Statistics Problem Set - modified July 25, _. d Q w. i n

Transcription

1 Statitic Problem Set - modified July 5, 04 x i x i i x i _ x x _ t d Q w

2 F x x t pooled calculated pooled. f d x x t calculated / /.. f d Kow cocept of Gauia Curve Sytematic Error Idetermiate Error t-tet q-tet Propagatio of Ucertaity

3 The awer are attached to thi documet. ] What i the molar cocetratio of 0.78 % (w/v) NaCl(aq) (MW = 58.4 g/mol). ] How may milliliter of M HCl are required to make 00.0 ml of 5.00 mm HCl? (5 poit) 3] The ma of 37.% (m/m) HCl(aq) (d =.9 g/ml, MWHCl = 36.46) required to make.00 L of.00 M HCl i 3 4] A 00.0 ml ample wa diluted to.00 L. A ubequet aalyi revealed that the cocetratio of aalyte revealed i the diluted ample wa M. What i the cocetratio of thi aalyte i the origial udiluted ample? 4 5] The cocetrated HCl(aq) i 37.% (m/m) HCl(aq) (d =.9 g/ml, MWHCl = 36.46). What i the molarity of thi olutio? 5 6] What volume of M KSO4 i required to make 50-mL of M olutio? 6 7] What i the volume of 0.33 M HCl(aq) required to make a olutio of mL of M HCl(aq)? 7 8] The Fe cotet of a meteorite wa foud to be: 8.6%.9% 3.0%.7%.5% Fid the 50% ad 95% cofidece limit for the Fe aalyi. Decribe i word what the cofidece iterval cocept mea. 9] Decide whether ay of the followig value ca be rejected at the 90% cofidece level: 9 0.7, 0.4, 0.95, 0., 0., 0.3 0] Expre the awer for the followig operatio the proper umber of igificat digit ad abolute ucertaity: ( )/(7. 0.6) = 0 ] Expre the followig operatio i the correct umber of igificat figure.e e-4 = ] What i the propagated error from the followig operatio?

4 (3.56e-6 ± e-8) (9.3e-6 ± 4e-9) = 3.8e- ± 3] You have obtaied the followig value for the aalyi of Cu i a ore ample..53%.47%.5%.99%.49%.54% Uig valid tatitical method how how oe of the value ca be rejected. 3 4] The aalyi of phophate i fertilizer wa made uig a reliable method. Seve meauremet were coducted. The mea value of phophate i the ample i.7 mg/g with a tadard deviatio, of 0.7. Expre the ample cocetratio (with ucertaity) aumig a 95% cofidece level. 4 5] Uig the iformatio from problem 5 etimate the chace that the true mea will be.0 mg/g or greater. Hit you will eed to calculate z for thi oe, ad thik about the light bulb example from lecture. 5 6] A blood ample wa et to two differet lab for choleterol aalyi. The reult are: Lab x = mg/dl = = 0 Lab x = 33 mg/dl = 4 = 0 Are the two tadard deviatio differet igificatly differet at the 95% cofidece limit? 6 7] You have carefully followed a aalytical procedure with = 6 ad foud a mea of 6.37 mm with = Meawhile, Joe Cutcorer ued a modified procedure with = 4, x = 6.87 mm with = 0.. Aumig that the tadard deviatio are ot tatitically differet from each other, doe Joe method have a ytematic error, i.e. tatitically differet at the 95% cofidece limit? 7 8] Expre the awer for the followig calculatio the proper umber of igificat figure: (.77± ±0.05) = 8 9] Stadard deviatio i expreio of 9 a) preciio b) backgroud c) eitivity d) accuracy e) dyamic rage 0] The aalyi of M (m/m) wa coducted o a Martia rock ample. The followig value were obtaied: % 4.8% 5.% 4.9% 5.8% 4.99% Which of the value ca be dicarded?

