Chapter 5-6 Experimental Error

Chapter 5-6 Expermental Error Contents n Chapter 5-6. Sgnfcant Fgures. Sgnfcant Fgures n Arthmetc 3. Error Systematc Error Random Error 4. Propagaton of Uncertanty from Random Error 5. Propagaton of Uncertanty: Systematc Error 6. Comprehensve Examples

. Sgnfcant Fgures ) Statement of sgnfcant fgures The dgts n a measured quantty, ncludng all dgts known exactly and one dgt (the last fgure) whose quantty s uncertan. The more sgnfcant dgts obtaned, the better the precson of a measurement. The concept of sgnfcant fgures apples only to measurements. The Exact values (e.g., km = 000 m) have an unlmted number of sgnfcant fgures. 3 ) Recordng wth sgnfcant fgures 0 3 Recordng:.4 or.5 or.6 sgnfcant fgure, certan, uncertan 0 3 Recordng:.5 or.5 or.53 3 sgnfcant fgure, certan, uncertan 4

3) Rules for Zeros n Sgnfcant Fgures Zeros between two other sgnfcant dgts are sgnfcant e.g. 003 no. of sg. fg.: 5 A zero precedng a decmal pont s not sgnfcant e.g., 0.003 no. of sg. fg.: 5 Zeros precedng the frst nonzero dgt are not sgnfcant e.g. 0.00003 no. of sg. fg.: 5 Zeros at the end of a number are sgnfcant f they are to the rght of the decmal pont e.g. 0.00300 no. of sg. fg.: 7 5 Zeros at the end of a number may or may not be sgnfcant f the number s wrtten wthout a decmal pont e.g. 9500 no. of sg. fg.: N/A Scentfc notaton s requred: e.g. 9.5x0 4 no. of sg. fg.: 3 e.g. 9.50x0 4 no. of sg. fg.: 4 e.g. 9.500x0 4 no. of sg. fg.: 5 6

. Sgnfcant Fgures n Arthmetc ) Rules for roundng off numbers ) If the dgt mmedately to the rght of the last sg. fg. s more than 5, round up. ) If the dgt mmedately to the rght of the last sg. fg. s less than 5, round down. ) If the dgt mmedately to the rght of the last sg. fg. s 5 followed by nonzero dgts, round up. v) If the dgt mmedately to the rght of the last sg. fg. s 5 round up f the last sg. fg. s odd. round down f the last sg. fg. s even. v) If the resultng number has ambguous zeroes, t should be recorded n scentfc notaton to avod ambguty. 7 Examples: 35.76 n 3 sg. fg. s 35.8 35.74 n 3 sg. fg. s 35.7 4.58 n 3 sg. fg. s 4.3 4.35 n 3 sg. fg. s 4.4 (roundng to even dgt) 4.5 n 3 sg. fg. s 4. (roundng to even dgt) 3,05 n 3 sg. fg. s.3 x 0 4 8

Zero s an even number An even number s a number that s exactly dvsble by. That means that when you dvde by two the remander s zero. Ths one makes me smle. Zero s an even number. All whole numbers and postve and negatve ntegers are ether ODD or EVEN. Just remeber the number lne f you have any doubts.... -3 - - 0 3... Go back to the rules on odd and even. (3rd grade I thnk) O= odd and E= even E + E = E - + = 0 O + O = E - + = 0 O + E = O - + = I hope ths settles ths. 9 After some further dggng, we turned up two other sources that argued that 0 falls nto the realm of even numbers. Dr. Pete of the Math Forum nssts that every nteger must be even or odd, and states, "An nteger 'n' s called *even* f there exsts an nteger m such that n = m...from ths, t s clear that 0 = ()(0) s even." The Straght Dope echoes these fndngs n a column on the subject. 0

) Addton and Subtracton: The reported results should have the same number of decmal places as the number wth the fewest decmal places Example : 49.46 m + 7.3 m 9. m.76 m Ans:. m Example : MW of KrF 8.998403 (F) + 8.998403 (F) + 83.798 (Kr).7948064 Ans:.795 g/mol Note: To avod accumulatng round-off errors n calculatons: ) Go through all calculaton by calculator, then roundng on the fnal answer OR ) Retanng one extra nsgnfcant fgure (a subscrbe dgt) for ntermedate results, then roundng on the fnal answer

Example 3: Wrte the answer wth correct number of dgts:.43x0 6 + 6.5x0 4.7x0 5 =?.43 x 0 6 + 0.065 x 0 6 0.7 x 0 6.3744 x 0 6 Ans:.374x0 6 3 3) Multplcaton and Dvson: The reported results should have no more sgnfcant fgures than the factor wth the fewest sgnfcant fgures Example :.87 m 0.76 m =?.87 m x 0.76 m =.3974 m =.39 m Ans:.39 m Example : (4.379 x 0 )(3.6x0 9 ) =.554444x0 6 =.6x0 6 4

