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and Statistics Analytical Chemistry
Random noise is part of any analytical measurement. Precise and Accurate answers demand a statistical analysis of the data. 1. How much analyte is present? 2. Is one test protocol better than another? 3. Are the results from this lab the same as from another? 4. Are the current results consistent with previous results?
CHEM*3440
Chemical Instrumentation
Topic 2
Statistics for Analytical Methods
What You Should Know
Calculate Mean Standard Deviation Work well with a spreadsheet program Find the slope and intercept of a linear least squares t to a collection of data points with a spreadsheet program
What We Want To Learn
Answer How much analyte is present? Calibration Curve Standard Addition Internal Standard Calculate correct condence limits. Understand when to use each experiment. Recognize when experimental data is awed due to instrument failure.
Linear Least Squares Fits
What is meant by linear? What is meant by least squares? How do we determine the goodness-of-t? What is the variance-covariance matrix? How do we derive error limits or condence intervals for interpolated data? What are the dangers of extrapolation?
Linear
Several meanings 1. Polynomial where the highest power of the independent variable is 1. 2. In a multivariable function, each term depends on only one of the variables. 3. In a multiparameter function, each term depends on only one of the parameters In our case, it is the last description that is applicable.
Linear cont 1
Typical equation is y = mx + b This is linear in polynomial power, variable, and parameters. Can also perform Linear Least Squares analysis on y = ax3 + bx2 + cx + d It is still linear in variables and parameters. Cannot perform Linear Least Squares on y = ae-bx since its is not linear in parameters.
Least Squares
Given a set of data points, nd equation for linear model (always linear in parameters, often linear in polynomial power) that minimizes the error between the points and the line.
20 15 10 5 0 0 3.75 7.50 11.25 15.00
Least Squares cont 1
Should minimize error in both x and y. Demands iterative solution. Instead, we assume error in x is negligible and ascribe the error to the dependent variable y.
Matrix Formulation of Least Squares
Measured data come in a set of ordered pairs (xi, yi). Fit the data to a selected model. Can t to any linear function, but most common to t to a straight line: y = mx + b. Assume error is all in y (the dependent variable). For each xi there is a measured value, yi, and a t value yi,t = mxi + b. The difference between the measured yi and the tted yi,t is the error for the point, ei. Strategy 1: Choose m and b so that the sum of the ei"s is 0. (Won"t work.) Strategy 2: Choose m and b so that the sum of squares of errors, ei2, is a minimum. (Works!) Write out the expression for the sum of the square of the errors, differentiate with respect to m and b, set derivatives to 0, leads to the following matrix equation:
Error in y.
Error in x and y.
A
f
i i 2 i
p=
y
i
/y n /x b / x / x p dmn = e / x y o
i i
Matrix cont 1
Task is to nd the parameter vector p. Invert the matrix and multiply both sides.
Variance of the Fit
How well does the data follow the model equation (a straight line)? The sum of the squares of the errors divided by the degrees of freedom.
b 1 / x 2 - / x i / yi i d n = p = A- 1 y = D e oe m / x i yi o - / xi n w h e r e D = d e t (A) = n / x
2 i
- _ / xi i
2
s2fit =
/ ^y - y
i
o
i , fit
h2
=
/ ^ y - mx - b h
i i
2
n-2
Form the sums, multiply out the two equations to obtain the two parameters for the straight line, b (the y-intercept) and m (the slope).
Degrees of freedom are the number of data points minus the number of parameters determined (in this case, 2). ! = n - p R2, the correlation coefcient is NOT a meaningful goodness-of-t measure. Use variance of t as evaluation.
An excellent article presents this material: C. Salter, J. Chem. Ed. 77, 1239-43 (2000)
Variance-Covariance Matrix
