dcvmuom Q Z deviations of experimental points from line2 Q2y abx2 2 2921 2 29b Next the partial derivatives of the latter relation are taken rst with respect to a an CD with PC 1116 I ll ual It 26101 Older 0 d til 65 Ct to 5 65111 g expressions are set Cq nd the a and b Values WlllCll W1 HUHHIIIZC Q 39 50 2 z y a M 0 2 30a 3Q ab 22xy abx 0 2 3la These expressions can be rearranged as follows 2 M b 2 x 2 3010 nya2xb2x2 2 31b vhcartc n is Eh number of xy pairs Finally we obtain solutions for a and b by solvin qu ion for a in terms of b and x and by substitutin the r It 139 39g equation 2 3lb g cm or a m 7li2 112x b2xy niy zxz mtquot2 2 30c 2 32a wherx z xn andj Zyn just as in onedimensional antilysis th he form of equation 2 29b shows clearly that it is only deviations parallel to e axts which are minimized The x values are regarded as xed This is the o 39 39 of the requirement mentioned above that the input standards be ver car prepared When it happens that signi cant errors can occur on either axis yth e 1 y Is referred to the excellent and simple explanation given b Y rk quot y 6 ma er listed at the end of this chapter y o m the paper TREATMENT OF ANALYTICAL DATA 32 1 Method of Calculation The data are presented as General expressions for the sums of squares we de ne the following n pairs of x and values For convenience in calculation relationships ZU2 Zx 22Zx2 n 22x2 220271 2 33 ZVZEZ j22 2 quot5 22J2 Eff391 234 EUVEZx fy j2xy rifj2xy ZxZn 2 35 Solutions for a and b Substitution of equations 233 and 2 35 in tion 2 323 gives a convenient expression for the calculation ofb equa UV slope b 27 2 32b After I has been calculated it is substituted in equation 2 30c from Which a can be found Uncertainties Con dence limits for b It often happens that the slope ofa regression line is predicted by theoretical considerations to have some given value Occasionally a set of observations which should de ne a flat straight line b 0 indicating that is independent ofx is examined to see ifsomc nite value ofb is found In either case ave a method for the assignment of confidence limits for b it is vital to h bout the regression The rst quantity required is termed the variance 2 de ned as follows V2 b2 U2 xv Lagd 236 I 2 Then the ofthe a 39 m 39 or a quot 39 39 quot ofb is given by 2 s 2 U 2 37 Con dence limits for b are assigned in the usual way CL ital st 238 where p n 2 Con dence limits for a regression estimate In practice the regression line is used to determine some estimate xiv of the chemical input which is responsible for an observed instrument outputyki The variance ofan xk value which has been determined by the observation of m outputs is given by the expression 5 1 l where for clarity the following points should be stressed iii b2 2 U2 2 39 TREATMENT OF ANALYTICAL DATA 33 1 taquot 1 xk is the unknown chemical input corresponding to the observed instru ment output For example in Figure 2 6 xk is the ironII concentration correspond ing to a particular observed solution absorbance 2 y39 is ijm the average instrument output obtained from m observations of the unknown chemical input Frequently only a single reading is taken and there fore m l 3 The quantities 542 5 72 and U2 all relate to the instrument calibration data and have the same values as in equations 2 29 through 2 36 Notice that 1 becomes large as gets farther away from This is nothing but a mathematical expression of the sensible fact that the uncertainty in any regression estimate is lowest near the center of the calibration data and that extrapolations are a particularly Chaney business The con dence