Stochometrc coeffcents n a chemcal formula, and unt converson factors etc, have an nfnte number of sgnfcant fgures. Example: Results of four measurements: 36.4 g, 36.8 g, 36.0 g, 37. g. What s the average? Soluton: (36.4+36.8+36.0+37.)/4 = 36.6 Ans: 36.6 g 5 4) Logarthms and Antlogarthms ) Mathematc terms For exponental expresson, e.g. 4. x 0 : 4. s called the coeffcent( 係數 )), s called the exponent( 指數 ). For logarthm operaton, e.g. log (4. x 0 ) =.38 s called the characterstc(( 對數的 ) 首數 ), 38 s called the mantssa ( 假數,( 對數的 ) 尾數 ). 6

) Logarthms operaton Example : log(56.7 x 0 6 ) =? Soluton: log(5.67 x 0 7 ) = 7 + log(5.67) = 7.754 3 sg. fg. Example : log(0.00735) =? Soluton: log(0.00735) =.5630 4 sg. fg. Mantssa reman equal number of sg. fg. as the orgnal dgt n coeffcent 7 Example 3: Percent transmttance (%T) s related to the absorbance (A) by the equaton: A = log (%T/00). What s correct dgts of A f %T s 7.9. Soluton: A = log (%T/00) = log (0.79) = ( 0.37) = 0.37 Example 4: The ph s defned as ph = log [H 3 O + ]. If the [H 3 O + ] s 3.8 x 0 M, what s the ph of the soluton? Soluton: ph = log [H 3 O + ] = log (3.8 x 0 ) = (.4) =.4 8

) Antlogarthms operaton Example : antlog(.67) =? Soluton: antlog(.67) = 0.67 = 0 0.67 x0 = 4.69 x 0 Example : What s the correct dgts of %T f the absorbance s 0.93? Soluton: Antlog ( 0.93) = 0 0.93 = 0.7 %T = 0.7 x 00% =.7% Fnal dgt has same number of sg. fg. as number of dgts n mantssa 9 Example 3: If the ph s 0.3, what s the [H 3 O + ]? Soluton: [H 3 O + ] = 0 ph = 0 0.3 = 5 x 0 Example 4: If the ph s.5, what s the [H 3 O + ]? Soluton: [H 3 O + ] = 0 ph = 0.5 = 3.0 x 0 3 0

3. Error ) Types of error Systematc (Determnate; 確定的 ) error Errors affectng the accuracy, whch can be detected and corrected. Ths error s always postve or always negatve. Random (Indetermnate, Statstc) error Errors affectng the precson (uncertanty), caused by uncontrolled (maybe uncontrollable) varables. Postve and negatve fluctuaton occur wth approxmate equal frequency. Gross error Caused by careless, lazness, or nepttude, can be rejected as outler by statstc method. ) Precson vs Accuracy Precson The measurement for how closely ndvdual measurement agree wth one another. Accuracy The closeness of the measurements to the expected true value

Accuracy how close a measurement s to the true value Precson how close a set of measurements are to each other accurate & precse precse but not accurate not accurate & not precse 3 3) Categores of systematc error ) Method Errors e.g. The relatonshp between the sgnal and the analyte are nvald e.g. Due to such as nterferents and contamnaton ) Measurement (Instrument) error e.g. Incorrect measurement n the equpment and nstruments. ) Personal (Operaton) error e.g. The bases ntroduced by the analyst. 4

4) Ways to detect systematc error ) Analyss of standard reference materal (SRM) ) Blank analyss, n whch contanng none of analyte ) Use dfferent analytcal methods to measure the same sample. v) Round robn experment (nterlaboratory comparson) 5 5) Accuracy evaluaton Absolute error: Error = x x : means of measured values :expected (reference) value Relatve error: E r x = 00% Recovery x = 00% 6

6) Absolute and percent relatve uncertanty Example: For a buret readng of.350.0 ml Absolute uncertanty = 0.0 ml Percent relatve uncertanty 0.0 ml.35 ml 00% % 7 4. Propagaton of Uncertanty from Random Error 8

) Addton and subtracton Example : The calbraton result of class A 0 ml ppet showed that the marker readng s 9.990.006 ml. When t s used to delver two successve volumes. What s the absolute and percent relatve uncertantes for the total delvered volume. Soluton: Total volume = 9.99 ml + 9.99 ml = 9.984 ml e y (0.006) 0.00848 %e y 9.984 Ans: (0.006) 00% 0.04 0.008 % 48 9.984(0.008) ml (absolute uncertanty) 9.984(4%) ml (percent relatve uncertanty) 9 Example : For a ttraton experment, the ntal readng s 0.05(0.0) ml and fnal readng s 7.88 (0.0) ml. What s the volume delvered? Soluton: Delvered volume = 7.88 ml 0.05 ml = 7.83 ml e y (0.0) (0.0) 0.0 0.0 %e 8 y 00% 0.6% 7.83 Ans: 7.83(0.03) ml (absolute uncertanty) 7.83(0.%) ml (percent relatve uncertanty) 8 30

