Statitic - Lyig without iig? "Lie, damed lie, ad tatitic" 954 Statitic - Lyig without iig? I North Dakota, 54 Millio Beer Bottle by the ide of the Road April 0 00 South Dakota' Pierre Capital Joural report (Mar. ) that "a average of 650 beer ca ad bottle are toed per mile of road aually." The tatitic i attributed to Dei W. Brezia, a activit agait drukdrivig. But how did he come up with hi data? Accordig to the Joural, Brezia traveled "highway acro the atio to determie whether the problem he perceived wa widepread. He made two trip to South Dakota, oe i 998 ad aother i 000." He couted "ca ad bottle i ditche i May of both year" ad claimed to have foud a average of "oe beer ca or bottle every 6 feet whe walkig radomly elected tretche of ditch." But the math appear a little blurry. The web ite of the South Dakota Departmet of Traportatio claim that the tate "ha 83,47 mile of highway, road ad treet." Aumig Brezia' etimate i correct, South Dakota appear to be world-cla litterbug, toig aide approximately 54,56,800 bottle or ca every year. Accordig to the Ceu Bureau there are 754,844 people i South Dakota. So, accordig to Brezia, the average reidet throw at leat 7 beer bottle or ca o the ide of the road every year. For more Check out www.stats.org
Statitic for Quatitative Aalyi Statitic: Set of mathematical tool ued to decribe ad make judgmet about data Type of tatitic we will talk about i thi cla ha importat aumptio aociated with it: Experimetal variatio i the populatio from which ample are draw ha a ormal (Gauia, bell-haped) ditributio. Normal ditributio Ifiite member of group: populatio Characterize populatio by takig ample The larger the umber of ample, the cloer the ditributio become to ormal Equatio of ormal ditributio: ( x) / y e
Normal ditributio Etimate of mea value of populatio = Etimate of mea value of ample = Mea = x x i x i Normal ditributio Degree of catter (meaure of cetral tedecy) of populatio i quatified by calculatig the tadard deviatio Std. dev. of populatio = Std. dev. of ample = ( x i i x) Characterize ample by calculatig x 3
Stadard deviatio ad the ormal ditributio Stadard deviatio defie the hape of the ormal ditributio (particularly width) Larger td. dev. more catter about the mea, wore preciio. Smaller td. dev. mea le catter about the mea, better preciio. Stadard deviatio ad the ormal ditributio There i a well-defied relatiohip betwee the td. dev. of a populatio ad the ormal ditributio of the populatio. (May alo coider thee percetage of area uder the curve) Amout of Data Stadard deviatio 68 % 95 % 99.7 % Total % of the data covered by ditributio 4
Example of mea ad tadard deviatio calculatio Coider Cu data: 5.3, 5.79, 6., 5.88, 6.0 M x = 5.86 M 5.8 M = 0.368 M 0.3 6 M Awer: 5.8 ± 0.3 6 M or 5.8 ± 0.4 M Lear how to ue the tatitical fuctio o your calculator. Do thi example by loghad calculatio oce, ad alo by calculator to verify that you ll get exactly the ame awer. The ue your calculator for all future calculatio. Lear to ue your calculator tatitical fuctio to calculate mea ad tadard deviatio. You ll ave yourelf a lot of work. http://www.willamette.edu/~mjaeba/help/ti-85-tat.htm http://www.ohloe.edu/people/jocoell/ti/ 5
Relative tadard deviatio (rd) or coefficiet of variatio (CV) rd or CV = x 00 From previou example, rd = (0.3 6 M/5.8 M) 00 = 6. % or 6% Stadard error Tell u that tadard deviatio of et of ample hould decreae if we take more meauremet Stadard error = x Take twice a may meauremet, decreae by.4 Take 4x a may meauremet, decreae by 4 There are everal quatitative way to determie the ample ize required to achieve a deired preciio for variou tatitical applicatio. Ca coult tatitic textbook for further iformatio; e.g. J.H. Zar, Biotatitical Aalyi 6
