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ch4 chem 210 - F ≤ table(9.28 The mean values are not...

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Kimberly Hargraves Chem 210 Chapter 4 Hw pg. 94 2-23-11 11) % gasoline 0.13 0.12 0.16 0.17 0.20 0.11 average 0.15 .90 C interval 0.02 .99 C interval 0.03 STD 0.03 t for .90 C -1.48 t for .99 C -3.36 13) dL=deciliter=0.1L Spooled= Sqrt (0.53) 2 (6-1) + (0.42) 2 (5-1)/6+5-2=.4842061086 T calculated= 14.57-13.95 /.4842061086×Sqrt (6×5)/6+5 T calculated=2.114 2.62 (t table) Yes, the trainee can be on her own.
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14) There is more than a 5% chance that the two sets of results lie within experimental error of each other, the results are not significantly different. (0.9869) t table(2.571). Sample Metho d 1 Method 2 Differe nce A 0.88 0.83 0.05 B 1.15 1.04 0.11 C 1.22 1.39 -0.17 D 0.93 0.91 0.02 E 1.17 1.08 0.09 F 1.51 1.31 0.2 average 0.05 STD 0.12 C level 0.95 n 6 t calculate d 0.9869 t table 2.571 16) The standard deviations are not significantly different because F calculated (2.43)
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Unformatted text preview: F ≤ table (9.28). The mean values are not significantly different because t calculated (1.55) t ≤ table (2.447). Quantit y Metho d 1 Metho d 2 mean 1.382 1.346 STD 0.025 0.039 number 4 4 F calculate d 2.43 F table 9.28 t calculate d 1.55 t table 2.447 21) The difference is significant in both cases because t calculated ≥t table. ppm 1 ppm 2 98.6 98.6 98.4 98.4 97.2 97.2 94.6 94.6 96.2 96.2 mean 97 94.5 STD 1.5 mean 96.6 number 5 STD 1.6 C level 0.95 C level 0.95 t calculat ed 146.5 t calculat ed 131.6 t table 2.776 t table 2.776 23) The value 216 should be retained in the results because Q calculated ≤ Q table. Results 192 216 202 195 204 mean 201.8 STD 8.4 number 5 C level 0.9 Q calculat ed 0.5 Q table 0.6...
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