limits for the desired xk are given by the relation CL ilwpszk 240 where q n 2 g fm Example 2 16 Use the method of least squares to construct a calibration line for the spectrophotometric determination of ironII described earlier 1 Actual data which are plotted in Figure 2 6 appear in the table below Standard Solution Concentration c molesliter Observed Absorbance A 100 X 10 0114 200 x 10 5 0212 300 x 10 5 0335 400 X 10 5 0434 600 X 10 5 0670 800 X 10 5 0868 2 The calibration line must t an equation of the form j a bx where y is proportional to the absorbance and x is proportional to the concentration In order to simplify the arithmetic we will take 10A and x 1050 With 7 6 we obtain from equations 2 33 2 34 and 2 35 2 x 2400 2 00 z x2 13000 2 U2 34 zymss jamm 2wtw0m5 2V2amw7 zgtww zUVwh In these calculations we have carried as many signi cant gures as the calculator will allow This is done to minimize roundOHquot errors and is a required procedure It is not synthetic precision unless it is used in reporting sonic result 3 The coef cients a and b are calculated using equations 2 32b and 2 30c bZUV 2U2 a it21 62x 002245 109147 TREATMENT OF ANALYTICAL DATA m m The equation for the regression litie thus takes the form v e 002245 109 Substituting the coded values nix andy in the latter relation we obtain the desired equation for the regression line 101 002245 1001 x 10 c 205H x 107 9166 10 511 4 To determine the confidence limits we first obtain the vanancepbout gin regression according to equation 2436 Because the numerator of equation 2 6 is the dilTerenee between two large numbers many sigmhcant figures must be carried in the computation 2 2 2 5 2 w 8325 X 103 I n 2 The standard deviation of b can be calculated from equation 2 37 2 Sb 2 5quotquot 2596 x 10 x 0016 v2 U2 Next we determine con dence limits for b by using equation 2 38 with at 0025 and p 4 CL 311 th 0044 Therefore b 1091 1 0044 h I I As an example of the calculation of confidence limits on a regresston estimate imagine a That a single unknown sample gave an absorbancc of 0527 r b That five replicate samples gave an average absorbancc of 027 For either of the two situations 6 2058 x 107 916 X 1021 4810 y 10441 The standard deviation of the regression estimate is obtained from equation 2 39 where in case a m 1 and in case b m 5 In both Sliu dilOnS139k 104 5 27 Inserting appropriate values in equation 239 for each of the two cases we get for a and for b xx 8804 x 103 in 0003R Jr 2878 X 0 3 I 0053 The confidence limits for the regression estimates are both computed from equation 2 40 with q 4 c 481 1 026 X 10 51 c 481 1 015 x 10 5 a CL itn ozs rk 026 b CL 531032345 i015 Note that because of the coding the confidence limits for c are 10 5 times those for the regression estimates TREATMENT or ANALYTICAL DATA 35 i g u i amojftntothigig elfe these results in which the correctly calculated uncertainties n o a out per cent demonstrate that the m cthod of least 3 39 stitute for quality in the calibration da quarts Is no SUb ta themselves Man chem39 t 1 method of least squares on 39 y S S cam and use the y up to the pomt of calculatin b d om S 39 g an a and com letel i tep 4 above In this way they misuse the statistical method and freqiiientl dCCClVC themsel e y P y V S a d the S C r a ll 11 0 l b I 0 ting 00 m n Sig i cant gures In their REJECTION OF AN OBSERVATION Consider this set of analytical results 1525 1528 1530 1643 per cent It izgrifal to suspect that something is wrong with the fourth bbservation and 11 ptation to reJect it before calculating the mean or the standard deviation i b gry strotgrSucl