Answers what is the variance in the t parameters (m and b). Quanties the correlation between the parameters.
Functions of the Fit Parameters
Usually, determining m and b is just the start of an analysis, not the end. Different experiments will use m and b differently to derive a property of interest. Some function F(b,m) will provide the desired result (often the concentration of an unknown). Complete analysis demands that one not only provide the unknown concentration, but one must also assign condence limits to the result. Common mistake: We have the errors in b and m from the diagonal elements of the variance-covariance matrix. The function F uses b and m. Well-known error propagation rules can give a variance to the result. But this ignores covariance, which can either over- or underestimate the true error.
s2fit / x2 - / xi i V = s2fit A- 1 = D e o - / xi n
Diagonal elements are the variances of the parameters. Off-diagonal elements are the covariance between the parameters.
s2 = V11 = b
s2fit / x2 i D
s2fit n s2 = V22 = D m
Errors in Functions of the Fit Parameters
Correct error calculation. The variance in the function F(b,m) is given by
Example: Interpolation x(y)
Calibration curve y(x). x could be analyte concentration, y detector response. Consider case of the curve being a rst-order polynomial. For a selected ysel value, the corresponding interpolated x value, xint, is given by inverting the LSQ-t equation of the calibration curve.
s / x - / xi o d F s2 = d T Vd F = D d T e F F F - / xi n
2 fit 2 i
The elements of the vector dF are the partial differentials of F with respect to the t parameters. The vector dFT is just the transpose of dF.
y = mx + b
(
x int =
y sel - b m
2b p dF = f 2F 2m
This is the key equation in error analysis.
2F
Find partial derivative vector for this function F(b,m) = (ysel - b)/m.
y -b y se b 2 a selm k 2 a ml - m k 2F 1 = = =- m 2b 2b 2b y -b 2 a selm k ^ y sel - b h 2F = =m2 2m 2m J N 1 2F O 2b p = K - m ` dF = f K ^ y sel - b h O 2F KO m2 2m L P
Example: Interpolation x(y) cont1
Find the variance in the interpolated result s2xint. Solve the matrix equation.
Three Analytical Experiments
The three most important analytical experiments are: 1. Calibration Curve Used when experimental conditions are well controlled, in a laboratory or factory environment where the solution containing the analyte is known and controlled. 2. Internal Standard Used for experiments where the detection volume and efciency is difcult to control and reproduce. Chromatography and mass spectrometry are common examples. 3. Standard Addition Used when sample matrix is complex with unknown interferences. Environmental samples are of this nature. Find the unknown concentration and determine the variance in the result so that condence limits can be assigned to the answer.
s2x
int
J 1N s2fit y sel - b m / x2 - / xi K - m O 1 i T = d F Vd F = D c- m - m2 e o K ysel - b O - / xi n KO m2 P L
Remember the rules for matrix multiplication. Solve to obtain
s2fit n ^ y sel - b h2 y -b s2x = D m2 c - 2 selm / xi + / x2 m i m2
int
Recall that the interpolated x-value is xint = (ysel -b)/m so that
s2fit s2x = D m2 _ nx2nt - 2x int / xi + / x2 i i i
int
This expression includes the effect of correlation between the t parameters.
Calibration Curve - Experiment
Form a series of solutions of known concentration of the analyte. Choose a concentration range to include that expected for any unknown. Measure their response with a particular instrument. Each solution measured once or several times; overall there are n measurements to which a straight line will be t. Use LS algorithm to nd the parameters b and m of the line. Compare variance of t with that expected for this instrument. Graph the data to make sure it is linear throughout the range studied. Measure unknown several times (p times).
Calibration Curve - Variance
Slight variation to variance equation; measurement of y introduces additional source of error. Variance is square of error. Variance in xcc arising from the measurement of y is
2x 2 c CC m s2y, meas 2y
The variance of the measured average is
