CHEM100B_Homework Solutions_Stats Solutions

# CHEM100B_Homework Solutions_Stats Solutions -...

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Skoog/West/Holler/Crouch Appendix 1 Principles of Instrumental Analysis 6th ed. Revision 1.0 February 13, 2008 Page 1 APPENDIX 1 PROBLEM SOLUTIONS a1-1 For A, x i 61.45 61.53 61.32 1 184.30 x = (a) x = 184.30/3 = 61.43; there are 3 degrees of freedom for the mean since three data points are used in the calculation of x (b) () 1 N 1 2 = = x x s N i i = 22 (0.02) + (0.10) + ( 0.11) 2 3 1 = 0.106 0.11. . There are 3 – 1 = 2 degrees of freedom for s . (c) CV = 0.106 × 100%/61.43 = 0.17 % (d) s m = 0.106 = = 0.061 s 3 N The remaining sets are treated in the same manner with the results shown: A B C D (a) x 61.43 3.25 12.10 2.65 df 3 6 2 4 (b) s 0.11 0.02 0.06 0.21 df 2 5 1 3 (c) CV 0.17% 0.60% 0.47% 7.9% (d) s m 0.061 0.008 0.040 0.10

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Principles of Instrumental Analysis 6th ed. Appendix 1 Revision 1.0 February 13, 2008 Page 2 a1-2 For set A (a) Abs. error = t x x = 61.43 – 61.71 = –0.28 (b) Rel. error = –0.28 × 100%/61.71 = –0.45% Proceeding in the same way, we obtain A B C D (a) Abs. error –0.28 –0.03 0.13 -0.10 (b) Rel. error –0.45% –0.82% –1.1% –3.8% a1-3 (a) Rel. error - –0.5 × 100%/25 = –2.0% (b) Rel. error - –0.5 × 100%/100 = –0.5% (c) Rel. error - –0.5 × 100%/250 = –0.2% (d) Rel. error - –0.5 × 100%/500 = –0.1% a1-4 % Cu = Cu sample 100% = 4.8 m m × where m = mass Cu sample Cu 100 = = 20.8 4.8 m mm × (1) Rel. error = 3 Cu 0.5 10 g 100% m ×× Cu 0.050 = Rel. error in % m ( 2 ) Substituting Equation (1) into (2) gives sample 20.8 0.050 1.04 = = g % Rel. error % Rel. error m × Substituting the relative errors into this equation gives, (a) 10 g (b) 2.1 g (c) 1.3 g (d) 0.87 g
Principles of Instrumental Analysis 6th ed. Appendix 1 Revision 1.0 February 13, 2008 Page 3 a1-5 m s s N = (Equation a1-18) 1% 0.01% N = N = (1/0.01) 2 = 1 × 10 4 replicate measurements a1-6 RSD = 1.84 × 10 –3 × 100%/0.500 = 0.368% Applying Equation a1-18 0.100% = 0.368%/ N N = (0.368/0.100) 2 = 13.5 or 14 replicates a1-7 m Cr = 0.400 g × 18%/100% = 0.072 g Rel. error = 3 1.8 mg 10 g/mg 1000 ppt = 0.072 g × × –25 ppt a1-8 (a) For Sample 1, x = 1 / x N = (5.15 + 5.03 + 5.04 + 5.18 + 5.20)/5 = 5.12% () 1 1 2 = = N x x s N i i = 222 2 ( 0.03) + (0.09) + (0.08) + ( 0.06) + ( 0.08) 5 1 −− 2 = 0.08 In a similar manner, we find Sample Mean Standard Deviation 1 5.12 0.08 2 7.11 0.12 3 3.99 0.12 4 4.74 0.10 5 5.96 0.11 (b) 3 12 123 11 1 pooled t ( ) ( ) N NN ij k k xx x x s NN N n == = −+ + = +++ ∑∑ " "

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Principles of Instrumental Analysis 6th ed. Appendix 1 Revision 1.0 February 13, 2008 Page 4 s pooled = 0.0254 + 0.0281 + 0.0462 + 0.0326 + 0.046 = 0.106 5 + 3 + 4 + 4 + 5 5 0.11% (c) s pooled
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## This note was uploaded on 05/26/2010 for the course CHEM CHEM 100B taught by Professor Pomeroy during the Winter '10 term at UCSD.

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CHEM100B_Homework Solutions_Stats Solutions -...

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