Answers Problem Set 2 Chem 314A Williamsen Spring 2000

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1 Answers Problem Set Chem 314A Wllamsen Sprng 000 1) Gve me the followng crtcal values from the statstcal tables. a) z-statstc,-sded test, 99.7% confdence lmt ±3 b) t-statstc (Case I), 1-sded test, 95% confdence lmt, upper lmt, 7 measurements c) t-statstc (Case I), -sded test, 99% confdence lmt, both lmts, 15 measurements ±.9768 d) F-statstc, 1-sded test, 95% confdence lmt, upper lmt, 5 measurements numerator, 8 measurements denomnator 4.1 e) F-statstc, -sded test, 95% confdence lmt, 5 measurements numerator, 7 measurements denomnator 6.3 ) Assume that you are performng a quanttatve analyss where the uncertanty n the ntensty s %RSD. In makng the calbraton curve, you make a stock soluton by weghng out g of manganese dssolved n a lttle acd and water n a 1.00-L volumetrc flask. You then dlute ths stock soluton by a factor of 100 usng a 1.00-mL volumetrc ppette and a mL volumetrc flask to make a standard soluton. Usng ths data, compare the uncertanty of the standard soluton to the uncertanty of the ntensty measurement. Is there a problem? <*Note*: To answer ths queston, assume that the manufacturng tolerances of the glassware are the uncertanty n the volume measurements and the uncertanty of the balance s ±1 dgt n the last decmal place. I wll put a catalog and the Coyne book on reserve to help you fnd the tolerances. Perform a propagaton of error analyss.> s g g g g + s Uncertanty of balance ±1 n last decmal place. 0.3mL ml or 0.6%RSD Uncertanty n molecular mass (nsde cover of tetbook). Uncertanty of Class A volumetrc flasks (Coyne p. 103) Uncertanty of Class A volumetrc ppettes (Coyne p. 109) ml ml ml ml Yes, there s a problem. One of the assumptons of the formulaton of least squares analyss gven n most programs s that the uncertanty n the -values s much less than the uncertanty n the y-values. Ths s not the case here. As mentoned n the pre-laboratory lectures, the uncertanty n the concentraton gven could be greatly dmnshed f the glassware was calbrated or decreasng the number of steps used. 1

2 3) You are performng a quanttatve analyss to determne the concentraton of Ca n a sample of seawater. After determnng that the other consttuents of the seawater wll not nterfere wth the analyss, you make a calbraton curve from ntensty data measured whle analyzng samples of known Ca concentraton. Then you analyze your sea water sample. After analyzng the results, you determne that the uncertanty about your unknown concentraton determnaton s too large. a) State 4 ways n whch you could decrease the uncertanty. Measure at more standard concentratons Make more measurements of the unknown Improve your technque to reduce s y Have more standard concentratons near the ends of the calbraton curve Pck your standard concentratons so that the unknown wll be near the center of the calbraton lne b) Wrte the mportant equaton(s) that helped you n determnng how you can decrease the uncertanty and show how your changes affect the values n ths equaton. 1 a t 1 n N + 1 n + (y s y ) b ( ) s y s b y ( y y ˆ ) n Measurng more standard concentratons (n) wll decrease 1 n, wll decrease t, and wll decrease s y ( ) More standard concentratons near the edges of the calbraton curve wll ncrease the More measurements of the unknown wll decrease 1 N Pckng standard concentratons so the unknown wll fall near the center of the calbraton curve wll decrease y s y ( ) Improve your technque to reduce s y to reduce ( ) y ˆ y 4) One area of consderaton when performng a hypothess test s the settng of α- and β-error. a) What are α- and β-errors? Descrbe ths n words and n pctures. α-error s the chance of rejectng the null hypothess when the null hypothess s true, and β-error s the chance of acceptng the null hypothess when the research hypothess s true. Values: accept null hypothess Values: accept research hypothess β α