) Multplcaton and dvson Example : The quantty of charge Q = I x t. When a current of 0.50.0 A passes through the crcut for 0 s, what s the total charge? Soluton: Total charge 0.5 A x 0 s = 8 C 0.0 %ey ( 00) 0.5 6. e 7 y 8. 00 ( 0 00) (6. 7 ) (0.8 3 ) 6. Ans: 8() C (absolute uncertanty) 8(7%) C (percent relatve uncertanty) 7 % 3 3) Mxed operatons [.76( 0.03) 0.59( 0.0)] Example:?.89( 0.0) Soluton: [.76( 0.03) 0.59( 0.0)].89( 0.0) [.7( 0.03 ) 6.89( 0.0).7( 3. %).89(. %) 0.69 0.69 0 0 ( 3. 3 %) ( 0.0 0 ) Ans: 0.6(0.0) (absolute uncertanty) 0.6(3%) (percent relatve uncertanty) ( 0.03) (0.0) 0.03 ) (. ) 3. 3 ( 3. 6 3

4) The real rule for sgnfcant fgures 0.8( 0.00) Example:? 0.803( 0.00) Soluton: 0.8( 0.00) 0.803( 0.00) 0.8( 0. 0.803( 0..0.0 4 4 44 49 ( 0.3 48 ( 0.003.0( 0.004).0( 0.3%) %) %) %) 6 ) The frst uncertan fgure s the last sgnfcant fgure Ans:.0(0.004) (absolute uncertanty).0(0.3%) (percent relatve uncertanty) 33 5) Exponents and logarthms Example: The ph of a soluton s 3.70.03, what s the [H 3 O + ] of ths soluton? Soluton: [H 3 O + ] = 0 ph x ey y 0.306 ex y y = 0 3.7 =.9 x0 4 ey ey.306 0.03 0.07 4 0 y.9 0 e y y (0.07 0 ) (.9 0 4 y(e y ) =.9 (0. 3 )x0 4 y (%e y ) =.9 (6. 8 %)x0 4 )(0.07 Ans:.9 (0.)x0 4 M (absolute uncertanty).9 (7%)x0 4 M (percent relatve uncertanty) 0 ). 3 0 5 34

5. Propagaton of Uncertanty: Systematc Error For systematc error Propagaton of uncertanty = n x (uncertanty n ndvdual entty) Example: Atomc mass of C s.0070.0008 and of H s.007940.00007. What s molecular mass of C H 4? Soluton: C: (.0070.0008)=4.04 0.006 4H: 4(.007940.00007)=4.0376 0.0008 (4.040.006) + (4.0376 0.0008) =8.05360.00 6 = 8.0530.00 Ans: 8.0530.00 g/mol ( 0.006) (0.0008) 0.00 6 Uncertanty come from sotopc varaton, dfferent sources have dfferent AM of C and/or H,.e., systematc error rather than random error However, C and H are ndependent 35 6. Comprehensve examples Example : A uncalbrated 5 ml transfer ppet s used to delver four successve volumes. What s the fnal volume wth absolute uncertanty? (manufacture s tolerance for 5 ml transfer ppet s 0.03 ml) Soluton: 4 x (5.000.03) ml = 00.00 0. ml Ans: 00.0 0. ml Here 0.03 ml s systematc error Example : The marker of a calbrated 5 ml transfer ppet shows 4.990.006 ml s used to delver four successve volumes. What s the fnal volume wth absolute uncertanty? Here 0.06 ml s Soluton: random error (4.990.006)+(4.990.006)+(4.990.006)+(4.990.006) =99.964 0.0 Ans: 99.96 0.0 ml 4 (0.006) 0. 0 36

Example 3: Whch of the followng methods for preparng a 0.000 M soluton from a.0 M stock soluton provdes the smallest overall uncertanty? (a) A one-step dluton usng a ml ppet and a 000 ml volumetrc flask. (b) A two-step dluton usng a 0 ml ppet and a 000 ml volumetrc flask for the frst dluton and a 5 ml ppet and a 500 ml volumetrc flask for the second dluton. Class A tolerance Ppet.0000.006 ml 0.000.03 ml 5.000.03 ml Volumetrc flask 500.00. ml 000.00.3 ml 37 Soluton: ( a) (.0 M)(.000 0.006 ml) 0.000 e (000.0 0.3 ml) % e y 0.006 00.000 y M 0.3 00 000.0 0.6 0 % ( b) (.0 M)(0.00 0.03 ml)(5.00 0.03 ml) 0.000 e (000.0 0.3 ml)(500.0 0. ml) % e y 0. 0.03 0.00 0 % 00 0.03 00 5.00 y M 0.3 00 000.0 0. 00 500.0 Ans: (b) Two-step dluton provdes the small overall uncertanty 38

Chapter 7 Statstcs Contents n Chapter 7. The Nature of Random Errors ) Random Errors and Ther Mathematcal Equatons ) Central Lmt Theorem. Gausan (Normal) Dstrbuton ) Defne Gausan Dstrbuton ) Z Value Transformaton 3) Z value Applcatons 4) Constructng Gaussan Curve From Expermental Data 3. Confdence Interval (CI) 4. Q Test for Outler Rejecton 5. F Test for Comparng Standard Devaton

6. t Test for Decdng Determnate Error 7. The Least Squares and Calbraton Curve ) Method of Least Squares ) Lnear Calbraton Curve 3) Establshng Calbraton Curve 4) Uncertanty n the Regresson Calbraton 5) Example of Applyng Calbraton Curve 6) Unknown Analytcal Result wth Propagaton of Uncertanty 3. The Nature of Random Errors ) Random Errors and Ther Mathematcal Equatons Random (Indetermnate, Statstc) error: Errors affectng the precson (uncertanty), caused by uncontrollable varables. Postve and negatve () fluctuaton occur wth approxmate equal frequency. Populaton ( 母體 ): The set of nfnte objects n the system beng nvestgated. Sample ( 樣品 ): The fnte members of a populaton that we actually collect and analyze. 4