s2y, meas ^y =
/ - y h
i
2
p-1
The derivative has the value 1/m. The expression for the overall variance in the result xcc and the solution is
xCC
^ y - bh = m
2xC 2 T s2x = c 2yC m s2y, meas + dCC VdCC s2fit 1 1 2 = m2 ' p + D _ nxCC - 2xCC / xi + / x2 i 1 i
CC
Internal Standard - Experiment
Internal standard needs to be chemically similar to analyte but distinguishable by analytical technique. Best if an atomic isomer. Make a series of solutions with a known concentration of both analyte and internal standard. Keep standard concentration relatively constant while varying analyte concentration throughout the expected unknown concentration range. Measure the response of both analyte and internal standard in each solution. Form a response curve of the ratio of analyte to internal standard response vs. analyte concentration. This least squares response curve is then treated identically to the calibration curve.
Standard Addition - Experiment
Standard is same molecule as analyte. Two common approaches: 1. Form a series of solutions with same amount of unknown but increasing amounts of standard added. Solutions then are all diluted to the same volume. (Constant volume experiment.) 2. A single unknown solution is measured and then increasing amounts of standard are added, the signal being measured for each addition. Each measurement has a different solution volume, but only one container needs to be used. (Variable volume experiment.)
A good reference for this is M. Bader, J. Chem. Ed. 57, 703-706 (1980).
Constant Volume
n separate containers. Put volume Vx of unknown (whose as yet unknown concentration is Cx) in each. Add volume of standard Vs(2) whose concentration is Cs to second container (leave the rst container with only unknown in it so that Vs(1) = 0). Add a larger volume Vs(3) to third, etc. Dilute all solutions with solvent to an identical nal volume, Vf. Measure response Rn of each solution. Determine linear response, least squares t parameters b and m. 1. nx = VxCx 2. nx = VxCx 3. nx = VxCx 4. nx = VxCx etc. ns(1) = 0 ns(2) = Vs(2)Cs ns(3) = Vs(3)Cs ns(4) = Vs(4)Cs ntotal = nx ntotal = nx + ns(2) ntotal = nx + ns(3) ntotal = nx + ns(4) Canalyte(1) = nx/Vf Canalyte(2) = (nx + ns(2))/Vf Canalyte(3) = (nx + ns(3))/Vf Canalyte(4) = (nx + ns(4))/Vf
Constant Volume cont1
Instrument responds linearly to analyte concentration. k represents the responsivity of instrument.
nt n n Vx C Vs C R = kC analyte = k ; Votal E = k ;V x + V s E = k ; V x + V s E f f f f f kV C kV C kC kV C R = Vx x + Vs s = c V s m Vs + c Vx x m = mx + b = y f f f f kV C kC ` b = Vx x m= V s f f
Solve two expressions for k, equate, and solve for unknown concentration Cx.
bV f k =V C and x x bV f mV f `V C = C x x s bC s ` C x = mV x
mV f k= C s
Constant Volume cont2
What is the variance in this result? Find the partial differentials with respect to b and m, solve the matrix equation. Recall that x is the volume of added standard and y is the instrument response for that particular solution.
Constant Volume cont 3
Expand this matrix equation to nd the variance in the extrapolated result.
2C x 2 c bC s m Cs = = mV 2b 2b mV x x 2C x 2 c bC s m bC s = = - m2 V 2m 2 m mV x x
We know the variance of the t from the least squares routine that determines b and m. Recall the matrix equation for the variation of the result.
s
2 x, ext
J Cs N s2fit Cs / x2i - / xi oK mV x O bC s = D c mV - m 2 V m e K O x x - / xi n K- b2C s O L m Vx P
J Cs N s2fit Cs / x2i - / xi oK mV x O bC s s = D c mV - m 2 V m e K O x x - / xi n K- b2C s O m Vx P L J Cs N bC s 2 K mV / x 2 + m 2 V / x i O i s fit C s bC s x x = D c mV - m 2 V m K O x x K - C s / x i + nbC s O 2 mV x m Vx P L s2fit C2 bC2 bC2 nb 2 C 2 s s s 2 = D ' m2 V2 / xi + m3 V2 / xi + m3 V2 / xi - m 4 V2s 1
2 x, ext x x x x
s2fit C2 2bC2 nb 2 C 2 = m2 D 'V2s / x2 + mV2s / xi - m2 V2s 1 i x x x
We now know the unknown concentration and the variance in that value, so an answer with condence limits can be given.
Variable Volume