5 ] The aalyi of Pb i a drikig water wa repeated 6 time ad yielded a mea of 0.45 ppm with a tadard deviatio of 0.0 ppm. What are the limit for the cocetratio aumig a 95% cofidece level? ] The Rope-A-Dope fihig lie compay guaratee that their Jaw-Max ylo lie will haul i at leat a 80 lb gilled moter. Their chief tatiticia, Myro Kumber ha 00 ample of the Jaw-Max lie teted ad fid that the mea weight for lie breakage i 0 lb with a tadard deviatio of 60 lb. What are the chace that the hooked 80 pouder will get away if you were uig Jaw-Max ad ed up beig aother fihig tory? 3] Expre the awer for the followig calculatio the proper umber of igificat figure: (5 poit) (.75cm ±0.03 cm 4.8cm ±0.05 cm) = 3 4] What i the 95% cofidece iterval for 5 meauremet whoe average i 3.44 ad with a tadard deviatio of 0.04? 4 5] What i the relative populatio that lie above the value of 55. for a Gauia ditributio whoe mea i 33.8 ad with a tadard deviatio of.8? 5 6] Which of the followig value may be dicarded with 90% cofidece? ] Which of the followig data poit ca be dicard baed o oud tatitical priciple? 7.550,.493,.5766,.537,.69 8] The method of leat quare fit a lie (L) to a et of x,y data by 8 a) maximizig (xi - xl) b) miimizig (xi - xl) c) miimizig (yi - yl) d) maximizig (yi - yl) e) miimizig (yi - yl) 9] Detectio limit of ay itrumetal method i defied a 9 30] Sketch a plot of a calibratio curve. Label the axe ad the followig: 30 a) backgroud b) dyamic rage c) eitivity 3] Expre the awer for the followig calculatio the proper umber of igificat figure: 3

6 (.75cm ±0.03 cm 4.8cm ±0.05 cm) = 9] Part of the labelig of a cla A pipet i the letter TD. What doe thi mea? 3 a) The correct liquid delivery proce hould have etire cotet of the pipet hould be blow out with the pipet blub. b) The pipet hould be acid wahed betwee uage. c) The pipet i defective ad oly emi-quatitative d) The pipet i coated with a iert aget. e) The olutio delivery proce will leave behid a mall amout of liquid i the tip. 33] After roudig the correct repreetatio of the followig i = 34] The relative error for each meaured value i the followig operatio are repreeted below. What i the relative error i the i fial calculated value? 34 (. ± 0.%) (7. ± 0.3%) 35] The tadard deviatio for the mea value of the followig i , 36.8, ] A method for the determiatio of iro i drikig water wa coducted 5 time ad mea value wa foud to be.7 ppm with a tadard deviatio of 0.7 ppm. What i the 95% cofidece iterval that expree likelihood that true mea lie withi the calculated oe? 36 37] The detectio limit of ay itrumetal method i bet expreed a 37

7 Sigal c e a d b 0 Cocetratio 38] Which of the labeled feature i the curve above bet repreet the oliear regio? 38 39] Which of the labeled feature i the curve above bet repreet the liear rage? 39 40] Which of the labeled feature i the curve above bet repreet the backgroud? 40 4] Preciio ca i bet decribed a which of the followig? 4 a) cotrol b) q-tet c) accuracy d) mea e) reproducibility 4] Which of the followig value may be dicarded baed o oud tatitical priciple? ] The cla average for the 005 UI Chem 53 cla o the America Chemical Society fial exam wa.4 out of 60 correct with a tadard deviatio of 8.0. The atioal average i 9.5/60 with a tadard deviatio of 6.3. Aumig that Gauia tatitic are oberved for the UI grade ditributio, what i the percetage of UI tudet corig above the atioal average? 43 44] The aalyi for Ni i a meteorite ample wa coducted by two differet method yieldig the followig: 44 x tadard deviatio

8 Method A Method B If pooled i 0.49 do the method yield the ame reult to the 95% cofidece limit? 45] The molality of a olutio of HX i.56. What i the molarity of that olutio of the deity i meaured a.33 g/ml ad the MW of the olute i 88. g/mol? 45 46] Two method of aalye were compared. Method A had a mea of 3. with a tadard deviatio of 4.4. Method B had a mea of 4. with a tadard deviatio of 4.8. Both et of meauremet were doe 6 time. What i the F ratio ad are the tadard deviatio igificatly differet from each other at the 95% cofidece level? 46 a).9, o b).9, ye c) 0.840, o d) 0.840, ye e).09, o 47] What i the molar cocetratio of 0.65% (w/w) KCl (74.55 g/mol)? Deity =.00 g/ml 47 Awer 0.3 M 5.0 ml 3.00L*.00 mol/l*36.46g/mol*/0.37 = 97 g 4.00e-3 M *.00/0.000 =.00e- M 5. M L * mol/l * L / mol = 5.6 ml 7 80-mL 8 Firt calculate x =.5% The tadard dev. = 0.4 For = 5 or df= 4 fid the 50% cofidece limit i table 4-, 0.74 Plug ito the t-tet equatio above