7 The Gaussan Dstrbuton As shown n Fgure 4, neurotransmtters bnd to membrane protens of the muscle cell and open up channels that permt catons to dffuse nto the cell. Of the 9 on channel responses recorded for Fgure 4, 90 are n narrow range from.64 to.68 pa. The bar graph n Fgure 4 s typcal of many laboratory measurements. The smooth, bellshaped curve supermposed on the data n Fgure 4 s called a Gaussan dstrbuton. 5 P.83 Fgure 4 Fgure 7- (a) In the absence of neurotransmtter, the on channel s closed and catons cannot enter the muscle cell. (b) In the presence of neurotransmtter, the channel opens, catons enter the cell, and muscle acton s ntated. 6 P.83

Fgure 4 Fgure 7- Observed caton current passng through ndvdual channels of a frog muscle cell. 7 P.84 Mean and Standard Devaton A Gaussan dstrbuton s characterzed by a mean and a standard devaton. The arthmetc mean, x, also called the average, s the sum of the measured values dvded by the number of measurements. 8 P.84

9 Mean (Average) Standard devaton - For populaton: - For sample: x for sample for populaton, symbol: n x mean ) ( N x x s N N x N ( ) μ x σ, s 0, n If 0 t N k k N j j N pooled N N N N x x x x x x S...... ) ( ) ( ) ( 3 3 3 Pooled standard devaton: For several subsets of data, estmatng standard devaton by poolng (combnng) the data. Varance = s Relatve standard devaton (RSD) (also called coeffcent of varaton, CV): 00% RSD x s

Degree of freedom (dof) In statstcs, the number of ndependent observatons on whch a result s based. - For standard devaton: dof = n-, n s munber of measurement. - For pooled standard devaton: dof = N M, N: Total number of measurements M: Number of the subset. Degrees of Freedom: For a set of data ponts n a gven stuaton (e.g. wth mean or other parameter specfed, or not), degrees of freedom s the mnmal number of values whch should be specfed to determne all the data ponts. For example, f you have a sample of N random values, there are N degrees of freedom (you cannot determne the Nth random value even f you know N- other values). If your data have been obtaned by subtractng the sample mean from each data pont (thus makng the new sample mean equal to zero), there are only N- degrees of freedom. Ths s because f you know N- data ponts, you may fnd the remanng (Nth) pont - t s just the sum of the N- values wth the negatve sgn. Ths s another way of sayng that f you have N data ponts and you know the sample mean, you have N- degrees of freedom.

) Central Lmt Theorem ) Defnton For the probablty dstrbuton plot for the frequency of ndvdual values. The measurements subject to ndetermnate errors arse a normal (Gaussan) dstrbuton. The sources of ndvdual error must be ndependent. The ndvdual error must have smlar magntude (no one source of error domnates the fnal dstrbuton). 3 ) Smulated Central Lmt Theorem (Frequency of Combnatons of four equal-szed uncertanty, u) Combnaton of uncertanty Magntude of Combnaton Frequency of combnatons Relatve frequency +U +U +U 3 +U 4 +4U /6=0.065 -U +U +U 3 +U 4 +U -U +U 3 +U 4 +U 4 4/6=0.50 +U +U -U 3 +U 4 +U +U +U 3 -U 4 -U -U +U 3 +U 4 +U +U -U 3 -U 4 +U -U +U 3 -U 4 0 6 6/6=0.375 -U +U -U 3 +U 4 -U +U +U 3 -U 4 +U -U -U 3 +U 4 +U -U -U 3 -U 4 -U +U -U 3 -U 4 -U 4 4/6=0.50 -U -U +U 3 -U 4 -U -U -U 3 +U 4 -U -U -U 3 -U 4-4U /6=0.065 4

4 random uncertantes 5 0 random uncertantes Infnte number of random uncertantes - A Gaussan (normal) dstrbuton curve. - Symmetrcal about the mean. 6

. Gausan (Normal) Dstrbuton ) Defne Gausan Dstrbuton A bell-shaped probablty dstrbuton curve for measurements showng the effect of random error, whch encountered contnuous dstrbuton. The equaton for Gaussan dstrbuton: yf(x) e (x) x : ndvdual value y : relatve frequency (0 ~ ) : true value : standard devaton e ( x) / dx 7 Case : Same Mean, Dfferent SD y e ( x ) 8

Case : Dfferent Mean, Same SD y e ( x ) 9 Case 3: Dfferent Mean, Dfferent SD y e ( x ) 0

) Z value Transformaton For x-axs: transform x to z by: z x For y-axs: transform to f(z) by : y f ( z) e z / e z dz @008 59/90 Now, the Gaussan Curve Always Consstent z x y e z @008 60/90

