Chapter 7 solutions - Fundamentals of Analytical Chemistry 8th ed Chapter 7 Chapter 7 7-1 The mean of 5 measurements x is a better estimate of the true

Chapter 7 solutions - Fundamentals of Analytical Chemistry...

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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 7 Chapter 7 7-1 The mean of 5 measurements, x is a better estimate of the true value μ than any single measurement because the distribution of means is narrower than the distribution of individual results. 7-2 (a) By inspection of Table 7-1 CL = 99.7 % (b) CL = 90 % (c) CL = 68 % (d) CL = 95.4 % 7-3 (a) As the sample size, N , increases the confidence interval decreases in proportion to N . (b) As the desired confidence level increases the confidence interval increases. (c) As the standard deviation, s , increases the confidence interval increases in direct proportion. 7-4 For Set A: x i x i 2 3.5 6.25 3.1 9.61 3.1 9.61 3.3 10.89 2.5 12.25 Σ x i = 15.5 Σ x i 2 = 48.61 mean: x = 15.5/5 = 3.1
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 7 standard deviation: ( ) 37 . 0 1 5 5 / 5 . 15 61 . 48 2 = - - = s Since, for a small set of measurements we cannot be certain s is a good approximation of σ , we will use the t statistic for confidence intervals. From Table 7-3 , at 95% confidence t for 4 degrees of freedom is 2.78, therefore CI for μ = 3.1 ± ( )( ) 5 37 . 0 78 . 2 = 3.1 ± 0.46 = 3.1 ± 0.5 Results for Sets A through F, obtained in a similar way, are given in the following table. A B C D E F x 3.1 70.19 0.82 2.86 70.53 0.494 s 0.37 0.08 0.05 0.24 0.22 0.016 CI ± 0.46 ± 0.20 ± 0.08 ± 0.30 ± 0.34 ± 0.020 The 95% confidence interval establishes the range about the mean that we can be 95% confident that the true value lies within (in the absence of systematic errors). 7-5 If s is a good estimate of σ then we can use z at 95% confidence equal 1.96 and for set A, CI for μ = 3.1 ± ( )( ) 5 20 . 0 96 . 1 = ± 0.18 Similarly for sets B-F A B C D E F CI ± 0.18 ± 0.079 ± 0.009 ± 0.26 ± 0.15 ± 0.013 7-6 For set A: 6 . 0 5 . 2 5 . 3 1 . 3 5 . 2 = - - = Q and Q crit for 5 observations at 95% confidence = 0.710. Since Q < Q crit the outlier value 2.5 cannot be rejected with 95% confidence.
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 7 A B C D E F Q 0.6 0.86 0.81 0.33 0.95 0.33 Q crit (95%) 0.710 0.970 0.829 0.710 0.829 0.710 Decision Keep Keep Keep Keep Reject Keep 7-7 (a) 80% CL = 18.5 ± 1.28 × 2.4 = 18.5 ± 3.1 μ g Fe/mL 95% CL = 18.5 ± 1.96 × 2.4 = 18.5 ± 4.7 μ g Fe/mL (b) 80% CL = 18.5 ± 2 4 . 2 28 . 1 × = 18.5 ± 2.2 μ g Fe/mL 95% CL = 18.5 ± 2 4 . 2 96 . 1 × = 18.5 ± 3.3 μ g Fe/mL (c) 80% CL = 18.5 ± 4 4 . 2 28 . 1 × = 18.5 ± 1.5 μ g Fe/mL 95% CL = 18.5 ± 4 4 . 2 96 . 1 × = 18.5 ± 2.4 μ g Fe/mL 7-8 (a) 90% CL = 8.53 ± 1.64 × 0.32 = 8.53 ± 0.52 μ g Cu/mL 99% CL = 8.53 ± 2.58 × 0.32 = 8.53 ± 0.83 μ g Cu/mL (b) 90% CL = 8.53 ± 4 32 . 0 64 . 1 × = 8.53 ± 0.26 μ g Cu/mL 99% CL = 8.53 ± 4 32 . 0 58 . 2 × = 8.53 ± 0.41 μ g Cu/mL (c) 90% CL = 8.53 ± 16 32 . 0 64 . 1 × = 8.53 ± 0.13 μ g Cu/mL 99% CL = 8.53 ± 16 32 . 0 58 . 2 × = 8.53 ± 0.21 μ g Cu/mL
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 7 7-9 (a) 1.5 = N 4 . 2 96 . 1 × For a 95% CL, N = 9.8 10 (b) 1.5 = N 4 . 2 58 . 2 × For a 99% CL, N = 17 7-10 (a) 0.2 = N 32 . 0 96 . 1 × For a 95% CL, N = 9.8 10 (b) 0.2 = N 32 . 0 58 . 2 × For a 99% CL, N = 17 7-11 For the data set: 22 . 3 = x and s = 0.06 (a) 95% CL = 3.22 ± 3 06 . 0 30 . 4 × = 3.22 ± 0.15 meq Ca/L (b) 95% CL = 3.22 ± 3 056 . 0 96 . 1 × = 3.22 ± 0.06 meq Ca/L 7-12 For the data set: 24 . 7 = x and s = 0.25 (a) 90% CL = 7.24 ± 3 25 . 0 92 . 2 × = 7.24 ± 0.42 % lindane (b) 90% CL = 7.24 ± 3 28 . 0 64 . 1 × = 7.24 ± 0.27 % lindane 7-13 (a) 0.3 = N 40 . 0 58 . 2 × For a 99% CL, N = 11.8 12 (b) 0.3 = N 40 . 0 96 . 1 × For a 95% CL, N = 6.8 7 (c) 0.2 = N 40 . 0 64 . 1 × For a 90% CL, N = 10.8 11
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Fundamentals of Analytical Chemistry: 8 th ed. Chapter 7 7-14 The null hypothesis is that μ = 30.15% and the alternative hypothesis is that μ 30.15%.
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