This experiment only requires 1 container. Introduce volume Vx of unknown (with, as yet, unknown concentration Cx). Measure instrument response. Add a volume Vs(2) of standard of concentration Cs. Remeasure the response. Add more standard and remeasure.
Variable Volume cont1
The response of the ith addition of standard is
Vs (i) Cs Vx C x Ri = kC analyte (i) = k <V + V + V + V F x s (i) x s (i)
Ri is dependent variable; Vs(i) is independent variable. Cannot rewrite this equation to linearize the variables. Non-linear problem. Can sort of linearize the problem.
1. nx = VxCx 2. nx = VxCx 3. nx = VxCx 4. nx = VxCx etc.
ns(1) = 0 ns(2) = Vs(2)Cs ns(3) = Vs(3)Cs ns(4) = Vs(4)Cs
ntotal = nx ntotal = nx + ns(2) ntotal = nx + ns(3) ntotal = nx + ns(4)
Canalyte(1) = nx/Vx Canalyte(2) = (nx + ns(2))/(Vx + Vs(2)) Canalyte(3) = (nx + ns(3))/(Vx + Vs(3)) Canalyte(4) = (nx + ns(4))/(Vx + Vs(2))
Ri ^V x + Vs (i) h = k 6V x C x + Vs (i) Cs @ = kV x C x + kC s Vs (i)
Plot Ri(Vx + Vs(i)) against Vs(i). Such a plot would give a straight line with intercept b = kVxCx and slope m = kCs. With y(i) = Ri(Vx + Vs(i)) and x(i) = Vs(i) a LSQ analysis gives a value for m and b. Solve for unknown concentration Cx.
b = kV x C x b k =V C x x b m `V C = C x x s
and and so that
m = kC s m k=C s bC C x = mVs x
Variable Volume cont2
Variance-Covariance matrix arises from the linear problem. Can use it as if it were linear. Variance for b and m are found from the diagonal elements. Can nd an approximate variance for the result; neglect covariance.
2 sC = s2 + s2 b m
x
Variable Volume cont 3
Vs (i) Cs Vs (i) Cs Vs (i) Cs Vx C x Vx C R i = k < V + V + V + V F , k ; V x + V E = k ;C x + V E x s (i) x s (i) x x x
under the assumption that Vx >>Vs(i)
Can provide an answer with condence limits. BUT, don"t forget the many approximations (errors?) made in reaching the result.
or
Vx C n n R i = k ; V x + V i E = k ;C x + V i E x x x
Another approximate approach: perform experiment where the added standard volumes are very small compared to the unknown volume. A concentrated standard would allow small added volumes or, if possible, standard could be added as a solid. Problem becomes linear in these approximations.
where in this case we are adding moles of material directly (presumably by weighing methods). You should be able to easily see how these two equations are linear and can be solved to nd the unknown concentration and its variance, accurate, again within our approximations used. You should be able to prove that the solutions for Cx are, respectively
bC C x = mVs x
or
b C x = mV x
Standard Deviation
Variance is the squared error parameter which sums between separate sources of error which are normally distributed. The standard deviation is the square root of the variance and has the same units as the property being measured.
0.45 0.4
Confidence Interval
Must decide upon a condence level (e.g. 95%) Know the standard deviation s. Number of measurements is important; more measurements makes us more condent of the result. Find the Student"s t-factor (degrees of freedom and condence level). Gaussian Curve
Normal Error Curve
1 S.D.
68.3%
0.35 0.3
2 S.D.
95.5%
0.25 0.2 0.15
C.I. =!
ts n
3 S.D.
99.7%
0.1 0.05 0 -4 -3 -2 -1 0 1 2 3 4
Students t-Table
Degrees of Freedom 1 2 3 4 5 6 7 10 60 50% 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.700 0.679 0.674 90% 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.812 1.671 1.645 95% 12.706 4.303 3.182 2.776 2.571 2.447 2.365 2.228 2.000 1.960 99% 63.656 9.925 5.841 4.604 4.032 3.707 3.500 3.169 2.660 2.576 99.9% 636.578 31.598 12.924 8.610 6.869 5.959
Reporting a Result
From a series of measurements, nd the average value, xaverage. Use the matrix formulation for errors to nd the variance in the result, while accounting for covariance between the t parameters. The square root of the variance (s2) is the standard deviation (s). Select a condence level (C.L.). Find the appropriate Student"s t-value, given the C.L. and the degrees of freedom (usually n-1). Calculate the condence interval.
5.408 4.587 3.460 3.291
Answer = xaverage !
ts n
#