教戰守則. 經過 z transform, Gaussan curve 長得一模一樣, 此時 x-axs 的單位為 σ. 看到標示為 x, 單位為 σ 時, 代表已經經過 z transform 3. Relatve frequency (y) at mean (x), 約為 0.4 4. The entre area of the Gaussan curve s. @008 6/90 3) Z value Applcatons ) Area (probablty, percentage) n defned z nterval Area wthn σ Area wthn σ Area wthn 3σ Area wthn 3 3 e π e π e π e π z z z z / / / dz 0.686 / dz 0.9546 dz 0.9973 dz @008 6/90

) Sngle-sded ordnate and area for Gaussan error curve y e z @008 63/90 Example: The amount of asprn n the analgesc tablets from a partcular manufacturer s known to follow a normal dstrbuton, wth µ = 50 mg and σ = 5 mg. In a random samplng of tablets from the producton lne, what percentage are expected to contan between 43 and 6 mg of asprn? @008 64/90

Soluton: z left z rght 43 50 5 6 50 5.4.4.4.4 y(probablty) 0.49 y(probablty) 0.498 0.49 + 0.498 = 0.90 x 00%= 9.0% Ans: 9.0% @008 65/90 ) Calculate the area by spreadsheet 略 96_Ch04_Norm_Dst_ Area_Calculaton.xls @008 66/90

4) Constructng Gaussan Curve From Expermental Data ) General procedure Step : Raw data collecton Step : Arrange the data n order from lowest to hghest Step 3: Condense the data by groupng them nto cells Step 4: Pctoral representaton of the frequency dstrbutons Step 5: Estmatng the σ from s Step 6: Plottng relatve frequency versus x or z @008 67/90 Example: Replcate Data for the Calbraton of a 0 ml Ppet 96_Ch04_Trea t_randdstr_d ata.xls @008 68/90

Cont d 96_Ch04_Trea t_randdstr_d ata.xls @008 69/90 96_Ch04_Trea t_randdstr_d ata.xls @008 70/90

Relatve frequency versus x Number n range y (total ppet)(volume per bar) (50)(0.003) e (0.006) π s π (x 9.98) (0.006) e (x μ) σ Percentage n range (volume per bar) y e s π (0.003) e (0.006) π (x μ) σ (x 9.98) (0.006) 96_Ch04_Trea t_randdstr_d ata.xls X-axs 刻度為體積之間隔 @008 7/90 Matchng Hstogram wth Gaussan curve @008 7/90

Z Transformed Gaussan Curve 96_Ch04_Trea t_randdstr_d ata.xls @008 73/90 3. Confdence Interval (CI) The range of values wthn whch there s a specfed probablty (confdence level, CL) that the true value les. Remark : CL, e.g., 95% CL, same as 00 0.05 若看到. 0.4 8 時 : 藍色數字 : 均為 sgnfcant fgures, 最後一位 uncertan 下標方式紅色數字 : not sgnfcant fgures, 是避免計算誤差用, 當然也是 uncertan @008 74/90

) Confdence nterval nterpretaton µ=0000 and σ=000, 4 member each set, 00 set computer random generate data: x z n 空方塊 46 (error bar nclude the true value), 實方塊 sold 54 空方塊 89 (error bar nclude the true value), 實方塊 @008 75/90 ) Probablty/CI of Gaussan dstrbuton 0.40 CL (probablty) (z, unt: σ) CI (Area le between) Relatve frequency 0.30 0.0 0.0 0.00 68.6% 95.46% 99.74% -3 - - 0 3 z ( x ) 50% ± 0.67 80% ±.9 90% ±.64 95% ±.96 99% ±.58 38.30% ± 0.50 68.6% ±.00 86.64% ±.50 95.46% ±.00 98.76% ±.50 99.74% ± 3.00 99.95% ± 3.50 @008 76/90

) CI applcatons for Populaton Expresson for the Expected value x Stuaton: n > 0, s σ(known, use z values) z n Expresson for the Measured value x z n @008 77/90 Values of z for determnng Confdence nterval Confdence level, % z 50 0.67 68.00 80.9 90.64 95.96 96.00 99.58 99.7 3.00 99.9 3.9 @008 78/90

z Expressons of the expected value ( x ) n Example: What s the 95% confdence nterval for the amount of asprn n a sngle analgesc tablet drawn from a populaton where μ s 50 mg and σ s 5? Soluton: x = µ ±.96σ = 50 ± (.96)(5) = 50 ± 0 Ans: 95% of the tablets n the populaton contan 50 mg ± 0 mg of asprn. @008 79/90 z Expressons of the sngle measured value ( x ) n Example: The populaton standard devaton for the amount of asprn n a batch of analgesc tablets s known to be 7 mg of asprn. A sngle tablet s randomly selected, analyzed, and found to contan 45 mg of asprn. What s the 95% confdence nterval for the populaton mean? Soluton: µ = x ±.96σ = 45 ± (.96)(7) = 45 ± 4 Ans: At 95% confdence level, the populaton s mean, µ, les wthn the range of 45 mg ± 4 mg of asprn. @008 80/90

z x n Example: The populaton standard devaton for the amount of asprn n a batch of analgesc tablets s known to be 7 mg of asprn. Fve tablets are randomly selected, analyzed, and found to contan 45 mg of asprn. What s the 95% confdence nterval for the populaton mean? Soluton: (.96)(7) 45 45 6 5 Ans: At 95% confdence level, the populaton s mean, µ, les wthn the range of 45 mg ± 6 mg of asprn. @008 8/90 v) CI applcatons for Sample*** Expresson for the Measured value x ts n Values of student s t Defnton of t: x t s n : t z, And μ x @008 8/90