points are called outliers and though we have prosvidenca1 r examp e or t e sake ofillustration 39 agony particularly when the case is not so 51251323 iii23036 congdemblc data pomts would make the result coincide with some expected valif I onehor two any39human being is tempted to look around for some statistical Oil39Wliljuch39cilses anomt his whim but this should be attempted only as a last resort since th w l39c 39 to tests for the exclusion of outliers are reasonably unsatisfactory e Statlsucal The rst thing to do is go back and thoroughly recheck all the calculatio h39 h led to the offending result Second check the data Did the same sample bottrlls which weighed 15145 g for samples 1 2 and 3 weigh 14145 g for sample 49 Ifso U 1i m weight rnust be incorrectly recorded Is there a note that the fourth sam 12 cladth a peculiar green lump These questions are aimed at uncoverin sage nllc u e assignable cause for the large deviation If such a cause can be foundgit can C39flirCbu corrected in the case of some arithmetic blunder or used as a f 1 e1 er 6 for excluding the outlier per CC V gOOd reason I Sometimes no clear cut assignable cause is found and the ex eriment k himself Don t I remember spilling a bit ofthat solution P or I sefm t er a1 5 that the analytical balance was acting up This is only more eiridence of 261116 er ity but in any real situation it has to be met with coldhearted honest A tr turfalri Sincerity of such suggestions can be made if the experimenter will ask himselfth ii 6 he is willing to resolve at that moment to discard any future result which mi hteli er been simtilagly affected regardless qfit value Another useful policy to considger is iii romise o 39 39 39 phe lanaIYSis iic1naiiglisample which might have been Similarly a ected before When there are only three or four observations in a sample it is nearl im 05 39b1 to provide a useful statistical test for the exclusion of outliers When ther arep vSI c more it is sometimes helpful to practice the technique of interior avera i c 3quot which both the highest and lowest data points in the sample are discarded g ng m SUMMARY quot t this chapter we have attempted to provide a solid foundation for the under s an ing and use of statistical techniques in chemical experimentation The discus suilnshafre not mathematically rigorous but neither are they useless trivializations ylvv 1ch ore the reiader into memorizing statistical formulas if anything is to be gained e ope t e rea er will avoid memorization a I nd instead Work t cardmal pomts o understand a few TREATMENT OF ANALYTICAL DATA 3 l The differences betwecn a sample and a universe between 5 and 039 between i and ii 2 The gen 3 The use 4 For any 5 Why th on anything r t 2 sum of deviations2 era 123 ion 539 of the normal distribution tables w fx 7 1 29w2 aw2 aw2 Vquot Vz V Vz 6x 6 u a e tdistribution is needed and the generalization confidence limits itaw39y nvlhi ng 6 The principle ofregression analysis and the concept that poor data cannot be improved by its use 7 The fact that outliers must have some assignable cause the idea is to find it not to throw away data QUESTIONS AND PROBLEMS 3 FO Equot S l 9 5quot De ne and illustrate each of the following terms signi cant gures precision accuracy systematic error random error average deviation confidence limits absolute uncertainty relative uncertainty 1 test Assuming that the following quantities are all determined to within i1 digit in the least signi cant gure express the relative uncertainty in each in parts per thousand or where appropriate in parts per million a 0104 g b 12043 m1 c 