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/z to6tf1,);nx 1C.>tf )tLq -q -=(-,-JASaJ At4r,o?3tFrom Section 6.7 Exercises; Precalculus by Coburn, 2nd Edition5'l+ 7-g".-tyc-.>f x : Z 2 |',nXcoS/ : + Lv-zL+ztl\<.? t/, + flLv+O xrt', a./_*+T,Lz-l"T'rfu- +LTILi7-cfw_2,L11,5.( a-7

Kent State - MATH - 3450:149

jtc-JAtu-7.S:,^ t?'9-,lo^ 3oo laALAr,tOFrom Section 7.1 Exercises; Precalculus by Coburn, 2nd Editiona)bl1- t;.92DZi3= Cs.-(=- ( ta,4 | b2,.4);46.7 : L*_.-_r'bx"\'76.Lt8,1 Srnll" L.tt7-7,",3;^C4 s.64z.uLl &.< si,^ tq "4e,z

Kent State - MATH - 3450:149

qJAS'phtran,olo(atcL-UtFrom Section 7.2 Exercises; Precalculus by Coburn, 2nd EditionLcfw_t c*,,+Lu-.LU)&)r-tn-( s) ( 6)- s' ) ("' -4Lv-'7"q- + q"5- z(3 7 1.5) c.-9 7e"/2c-.-t"-zQ,"71"- u(gt)r3P"(' ,< S-9in 5\5?.>: \r-o IO,1'1

Kent State - MATH - 3450:149

cfw_o-/"1 .>"/'t1rt'16.)?.g.6-! c u!.*rs ?. t l+k) #'5 vcfw_/\q, 'z.f , 1,br ?/, ,t/ , urcfw_.\ 3X 1y =!'l /27/t'From Section 8.1 Exercises; ri,fS "/ ,"/, G/, ez,-/, ,/Edition Precalculus by Coburn, 2nd(z z)0<_ ,G ltt -rt i,t. -t?i-/)

Kent State - MATH - 3450:149

/46lp-2,",1)J"-r"t*, p.cfw_"^-o?l LFrom Section 8.2 Exercises; Precalculus by Coburn, 2nd Editionz-u-Z:*!I -:31cfw_ -z!t): -[d+y-'-.?.2,':-tr=y-)*3-e*-..cfw_.(,:-)= *r7)23--3(t-) rz(0-l:-?*,t.?,-v- -z,-1Y,t). -- _1-*:-t:.:.t- -.-,

Kent State - MATH - 3450:149

From Section 8.3 Exercises; Precalculus by Coburn, 2nd Editionfoq,;- !n.o?/8VZ?4 /z/tcU+c)Q-'z) cfw_+3/--zIiAi:?-fui1(*:ul -*lstuL'* j-7*lcfw_- L1J)lt' rL l+z d-ApVz!il]--/-g'-'(r.-71(r't"s*+7/ -b(t-")st , t?/t - jx '7tx'- ?x

Kent State - MATH - 3450:149

.A-e,-'From Section 10.1 Exercises; Precalculus by Coburn, 2nd Editioni-l'i,*','f,.Q.'^teL;4,z-:.JLi-f-i-J'/.,.l.n.J:-g:o,)t5 (esLr-II Ii iIl itII !ajlA^ - (a^-t41, - (z - t?_cfw_za^-(a+7T+\> llL\q6n-'cfw_rtuIa56l1't,

Kent State - MATH - 3450:149

l()/'l-.r' l/Jlo!yer-C*Jcr"p.uiI cfw_ion*, 4*oA.)From Section 10.2 Exercises; Precalculus by Coburn, 2nd Editionf-l1ta;/_cfw_.'t-3 l, il-b,!:8-.r?-2,,L ')L7 1 -, bO UD 4I c q-'.&."g.$.t'c117mn- Jl'an',.kL-t6- ,cfw_ l0*\ + l] , tL+ (

Kent State - MATH - 3450:149

l9 lc)10. aJa,wnFrom Section 10.3 Exercises; Precalculus by Coburn, 2nd EditiontGt (- a(4eCLt 4.-b- --g-:-:'b\ ;z t <.+ -zt( 49Aj _- 3 K-33. -Gcfw_:a-^tFl-20, _ 4"t_-= V ?_ ,.-C, \J.54?"9:-6, J1.i, S- ,.5-:-v-1',a,16t LS -Ngit-Gnp