Expressons of N measured values and σ s unknown x ts n Example: What s the 95% confdence nterval for the data n followng table? Masses of seven US pennes n crculaton Penny Mass (g) 3.080 3.094 3 3.07 4 3.056 5 3. 6 3.74 7 3.98 @008 83/90 Soluton: x = 3.7 g s = 0.05 t =.45 (95% confdence level) (.45)(0.05) 3.7 3.7 0.047 g 7 Ans: At 95% confdence level, the populaton s mean, μ, les wthn the range of 3. ± 0.05 g. @008 84/90

教戰守則. Large sample use z value (A) Reportng expected value: (B) Reportng measured value: x x z n z n. Small sample use t value Reportng measured value: x ts n @008 85/90 4. Q Test for Outler rejecton 一種統計檢定, 據以判斷被懷疑的數據 ( 離群值 ) 是否應予以刪除 (gross error) Procedure for Q-test (Dean and Dxon, 95) Step : Calculate the range (w = 最大值最小值 ) Step : Calculate questonable result (x q ) 與 the nearest neghbor (x n ) 之間的差值 Step 3: Calculate expermental Q value (Q exp ) xq - xn Step 4: Q exp w 查表找出 crtcal values for rejecton quotent (Q crt ) @008 86/90

Crtcal values for rejecton quotent (Q crt ) Number of Confdence Level Observatons (N) 90% 95% 99% 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 e.g. 95% confdence level same as α= 0.05 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 Step 5: Q exp > Q crt : 刪除 Q exp < Q crt : 保留 @008 87/90 Example: The followng masses, n grams, were recorded n an experment to determne the average mass of a U.S. penny: 3.067, 3.049, 3.039,.54, 3.048, 3.079, 3.094, 3.09, 3.0 Determne f the value of.54 g s an outler at 95% confdence. Soluton: Place the masses n order from smallest to largest:.54, 3.039, 3.048, 3.049, 3.067, 3.079, 3.094, 3.0, 3.09 95% confdence, n = 9, Q crt = 0.493 Q exp > Q crt Ans:.54 s assumed an outler, and can be rejected. @008 88/90

5. F Test for Comparng Standard Devaton 一種統計檢定, 據以判斷兩組數據之標準偏差 ( 精密度 ) 的差異是否顯著 Procedure for F-test Step : Calculate the F exp F exp S S S 較大者作為分子 S (Numerator) S 較小者作為分母 S (Denomnator) 所以 always F exp > Step : 查表找出 crtcal values for F (F crt ) @008 89/90 Crtcal values for F (F crc ) at 95% confdence level (Numerator, 分子 ) (Denomnator, 分母 ) Step 3: F exp > F crt : 95% 信賴度下兩組數據之標準偏差之差異顯著 F exp < F crt : 95% 信賴度下兩組數據之標準偏差之差異不顯著 @008 90/90

Example Degree of freedom σ or s Reference method 0. Method 0.5 Method 0. S (0.) (a) Ref F.96 S (0.5) 分子自由度, 分母自由度, 查表得 F crt =.30 95% 信賴度下 method 與 reference method 的標準偏差之差異不顯著 @008 9/90 S (b) Ref (0.) F 3.06 S (0.) 分子自由度, 分母自由度, 查表得 F crt =.30 95% 信賴度下 method 與 reference method 的標準偏差之差異顯著 (c) S F S (0.5) (0.).56 分子自由度, 分母自由度, 查表得 F crt =.69 95% 信賴度下 method 與 method 的標準偏差之差異不顯著 @008 9/90

6. t Test for Decdng Determnate Error 一種統計檢定, 據以判斷分析數據與真值的信賴區間 (confdence nterval) 分佈的差異是否顯著, 或判斷兩組數據之高斯分佈曲線分佈的差異是否顯著兩組數據具有顯著的差異兩組數據並無顯著的差異兩組數據差異是否顯著? @008 93/90 ) t test methods Classfcaton (A) Comparson an expermental mean wth a true value (B) Comparson of two expermental mean (unpared t- test) 比較兩組數據的差異, 兩組數據來自於兩個分析者或兩種分析方法或兩組樣品兩組數據的 N 值不要求相同 (C) Comparng multple samples (pared t-test) N 個樣品, 分別以方法分析一次, 另以方法分析一次, 據以比較兩種方法是否具有顯著性的差異 @008 94/90

(A) Comparson an expermental mean wth a true value Procedure Step : Step : Calculate the t (exp) x N t(exp) s 查表找出特定信賴度下, 分析數據 dof (N-) 的 t (theo) Step 3: t (exp) > t (theo) : 分析值與真值差異顯著 ( 系統誤差 ) t (exp) < t (theo) : 分析值與真值差異不顯著 @008 95/90 Example: (Chrstan, Example -7) A NIST SRM wth a reference value of.7 ppm of Cu. New method for 5 replcates, obtan a mean of 0.8 ppm and ± 0.7 ppm SD, do the data ndcate bas at 95% confdence level? @008 96/90