1007825 atomic mass units the mass ofa 1H atom d 569354 atomic mass units the mass ofa 57Fe atom e 456 x 109 yr the age ofthe earth f 5730 yr the halflife of14C Are equations 2 1 and 2 3 entirely equivalent If not why not What about equations 2 23 and 2 3 Construct a simple algebraic proofofthe equality of equations 2 2a 2 2b and 2 2c 39 The following results were obtained for replicate determinations of the per centage of chloride in a solid chloride sample 5983 6004 6045 5988 6033 6024 6028 5977 Calculate a the arithmetic mean b the standard deviation and c the relative standard deviation in per cent Assuming the sample of problem 5 was pure sodium chloride calculate the absolute and relative errors of the arithmetic mean Working from equation 2 3 prove that for It sets of duplicates 2 2 w22k where w is the difference between the two measurements in a dupli cate pair Five students carry out the same analytical procedure on five different trace iron samples The results are given below Assuming an absence of systematic errors calculate the standard deviation a of the procedure and b of the mean of three measurements on a single sample Iron concentrations in ppm student 1 sample C 43 48 47 39 student 2 sample A 38 38 42 40 41 student 3 sample D 51 54 57 58 student 4 sample B 35 36 38 41 student 5 sample B 45 48 52 54 Drill problems on the use of the normal distribution tables Given 3 nor mally distributed universe ol qiiantitative observations what is the probability that a some observation will fall more than 080 above the mean b some observation will fall more than 50 below the mean c some observed x will be in the range it 046 S x 3 i 130 d an observation will fall in the range l60 lt u S 024 TREATMENT OF ANALYTICAL DATA 3397 Table I Y I X 51L 4 393 AREAS OF THE STANDARD NORMAL DENSITY FUNCTION p AZ IS THE AREA UNDER THE CURVE BETWEEN 0 AND 2 2 42 z 142 z Mz 2 42 z AZ z 42 2 Mz 00 0000 60 2257 1 20 3849 180 4641 240 4918 300 4987 360 4998 01 0040 61 2291 121 3869 1 11 4648 241 4920 301 4987 3 61 4998 02 0080 82 2324 122 3888 1 82 4658 2 42 4922 3 02 4967 3 62 4999 03 0120 63 2357 123 3907 183 4864 249 4925 303 4988 3 68 4999 i 04 0160 84 2389 124 3925 184 4871 244 4927 304 4988 3 64 4999 05 0199 85 2422 125 3944 185 4678 245 4929 305 4989 365 4999 I 06 0239 66 2454 1 26 3962 186 4686 2 48 4931 3 06 4989 3 66 4999 f 07 0279 67 2486 127 3980 187 4693 2 47 4932 3 07 4989 3 67 4999 08 0319 68 2517 128 3997 188 4700 248 4934 308 4990 368 4999 09 0359 89 2549 129 4015 189 4708 2 49 4938 3 09 4990 3 89 4999 10 0398 70 2580 130 4032 190 4713 250 4938 310 4990 370 4999 11 0438 71 2811 131 4049 191 4719 251 4940 311 4991 371 4999 12 0478 72 2642 132 4066 192 4726 252 4941 312 4991 372 4999 13 05171 1z 73 2873 133 4082 193 4732 253 4943 313 4991 373 4999 14 r0557 quot 74 2703 1 34 4099 194 4738 254 4945 314 4992 374 4999 i 15 0598 75 2734 135 4115 195 4744 255 4948 315 4992 3 75 4999 16 0636 76 2764 136 4131 1 96 4750 258 4948 316 4992 3 76 4999 17 0675 77 2794 137 4147 197 4756 257 4949 317 4992 3 77 4999 18 0714 78 2823 138 4162 1 98 4762 2 58 4951 318 4993 378 4999 19 0753 79 2852 139 4177 199 4787 2 59 4952 319 4993 379 4999 g 20 0793 80 2881 1 40 4192 200 4772 2 80 4953 3 20 4993 3 80 4999 39 21 0832 81 2910 141 4207 201 4778 261 4955 3 21 4993 381 4999 22 0871 82 2939 142 4222 202 4783 262 4956 322 4994 3 82 4999 g 23 0910 83 2967 143 4238 2 03 4788 263 4957 4994 383 4999 24 0948 84 2995 144 4251 2 4 4793 64 4959 3 24 4994 384 4999 1 T 25 0987 85 3023 1 45 4265 205 4798 265 4960 3 25 4994 3 85 4999 26 1026 86 3051 146 4278 206 4803 266 4961 3 26 4994 3 86 4999 