Answers Problem Set 2 Chem 314A Williamsen Spring 2000

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Experimental t-values: Sample Number t-statistic 1 1.54 2 1.60 3 1.18 4 1.60 Critical Value: t(0.95,7)=±2.36 Decision: Because all of the experimental t-statistics are within the critical limits, accept the null hypothesis. The spectroscopically obtained values differ from the calibrated values by random variation at the 95% confidence limit with 7 degrees of freedom. Therefore, the spectroscopic method appears to accurately determine the value. 8) Statistics for Analytical Chemists p. 80 Problem 7. Answer Chapter 3 p. 179. You would give the complete hypothesis test, I will only give you the important values. a) F-test (compare variances) Experimental F: F=112 Critical value: F(0.95,7,7)=3.787 The decision would be that the longer boiling point's variation is greater than can be explained by random variation at the 95% confidence limit with 8 measurements. b) t-test (Case II) (compare means) Experimental t: t=1.8 Critical value: t(0.95,14) )=±2.14
11) Mermet et al. Analytical Chemistry p 733 Problem 19. Hypotheses: H 0 : s A =s B H 1 : s A ≠s B F=6.13 Because the calculated F-value is larger than the critical F-value, reject the null hypothesis. The only statement that is true is that the hypothesis of method A being less precise than method B can not be rejected. You did not directly test this hypothesis (that would be a 1-sided test), but because the variances were found to be different, this hypothesis is not ruled out. 12) Mermet et al. Analytical Chemistry p 733 Problem 20. Bullet 1: Test: F-test Level of significance: 95% confidence limit Hypotheses: H 0 : s field ≤s laboratory H 1 : s field >s laboratory Experiment F: 14.0 Decision: Because the experimental F-value is greater than the critical value, reject the null hypothesis. Therefore at the 95% confidence level with 7 degrees of freedom for each measurement, the precision of the laboratory method is significantly greater than the field method. Bullet 2: Test: t-test (Case II) Level of significance: 95% confidence limit Hypotheses: H 0 : 5.50=4.80 H 1 : 5.50≠4.80 Experimental t: 4.25 Critical t: t(0.95,14)=2.145 Decision: Because the experimental t-value is greater than the critical t-value, reject the null hypothesis. Therefore at the 95% level of significance with 14 degrees of freedom, the means of the two methods do differ significantly. Last Revision: 4 March 2000 EJW 5
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Page 3: ) How is the β-error set The β-er

Answers Problem Set 2 Chem 314A Williamsen Spring 2000

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