Soluton: t (exp) x s N 0.8.7 0.7 5.9 At 95% confdence dof = N- = 5- = 4, t (theo) =.776 t (exp) > t (theo) Ans: at 95% confdence level, 差異顯著, 新方法可能存在系統誤差 @008 97/90 Example: A student check her procedure by analyzng a sample known to contan 98.76% (w/w) Na CO 3. Fve replcate determnatons of the %w/w Na CO 3 n the standard were made wth the followng results: 98.7%, 98.59%, 98.6%, 98.44%, 98.58% Is the mean for these fve trals sgnfcantly dfferent from the accepted value at the 95% confdence level? Soluton: 95% confdence, dof = 5 = 4, t theo =.78 t exp > t theo Ans: At the 95% confdence, ther dfference s sgnfcant (may be affected by a determnate source of error) @008 98/90

(B) Comparson of two expermental mean (unpared t-test) 使用時機 : - 比較兩個分析者之兩組數據的差異 - 比較兩種方法之兩組數據的差異 - 比較兩個樣品之兩組數據的差異需要先比較兩組數據之標準偏差是否具有顯著性的差異 (F-test), 再使用適當的 t-test 檢定 @008 99/90 a) 精密度無顯著差異之 unpared t-test Procedure Step : F-test 精密度無顯著差異 Step : 計算 pooled standard devaton (S p ) S p ( x Set x ) n n Set ( x j x ) = S ( n ) S n n ( n ) Step 3: 計算 t (exp) t (exp) x x S p n n n n @008 00/90

Step 4: 依選定之 confdence level (e.g., 95%) 以及 degree of freedom (= n + n ) 查表取得 t (theo) Step 5: t (exp) > t (theo) : 兩組數據差異顯著 ( 系統誤差 ) t (exp) < t (theo) : 兩組數據差異不顯著 @008 0/90 Example: 以 New method (n= 6) 和 reference method (n=5) 分析同一個樣品分之 Fe(III) 含量結果如下 : New method (%) Reference method (%) 0.0 8.89 0.50 9.0 8.65 9.00 9.5 9.70 9.40 9.40 9.99 - At 95% confdence level,new method 是否與 reference method 具有 sgnfcant dfference? @008 0/90

Soluton: F-test F cal New method Reference method Mean 9.65 9.4 (0.673) (0.34) SD 0.673 0.34 N 6 5 4.3 分子自由度為 6-=5, 分母自由度為 5-=4, 查表得 F crt = 6.6 F cal < F crt, At 95% confdence level, 兩種方法之精密度無顯著差異 @008 03/90 S t p (exp) S x ( n x S p ) S n n n n ( n n n ) = (0.673) 9.65 9.4 = 0.546 (6 ) (0.34) 6 5 6 5.3 (5 ) = 0.546 95% confdence level and dof = n + n = 6+5 = 9 t (theo) =.6 t (exp) < t (theo) Ans: At 95% confdence level, 兩種分析方法之分析結果無顯著差異 @008 04/90

b) 精密度有顯著差異之 unpared t-test Procedure Step : F-test 精密度有顯著差異 Step : 計算 t (exp) t (exp) s x x / n s / n Step 3: 計算自由度 (ν) s / n s / n s / n s / n n n 四捨五入至整數 @008 05/90 Step 4: 依選定之 confdence level (e.g., 95%) 以及計算得到之 degree of freedom (ν) 查表取得 t (theo) Step 5: t (exp) > t (theo) : 兩組數據差異顯著 ( 系統誤差 ) t (exp) < t (theo) : 兩組數據差異不顯著 @008 06/90

Example: The %w/w Na CO 3 n soda ash can be determned by an acd base ttraton. The results obtaned by two analysts are shown here. Determne whether the dfference n ther mean values s sgnfcant at 95% confdence. @008 07/90 Soluton: 分子自由度為 6 =5, 分母自由度為 6 =5 查表, 於 95% confdence level, F crt = 7.5 F exp > F crt, 標準偏差有顯著差異 @008 08/90

dof = 5 and 95% confdence t theo =.57 t exp > t theo Ans: Dfference of the results between two analysts are sgnfcant at the 95% confdence level. @008 09/90 (C) Comparng multple samples (pared t-test) 使用時機 : 不同來源的 N 個樣品, 分別以已知方法分析一次, 並以另一新方法分析一次, 據以比較兩種方法是否具有顯著性的差異 Procedure Step : 計算個別樣品兩種分析結果之差異 : d = A -B Step : 計算 d 之平均值 : d d N Step 3: 計算所有 d 之標準偏差 : S d ( d N d) @008 0/90

Step 4: 計算 t (exp) : t(exp) N Sd Step 5: 依選定之 confdence level (e.g., 95%) 以及 dof (= N-), 查表取得 t (theo) d Step 6: t (exp) > t (theo) : 兩組數據差異顯著 ( 系統誤差 ) t (exp) < t (theo) : 兩組數據差異不顯著 @008 /90 Example: 6 個檢體分別以 new method 和 reference method 分析 blood urea ntrogen (BUN) 之結果如下 : Sample New method, mg/dl Reference Method, mg/dl Dfference, d A 0. 0.5 0.3 B.7.9 0.8 C 8.6 8.7 0. D 7.5 6.9 0.6 E. 0.9 0.3 F.5. 0.4 At 95% confdence level,new method 是否與 reference method 具有 sgnfcant dfference? @008 /90