Variace Ued i may other tatitical calculatio ad tet Variace = From previou example, = 0.3 6 = (0.3 6 ) = 0. 9 (ot rouded becaue it i uually ued i further calculatio) Aother way to expre degree of catter or ucertaity i data. Not a tatitically meaigful a tadard deviatio, but ueful for mall ample. Uig previou data: Average deviatio d i ( x x ) 5.35.8 5.795.8 6.5.8 5.885.8 6.05.8 d 5 d.5 0. or 0. M 0 5 Awer 5.8 0. M or 5.8 0. M : 5 i 7
Relative average deviatio (RAD) d RAD 00 x d RAD 000 x Uig previou data, ( a percetage ) ( a part per thouad, ppt) RAD = (0. 5 /5.8 ) 00 = 4. or 4% RAD = (0. 5 /5.8 ) 000 = 4 ppt 4. x 0 or 4 x 0 ppt ( 0 / 00 ) Some ueful tatitical tet To characterize or make judgmet about data Tet that ue the Studet t ditributio Cofidece iterval Comparig a meaured reult with a kow value Comparig replicate meauremet (compario of mea of two et of data) 8
From D.C. Harri (003) Quatitative Chemical Aalyi, 6 th Ed. Cofidece iterval Quatifie how far the true mea () lie from the meaured mea, x. Ue the mea ad tadard deviatio of the ample. x where t i from the t-table ad = umber of meauremet. Degree of freedom (df) = - for the CI. t 9
Example of calculatig a cofidece iterval Coider meauremet of diolved Ti i a tadard eawater (NASS-3): Data:.34,.5,.8,.8,.33,.65,.48 M DF = = 7 = 6 x =.3 4 M or.3 M = 0. 7 or 0. M 95% cofidece iterval t (df=6,95%) =.447 CI 95 =.3 ± 0.6 or.3 ± 0. M 50% cofidece iterval t (df=6,50%) = 0.78 CI 50 =.3 ± 0.05 M x t Iterpretig the cofidece iterval For a 95% CI, there i a 95% probability that the true mea () lie betwee the rage.3 ± 0. M, or betwee. ad.5 M For a 50% CI, there i a 50% probability that the true mea lie betwee the rage.3 ± 0.05 M, or betwee.5 ad.35 M Note that CI will decreae a i icreaed Ueful for characterizig data that are regularly obtaied; e.g., quality aurace, quality cotrol 0
frequecy Nitrate Cocetratio (g/ml) Trial Trial Trial 3 Trial 4 Trial 5 Trial 6 Trial 7 Trial 8 Trial 9 Trial 0 0.5 0.5 0.5 0.5 0.5 0.49 0.5 0.53 0.5 0.47 0.5 0.5 0.53 0.48 0.49 0.5 0.5 0.49 0.49 0.5 0.49 0.48 0.46 0.49 0.49 0.48 0.49 0.49 0.5 0.47 0.5 0.5 0.5 0.48 0.5 0.47 0.5 0.5 0.49 0.48 0.5 0.5 0.5 0.53 0.5 0.5 0.5 0.5 0.5 0.5 0.506 0.504 0.50 0.496 0.50 0.49 0.506 0.504 0.5 0.486 mea average 0.4998 tdev 0.0647 mg/ml frequecy 0.53 3 0.5 5 0.5 3 0.5 0 0.49 0 0.48 5 0.47 3 0.46 Let Graph the Data! itrate cocetratio 4 outlier 0 8 6 ± 4 ± 0 0.44 0.46 0.48 0.5 0.5 0.54 g/ml
Cofidece Iterval Exercie x t m t Calculate the 95, 98 ad 99 % cofidece iterval For the itrate cocetratio data 95 % 0.500 ± 0.005 98 % 0.500 ± 0.006 99 % 0.500 ± 0.006 50 % 0.500 ± 0.00 0.500 ± 0.006 0.500± 0.006 0.500 ± 0.005 0.500 ± 0.00
Tetig a Hypothei (Sigificace Tet) Carry out meauremet o a accurately kow tadard. Experimetal value i differet from the true value. I the differece due to a ytematic error (bia) i the method - or imply to radom error? Aume that there i o bia (NULL HYPOTHESIS), ad calculate the probability that the experimetal error i due to radom error. Figure how (A) the curve for the true value ( A = t ) ad (B) the experimetal curve ( B ) Comparig a meaured reult with a kow value Kow value would typically be a certified value from a tadard referece material (SRM) Aother applicatio of the t tatitic t calc Will compare t calc to tabulated value of t at appropriate df ad CL. df = - for thi tet kow value x 3
Comparig a meaured reult with a kow value--example Diolved Fe aalyi verified uig NASS-3 eawater SRM Certified value = 5.85 M Experimetal reult: 5.7 6 ± 0. 7 M ( = 0) kow value x 5.85 5.76 t calc 0.7 (Keep 3 decimal place for compario to table.) 0.674 Compare to t table ; df = 0 - = 9, 95% CL t table(df=9,95% CL) =.6 If t calc < t table, reult are ot igificatly differet at the 95% CL. If t calc t table, reult are igificatly differet at the 95% CL. For thi example, t calc < t tet, o experimetal reult are ot igificatly differet at the 95% CL. THE NULL HYPOTHESIS IS MAINTAINED ad o BIAS at the 95 % cofidece level. Comparig replicate meauremet or comparig mea of two et of data Aother applicatio of the t tatitic Example: Give the ame ample aalyzed by two differet method, do the two method give the ame reult? t calc pooled x x Will compare t calc to tabulated value of t at appropriate df ad CL. df = + for thi tet pooled ( ) ( ) 4