3 b) How s the β-error set? The β-error s ndrectly set by the α-error (whch s drectly set) and by the number of measurements. 5) When determnng the correct t-value to use n calculatng the confdence lmt of a number of measurements, only one degree of freedom s lost, whle when performng the calculaton to determne the correct t-value to use n calculatng the confdence lmt about a regresson lne, two degrees of freedom are lost. Why? One degree of freedom s lost when calculatng the confdence lmt about a measurement because the sample mean s calculated. Two degrees of freedom are lost about the regresson lne are lost because two statstcs, the y-ntercept and the slope, are calculated. 6) Most of the data sets that you wll acqure durng your lfetme wll only be a sample of a populaton. Although practcal reasons often lmt you to small number of ponts due to tme constrants, statstcal theory tells us that under most condtons t s better to acqure more ponts. State why we would want to obtan more data n each of the followng cases. Use dagrams or epressons to support your reasonng, f applcable. a) Your data has a Posson dstrbuton, but you would lke to analyze your data by performng an F-test. One of the assumptons of the F-statstc s that the data s obtaned from a normal dstrbuton. The F- statstc s not very robust so even data that s obtaned from an almost normal dstrbuton s suspect. Although the data on whch you wsh to work follows the Posson dstrbuton, you can stll use the normal dstrbuton statstcs f the sample s large enough (n>30). The central lmt theorem tells us that no matter what the underlyng dstrbuton s, the data becomes more normally dstrbuted as more data s taken. b) You are performng a hypothess test and wsh to reduce both the α- and β-error smultaneously. One part of a hypothess test s to set the α-level, whch also sets the β-level. Wthout takng more data, decreasng the β -level wll ncrease the α -level and vce-versa. Takng more data wll narrow the dstrbutons (decrease uncertanty) and therefore lmt the values wthn the dstrbuton that are outsde the set threshold(s). c) You are performng a hypothess test usng the t-statstc and wsh to accept the research hypothess at the 95% confdence lmt. The research hypothess s accepted f the calculated t-value s larger than the absolute value of the crtcal t-value. As the sample sze ncreases, the crtcal t-value decreases. Also as more data s acqured, the calculated t-value ncreases as t s proportonal to n. Therefore, f there really s a bas, the calculated t-value wll eventually become larger than the crtcal t-value. d) You can calculate the sample mean and wsh to decrease the confdence lmt of your estmate of the populaton mean. ts The confdence lmt of an estmate of the mean s gven as. Therefore, as the number of data n ponts (n) ncreases, the confdence lmts and uncertanty decrease. 7) Statstcs for Analytcal Chemsts p. 80 Problem 6. Answer p Queston: Is ths spectroscopc method accurate? Statstc: t-test (Case I) Level of sgnfcance: 95% confdence lmt Hypotheses (General form): H 0 : µ certfed spectroscopc H 1 : µ certfed spectroscopc 3

4 Epermental t-values: Sample Number t-statstc Crtcal Value: t(0.95,7)±.36 Decson: Because all of the epermental t-statstcs are wthn the crtcal lmts, accept the null hypothess. The spectroscopcally obtaned values dffer from the calbrated values by random varaton at the 95% confdence lmt wth 7 degrees of freedom. Therefore, the spectroscopc method appears to accurately determne the value. 8) Statstcs for Analytcal Chemsts p. 80 Problem 7. Answer Chapter 3 p You would gve the complete hypothess test, I wll only gve you the mportant values. a) F-test (compare varances) Epermental F: F11 Crtcal value: F(0.95,7,7)3.787 <1-sded test because the queston asks whether the longer bolng tme has greater varaton> The decson would be that the longer bolng pont's varaton s greater than can be eplaned by random varaton at the 95% confdence lmt wth 8 measurements. b) t-test (Case II) (compare means) Epermental t: t1.8 Crtcal value: t(0.95,14) )±.14 <Note: error n crtcal value n tet; -sded test because the queston does not ask for a "sde" nor s there anythng nherent n the data that suggests that one sde of the dstrbuton s mpossble to be consdered. The decson would be that there s no statstcally sgnfcant dfference between the means at the 95% confdence lmt wth 7 degrees of freedom. 9) Mermet et al. Analytcal Chemstry p 733 Problem 16. Hypotheses: H 0 : 5.65%5.60% H 1 : 5.65% 5.60% Decson: Use t-test (Case I) t1.56 Crtcal value: t(0.95,4).776 The null hypothess s accepted at the 95% confdence level wth 4 degrees of freedom. Statements: The statements that are true are that there s no evdence of bas, that there s a 95% probablty that the null hypothess s true, and that the null hypothess can not be rejected (wth the evdence at hand). 10) Mermet et al. Analytcal Chemstry p 733 Problem 17. Test: t-test (Case II) Level of sgnfcance: 95% Hypotheses: H 0 : Jaffe on par H 1 : Jaffe on par Epermental Value: t0.57 Crtcal value: t(0.95,18).101 Decson: At the 95% confdence lmt wth 18 degrees of freedom, the null hypothess s accepted so there s no evdence of bas under these condtons. 4

5 11) Mermet et al. Analytcal Chemstry p 733 Problem 19. Hypotheses: H 0 : s A s B H 1 : s A s B F6.13 Because the calculated F-value s larger than the crtcal F-value, reject the null hypothess. The only statement that s true s that the hypothess of method A beng less precse than method B can not be rejected. You dd not drectly test ths hypothess (that would be a 1-sded test), but because the varances were found to be dfferent, ths hypothess s not ruled out. 1) Mermet et al. Analytcal Chemstry p 733 Problem 0. Bullet 1: Test: F-test Level of sgnfcance: 95% confdence lmt Hypotheses: H 0 : s feld s laboratory H 1 : s feld >s laboratory Eperment F: 14.0 Decson: Because the epermental F-value s greater than the crtcal value, reject the null hypothess. Therefore at the 95% confdence level wth 7 degrees of freedom for each measurement, the precson of the laboratory method s sgnfcantly greater than the feld method. Bullet : Test: t-test (Case II) Level of sgnfcance: 95% confdence lmt Hypotheses: H 0 : H 1 : Epermental t: 4.5 Crtcal t: t(0.95,14).145 Decson: Because the epermental t-value s greater than the crtcal t-value, reject the null hypothess. Therefore at the 95% level of sgnfcance wth 14 degrees of freedom, the means of the two methods do dffer sgnfcantly. Last Revson: 4 March 000 EJW 5

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