9 _ t 0.74(0.4) x % 5 Thi calculatio tell u that there i a 50% chace that the true mea, lie withi the iterval.5% 0.% Now fid the 95% C.L. = 5, d.f. = 4, 8.60 _ t.776(0.4) x % Note that the C.L. iterval wide. 9 d Q w = 6 ue Table 4-6 Qtable = 0.56 Q > Qtable o the poit ca be dicarded e e-3 =.74e-3 ue leat umber of decimal poit For / i et = (%e + %e ) / %e = 00*(e-8/3.56e-6) = 0.6% %e = 00*(4e-9/9.3e-6) = 0.04% %et = (0.6% %) / = 0.6% 0.6% of 3.8e- = e-3 3 Q = / = Qtable = 0.56 Q > Qtable o it ca be rejected 4.7(.447*0.7/7 / )=.70.6mg/g 5 z=(.0-.7/0.7)=.8 Table 4- z=.8;area=.4974 chace = = 0.006, 0.6% chace 6 F=4 / =.6 F-Table = 3.8 o they are ot tatitically differet from each other 7 Spooled=(0.37 *5+0. *3/6+4-) / =0.33

10 t=( /0.33)(6*4/6+4) / =.4 ttable@95%=.306 o they are differet from each other ± Preciio 0 Q=(5.8-5.)/( ) = 0.57 df =5 Qtable = 0.56 < 0.57 the umber ca be rejected /- (.57*0.0/6 / ) = / ppm z = [x x-bar]/ = [80 0]/60 = z 0.7 area = 0.58 Area from 0 to 80 i = 0.4 4% 3 (.75 cm ±0.03 cm 4.8cm ±0.05 cm) = (.75 cm ±.% 4.8 cm ±.%) =.77 ± (.% +,% ) / =.8 cm ±.6% =.8 cm ± 0. cm 4 ±.776 (0.04)/(5) / 5 z = ( )/.8.80 ue table 4- area = above = % 6 Q = / = 0.75 Qtable = 0.64 for = 5 o 9.77 ca be dicarded 7.550,.493,.5766,.537,.69 Q = / = Q = / = {Q-table for = 5} = 0.64 all poit mut be retaied. 8 Awer c 9 yd.l. = yblak + 3 where = td.dev. ee dicuio i the text. See alo Figure 5- (8 th ed.) For a calibratio curve mi. detect. Coc. = 3/m where m = lope

11 30 ee book ad lecture ote 3 (.75 cm ±0.03 cm 4.8cm ±0.05 cm) = (.75 cm ±.% 4.8 cm ±.%) =.77 ± (.% +,% ) / =.8 cm ±.6% =.8 cm ± 0. cm 3 The olutio delivery proce will leave behid a mall amout of liquid i the tip e 0.% 0.3% % 35 Mea value = (xi x-bar) Σ _ 36 t.776(0.7) x ppm 5 37 See yd.l. = yblak c 39 b 40 d 4 e Q calc Qtable > Qcalc o we caot dicard ay of the data

12 43 firt calculate z z = x-u/ i thi cae = 8.0, x = 9.5 & u =.4 z = [9.5-.4]/8.0 = (5 poit) from table 4- z = 0.4 correpod to a relative area of 0.554, thi would mea that fractio of UI tudet corig above the atioal average i or about 65% 44 Firt calculate F value Fcalc = 0.97 /0.0 = 3. for = 0.97 d.f. =3 for = 0.0 d.f. = 5 Ftable = 5.4 F-table > F-calc Calculate t by t calculated x x pooled d.f = 6+4- = * ue table 4- t-table =.306 therefore t-table > t-calc, we are ure the two reult are the ame with the 95% c.l mol * 88. g/mol HX = 37.5 g.56 mol HX/37.5g ol *.33 g/ml * 000 ml/l =.84 mol/l 46 A 47 Aume g of olutio * g = g KCl g KCl * (mol / g) * ( / L) = 8.7e- M