Comparig replicate meauremet or comparig mea of two et of data example Determiatio of ickel i ewage ludge uig two differet method Method : Atomic aborptio pectrocopy Data: 3.9, 4.0, 3.86, 3.99 mg/g Ewww! Method : Spectrophotometry Data: 3.5, 3.77, 3.49, 3.59 mg/g x = 3.94 5 mg/g = 0.07 3 mg/g x = 4 = 3.5 9 mg/g = 0. mg/g = 4 Comparig replicate meauremet or comparig mea of two et of data example pooled t calc ( ) ( ) (0.073) (4) (0. ) 4 4 (4) x x pooled 3.945 3.59 (4)(4) 5.056 0.0993 4 4 0.0993 Note: Keep 3 decimal place to compare to t table. Compare to t table at df = 4 + 4 = 6 ad 95% CL. t table(df=6,95% CL) =.447 If t calc t table, reult are ot igificatly differet at the 95%. CL. If t calc t table, reult are igificatly differet at the 95% CL. Sice t calc (5.056) t table (.447), reult from the two method are igificatly differet at the 95% CL. 5
Comparig replicate meauremet or comparig mea of two et of data Wait a miute! There i a importat aumptio aociated with thi t-tet: It i aumed that the tadard deviatio (i.e., the preciio) of the two et of data beig compared are ot igificatly differet. How do you tet to ee if the two td. dev. are differet? How do you compare two et of data whoe td. dev. are igificatly differet? t-tet ad the Law Clearly, the meaig of.083 ± 0.007 ad.0 ± 0.4 are very differet. A a pero who will either derive or ue aalytical reult, you hould be aware of thi warig publihed i a report etitled Priciple of Evirometal Aalyi : Aalytical chemit mut alway emphaize to the public that the igle mot importat characteritic of ay reult obtaied from oe or more aalytical meauremet i a adequate tatemet of it ucertaity iterval. Lawyer uually attempt to dipee with ucertaity ad try to obtai uequivocal tatemet: therefore, a ucertaity iterval mut be defied i cae ivolvig litigatio ad or eforcemet proceedig. Otherwie, a value of.00 without a pecified ucertaity, for example may be view a legally exceedig a permiible level of. L. K. Keith, W. Crummett, J. Deega Jr., R. A. Libby, J. K. Taylor, ad G. Wetler, Aalytical Chemitry, 55, 0 (983). 6
F-tet to compare tadard deviatio Ued to determie if td. dev. are igificatly differet before applicatio of t-tet to compare replicate meauremet or compare mea of two et of data Alo ued a a imple geeral tet to compare the preciio (a meaured by the td. dev.) of two et of data Ue F ditributio F-tet to compare tadard deviatio Will compute F calc ad compare to F table. F calc where DF = - ad - for thi tet. Chooe cofidece level (95% i a typical CL). 7
From D.C. Harri (003) Quatitative Chemical Aalyi, 6th Ed. F-tet to compare tadard deviatio From previou example: Let = 0. ad = 0.07 3 F calc Note: Keep or 3 decimal place to compare with F table. Compare F calc to F table at df = ( -, -) = 3,3 ad 95% CL. If F calc F table, td. dev. are ot igificatly differet at 95% CL. If F calc F table, td. dev. are igificatly differet at 95% CL. F table(df=3,3;95% CL) = 9.8 (0. ) (0.07 ) 3.70 Sice F calc (.70) < F table (9.8), td. dev. of the two et of data are ot igificatly differet at the 95% CL. (Preciio are imilar.) 8
9 Comparig replicate meauremet or comparig mea of two et of data- reviited The ue of the t-tet for comparig mea wa jutified for the previou example becaue we howed that tadard deviatio of the two et of data were ot igificatly differet. If the F-tet how that td. dev. of two et of data are igificatly differet ad you eed to compare the mea, ue a differet verio of the t-tet Comparig replicate meauremet or comparig mea from two et of data whe td. dev. are igificatly differet ) / ( ) / ( ) / / ( / / DF x x t calc