Soluton d 0.8 S d = 0.4 t (exp) d S d N 0.8 0.4 6.63 At confdence level 95%,dof = N- = 6- = 5 t (theo) =.57 t (exp) < t (theo) Ans: At 95% confdence level, 兩組分析數據無顯著差異 @008 3/90 教戰守則. Evaluatng outler: Q-test. Evaluatng precson: F-test 3. Evaluatng accuracy: t-test (a) Comparson an expermental mean wth a true value (b) Comparson of two expermental mean (unpared t-test) (c) Comparng multple samples (pared t-test) @008 4/90

7. The Least Squares and Calbraton Curve ) Method of Least Squares A mathematcal optmzaton technque that attempts to fnd a "best ft" to a set of data by attemptng to mnmze the total squares of resdual error, between the ftted functon and the data. A lnear least squares for example: Regresson lne (Calbraton curve) ŷ y Resdual error = y ŷ y : Expermental value ŷ : Calculated by y = mx + b Total squares of resdual error = Σ(y ŷ ) Assumptons n method of least squares: error n y >> error n x Uncertantes (standard devaton) n all y values are same @008 5/90 ) Lnear Calbraton Curve The lnear relatonshp between the amount of analyte and a method s measurng sgnal: S meas = kc A + b (.e., y = mx + b) (establshed by the method of least squares) S meas : Sgnal response by sample C A : Analyte concentraton k: Slope between S meas and C A b: y-ntercept of the lnear equaton @008 6/90

3) Establshng Calbraton Curve D n(x ) ( x slope (m) : m y ntercept (b) : b ) n(x (x y ) y Lnear equaton: y = mx + b ) x D (x D y y ) x @008 7/90 Example: Establsh a lnear calbraton curve by followng data: Soluton: x (concentraton) y (sgnal) 3 3 4 4 6 5 @008 8/90

D n(x ) ( x ) n(x y ) x m D b (x ) y (x D 5 y y 0.6538 ) x.3465 Ans: Lnear equaton s y = 0.6538x +.3465 @008 9/90 4) Uncertanty n the Regresson Calbraton s y (SD of y values): s y ( d ) n s m (SD of slope): s m s y D n s b (SD of ntercept): s b s y D ( x ) @008 0/90

Example: Calculate S y, S m, and S b of the prevous example. x y 3 3 Soluton: s y ( d ) n 0.96 4 4 6 5 s m s y D n 0.05439 s b s y ( x D ) 0.45 @008 /90 Usng Excel: 略 96_Ch04_Least_ Square.xls @008 /90

Excel Marco Functon: LINEST () Formula for cell E, then, ENTER 略 96_Ch04_LI NEST_Marco_ Functon.xls () Hghlght cells E3:F5, then, press F @008 3/90 Cont d (3) Press CONTROL + SHIFT + ENTER 略 (4) Wrte Labels for E~F4 @008 4/90

Excel Marco Functon: m, b, r 略 96_Ch04_LI NEST_Marco_ Functon.xls The coeffcent of determnaton (r ): r SS resd N SSresd SS tot SS tot [ y ( y y) ( b mx ] r 越接近, 越好. @008 5/90 5) Example of Applyng Calbraton Curve Followng are results of spectrophotometrc analyss of proten standard: ) Reject datum of (0.39) (t s an outler) @008 6/90

) Reject data of 5.0 μg proten standard (out of the lnear range) @008 7/90 ) 剩下 4 個數據建立 lnear equaton m = 0.063 0 b = 0.004 7 s m = 0.000 s b = 0.00 6 s y = 0.005 9 y = (0.063 0 )x + 0.004 7 @008 8/90

v) 計算 unknown sample 中分析物含量 Example: 已知 calbraton curve: y = (0.063 0 )x + 0.004 7 樣品分析結果 blank 吸收值 0.04,unknown 吸收值 0.406, 樣品中 proten 含量為多少? Soluton: (0.406 0.04) 0.0047 μg of proten 0.0630 8.4 μg @008 9/90 6) Unknown Analytcal Result wth Propagaton of Uncertanty Reportng analytcal result: µ = x ts x Choose t value: usng dof = n-, n s number of standards Calculatng s x by: s m : n : k : x y x : : s y m k absolute value of number of standards number of (y y) n m (x x) the slope replcate measurements of ndvdual value of standard ndvdual value of sgnal responses the unknown x : y : mean of standards mean of sgnal responses @008 30/90

Example: Informaton of calbraton standard followng: Four replcate analyss of an unknown sample gave a mean absorbance of 0.30. What s the 95% confdence of the proten n the sample? @008 3/90 Soluton: s y = 0.005 9, m = 0.063 0, k = 4 s x = 0. 3 At 95% confdence dof (= 4 ) =, t =.79 t x s x = (.79)(0. 3 ) = 0.5 0 0.30 0.04 x 8.4 0.063 Ans: 8. 4 0.5 0 μg 0 μg @008 3/90