Kent State - MATH - 3450:149

Quiz1l ^'\O :)NuQuiz 1 All work must be shown to receive credit.",JAS,-ALA*oPrecalculus January 15,2010 L. Brubaker2. (2 points) Solve the inequality indicated using a number line. Write yow answer in interval notation.g(x) = -x' +10x -/- -5=*

Kent State - MATH - 3450:149

J boLtcfw_ a z:(s,oa'l<-r r-\ '"eJ"Quiz 2Name:Quiz 2)fi9ouIPrc."1."t*AtauoJanuary 22,2010 L. BrubakerAll work must be shown to receive credit.l. (3 points) This function is not onerto-one. (a) Determine a domain restriction that preseryes all

Kent State - MATH - 3450:149

Quiz3Name:Quiz 3)Auul-l'lr-uP*."1."1."January 29,2010 L. BrubakerAll work must be shown to receive credit.2. (2 points) Find the domainof /(x) = t"r,[=)/*\2O x>-44- < 76-I:_ij t( >v/z\q-xr' zrl/, > -ul>p l*4D,. (- +,\)(l,rcfw_+ -,fr-c

Kent State - MATH - 3450:149

ID':(Quiz4Name:Quiz 4PrecalculusFebruary 5,2010 L. Brubakersteps. Do not use a calculator to do conversion.All work must be shown to receive credit. 2. (2 points) Convert the following. Show ALL a) 45J 45' 45" to decimal degrees.b) 142.207 5Ll t

Kent State - MATH - 3450:149

)ur(z-''t"I/'ii-fQuiz5 I /i-' l!,I'\h, l".aaIName: J A9o ^r ,/t_y'r".,0IQuiz 5PrecalculusFebruary 12,2010 L. Brubaker All work must be shown to receive credit. Do not use a calculator to do the graphing. lra) (2 points) Use a reference

Kent State - MATH - 3450:149

5r^ Oc-otocftita* A; irrauz( Qt-ol oc"et dtilb1in L.o t. (osz6 ". I(^" 09Ac o.,(D(d,tcfw_^e .:SPd.*c" OLa-ro+ t; s<czJ Quiz c c-o( ta )- 6 (9a'PQuiz6,^(*),-t'"dcotcfw_8l,t(-dt - -'ftu*", (-o) - _a_ " EJ45u"t I (A-r,\D Precalcu

Kent State - MATH - 3450:149

QutzTtra 'all; Quiz 7Nu*J /c*t*; ", JA-S aPrecalculusMarch 5,2010 L. BrubakerAll work must be shown to receive credit.' 9',< ( ! B) 7 ,:rT'I'ff:-;"' "[#)'"'(#).'"'(#)"'(#) ; t',':?)lT;T i .,a ! btd 4:rp,A_;l2. (4 points) Givend2g^'"'(.'+ ^ -+

Kent State - MATH - 3450:149

"a ,'r'?,,a2cfw_.t-| ,o.cfw_)GO<-oJ;/ v^1"Quiz8 LT-Name:r' Laq*t-.J Arrr 4 r,t nroPrecalculusQuiz 8%wvrlwvo .ot[.or-March 12,2010 L. BrubakerAll work must be shown to receive credit.n urqrvr' 1. (1 point) Find the exact value of the

Kent State - MATH - 3450:149

Quiz 9q)a.Y 6j,All work must be shown to receive credit.Quiz 9loName:JA fr'-'r Ac4, ls.Precalculus March 26,2010L. Brubakerl;,t,il.'J']ilI"#-t'n"'T,"l"T'*;:)Tl vangleB:45" sidea:32.8km t;* cfw_ (" 2,1 .q' 1i^ 45"Xtriangresexi'l:"il'iithcompret

Kent State - MATH - 3450:149

Quiz 10glo*"*",JAfu^' 4&'44-PPrecalculus April9 ,2010Quiz 10L. BrubakerAll work mustbe shown to receive credit. l. (5 points) Solve using elimination. State if a system is inconsistent or dependent. For systems with linear dependence, write the a