Flowchart for comparig mea of two et of data or replicate meauremet Ue F-tet to ee if td. dev. of the et of data are igificatly differet or ot Std. dev. are igificatly differet Std. dev. are ot igificatly differet Ue the d verio of the t-tet () Ue the t verio of the t-tet (ee previou, fully worked-out example) Oe lat commet o the F-tet Note that the F-tet ca be ued to imply tet whether or ot two et of data have tatitically imilar preciio or ot. Ca ue to awer a quetio uch a: Do method oe ad method two provide imilar preciio for the aalyi of the ame aalyte? 0
Outlier Dirupt the Mea Jauary 0 999 Statitic i the New I 984, accordig to Larry Goick ad Woollcott Smith, the Uiverity of Virgiia aouced that the mea tartig alary of it graduate from the Departmet of Rhetoric ad Commuicatio wa a very hefty $55,000 per year. But before you abado your computer ciece traiig for peech clae, you hould kow that the graduatig cla cotaied a igificat "outlier," or extreme data poit ot typical of the ret of the data et - Ralph Sampo, future NBA All- Star, who majored i peech. It would have bee better to lear the media alary, the data poit i the middle of the et. Evaluatig quetioable data poit uig the Q-tet Need a way to tet quetioable data poit (outlier) i a ubiaed way. Q-tet i a commo method to do thi. Require 4 or more data poit to apply. Calculate Q calc ad compare to Q table Q calc = gap/rage Gap = (differece betwee quetioable data pt. ad it earet eighbor) Rage = (larget data poit mallet data poit)
Evaluatig quetioable data poit uig the Q-tet--example Coider et of data; Cu value i ewage ample: 9.5, 0.7, 3., 9.7, 0.3, 9.99 mg/l Arrage data i icreaig or decreaig order: 9.5, 9.7, 9.99, 0.3, 0.7, 3. The quetioable data poit (outlier) i 3. gap (3. 0.7) Calculate Q calc 0.670 rage (3. 9.5) Compare Q calc to Q table for obervatio ad deired CL (90% or 95% i typical). It i deirable to keep -3 decimal place i Q calc o judgmet from table ca be made. Q table (=6,90% CL) = 0.56 From G.D. Chritia (994) Aalytical Chemitry, 5 th Ed.
Evaluatig quetioable data poit uig the Q-tet--example If Q calc < Q table, do ot reject quetioable data poit at tated CL. If Q calc Q table, reject quetioable data poit at tated CL. From previou example, Q calc (0.670) > Q table (0.56), o reject data poit at 90% CL. Subequet calculatio (e.g., mea ad tadard deviatio) hould the exclude the rejected poit. Mea ad td. dev. of remaiig data: 0.0 4 0.4 7 mg/l Q or G outlier tet? G calc quetioable _ value x reject if G calc > G table G (95 % cofidece) Number of Obervatio.463 4.67 5.8 6.938 7.03 8. 9.76 0.34.85.409 5.557 0 Q calc gap rage reject if Q calc > Q table Q (90 % cofidece) Number of Obervatio 0.76 4 0.64 5 0.56 6 0.5 7 0.47 8 0.44 9 0.4 0 3
No. of obervatio 90% 95% 99% cofidecelevel Rejectio of outlier recommeded if Q calc > Q table for the deired cofidece level. Note:. 3 0.94 0.970 0.994 4 0.765 0.89 0.96 5 0.64 0.70 0.8 6 0.560 0.65 0.740 7 0.507 0.568 0.680 8 0.468 0.56 0.634 9 0.437 0.493 0.598 0 0.4 0.466 0.568 The higher the cofidece level, the le likely i rejectio to be recommeded.. Rejectio of outlier ca have a marked effect o mea ad tadard deviatio, ep. whe there are oly a few data poit. Alway try to obtai more data. Q Tet for Rejectio of Outlier Q calc The followig value were obtaied for the cocetratio of itrite io i a ample of river water: 0.403, 0.40, 0.40, 0.380 mg/l. Should the lat readig be rejected? 0.380 0.40 (0.40 0.380) 0.7 But Q table = 0.89 (at 95% level) for 4 value Therefore, Q calc < Q table, ad we caot reject the upect value. Suppoe 3 further meauremet take, givig total value of: 0.403, 0.40, 0.40, 0.380, 0.400, 0.43, 0.4 mg/l. Should 0.380 till be retaied? Q calc 0.380 0.400 (0.43 0.380) 0.606 But Q table = 0.568 (at 95% level) for 7 value Therefore, Q calc > Q table, ad rejectio of 0.380 i recommeded. But ote that 5 time i 00 it will be wrog to reject thi upect value! Alo ote that if 0.380 i retaied, = 0.0 mg/l, but if it i rejected, = 0.0056 mg/l, i.e. preciio appear to be twice a good, jut by rejectig oe value. 4