27 1064 87 3078 1 47 4292 2 07 4808 267 4962 3 27 4995 3 67 4999 28 1103 88 3106 148 4306 201 4812 268 4963 3 28 4995 3 88 4999 29 1141 89 3133 149 4319 209 4817 269 4964 29 4995 3 89 5000 30 1179 90 3159 150 43327 210 4821 270 4965 3 30 4995 390 5000 31 1217 91 3186 1 51 4345 2 11 4826 271 4966 3 31 4995 3 91 5000 i 32 1255 92 3212 152 4357 212 4830 272 4967 332 4996 3 92 5000 1 33 1293 93 3238 153 4370 213 4834 273 4968 333 4996 393 5000 34 1331 94 3264 154 4382 214 4838 274 4969 334 4996 394 5000 35 1368 95 3289 155 4394 215 4842 275 4970 335 4996 395 5000 36 1406 96 3315 158 4406 2 16 4846 276 4971 3 36 4996 3 96 5000 37 1443 97 3340 1 57 4418 2 17 4850 277 4972 3 37 4996 3 97 5000 7 38 1480 98 3365 158 4430 218 4854 278 4973 3 38 4996 398 5000 39 1517 99 3389 159 4441 219 4857 279 4974 339 4996 399 5000 40 15 54 100 3413 160 4452 220 4861 280 4974 340 4997 400 5000 41 1591 1 01 3438 161 4463 2 21 4864 281 4975 341 4997 i 42 1828 1 02 3461 162 4474 2 22 4868 2 82 4978 342 4997 43 1664 103 3485 163 4484 223 4871 283 4977 343 4997 I 44 1700 104 3508 164 4495 224 4874 284 4977 344 4997 I 13 0390 45 1736 105 3531 165 4505 225 4878 285 4978 345 4997 46 1772 1 06 3554 166 4515 2 26 4881 2 86 4979 346 4997 47 1808 107 3577 1 67 4525 2 27 4884 2 87 4979 3 47 4997 46 1844 1 08 3599 168 4535 2 28 4887 2 88 4980 3 48 4997 gt 49 1879 109 3621 189 4545 229 4890 289 4981 349 4999 i 50 1915 110 3643 170 4554 2 30 4893 290 4981 3 50 4998 51 quotquot quot1950 111 3665 171 4564 2 31 4896 291 4982 351 4998 52 1985 1 12 3688 172 4573 232 4898 292 4982 352 4998 53 2019 113 3708 173 4582 2 33 4901 293 4983 353 4998 54 2054 114 3729 1 74 4591 2 34 4904 2 94 4984 3 54 4998 55 2088 115 3749 175 4599 235 4906 295 4984 355 4998 56 2123 116 3770 1 76 4808 2 36 4909 296 4985 356 4998 57 2157 1 17 3790 177 4816 237 4911 297 4985 3 57 4998 58 2190 118 3810 178 4625 238 4913 298 4968 358 4998 59 2224 119 3890 179 4833 239 4918 299 4986 359 4998 N 2 Vquot 294 39 5 9 1 9 1M X 39 4 M lt 7 39iquotquot i E139 3 N 39 4 A V 44 577 1 01 98 1 2 113lt1lt1 11 1 I V 6 A i 39P 7quot 7 1 BXOMmMaV TABLEII VALUES OF ta FOR GIVEN PROBABILITY VALUES m J Area 0 Degrees of Probability of a value larger than t a Freedom l 9 95 32911005 1 9005 84 1 3 6 31 12 71 3182 63 66 637 1g 2 1 2 92 4 30 696 9 92 316 3 1 2 35 3 18 454 5 84 129 2 4 1 2 13 278 375 4 60 861 5 148 2 01 257 336 4 03 686 h 6 i 144 1 94 a 45 314 371 596 7 142 1 90 2 36 300 350 540 8 140 186 231 290 336 504 9 1 38 183 2 26 282 325 478 10 1 37 181 2 23 276 3 17 459 1 11 3 1 36 1 80 220 2 72 311 444 12 g 136 1 78 218 2 68 306 432 13 1 35 177 216 2 65 301 422 14 z 1 34 176 214 2 62 298 414 15 i 1 34 175 213 2 60 2 95 407 3916 I 1 34 175 212 2 58 292 402 17 1 33 174 211 2 57 290 396 18 39 1 33 1 73 210 2 55 2 88 392 i 19 1 33 173 209 2 54 2 86 388 x 20 1 132 173 209 2 53 2 84 385 21 a 132 172 2 08 252 283 382 3 22 39 132 172 207 251 282 379 23 j 1 32 171 2 07 250 281 377 24 132 171 2 06 249 280 374 I 25 g 132 171 2 06 248 279 372 l 26 I 132 171 2 06 248 278 371 27 131 1 70 205 247 277 369 28 131 1 70 2 05 247 276 367 29 131 1 70 2 04 246 276 366 30 131 1 70 2 04 246 275 3 65 40 I 1 30 1 68 202 2 42 2 70 3 55 r 60 I 1 30 1 67 200 239 2 66 3 46 Propagation of Error Summary Functional Relation Propagation Rule R CMD 0R C39oM R M1eM2 0R2 cle2chz2 R M1 M2 me 6M26M22 R LnM UR6M R expM 6ROM V hw d me 6h26w2lt6d2 Table 5 5 The above expressions assume that C and D are constants M M1 and M2 are measure ments with error 0 M R is the result with error 0 R and the relative error is 6 g 6R0RR SQ Q ai 40


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