9 - Statistical comparison tests Comparisons between...

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1 Statistical comparison tests • Comparisons between different sets of data are required in chemical analysis in the following instances: – Comparing a result with a standard – Comparing the results of the same analysis by – Comparing the results of the same analysis by two or more different methods – Comparing two samples to see if they differ significantly – Comparing the results or the precisions obtained by two different laboratories • See experiment 4 in lab manual 1 Statistical comparison tests • The methods all rely on the "null hypothesis" principle – The assumption that there is "no difference" (null) is tested at various levels of confidence 2 Statistical comparison tests • Comparing a result with a "true" or standard value. Example : a method of analysis was tested for accuracy by performing it on a standard material, which contains 18.45% nickel N = 6; Mean = 18.22% Ni; S.D. ( s ) = 0.29 – Find the 90% and 99% confidence intervals – Find the difference between the mean and the standard result – Compare the confidence limit with the above difference 3

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2 Statistical comparison with standard Example : a method of analysis was tested for accuracy by performing it on a standard material, which contains 18.45% nickel N = 6; Mean = 18.22% Ni; S.D. ( s ) = 0.29 – 90% CI ( μ ) = x ± t * s / N = 18.22 ± 2.02*0.29 / 6 = 18.22 ± 0.5858/2.4494 = 18.22 ± 0.24 – 99% CI ( μ ) = x ± t * s / N = 18.22 ± 4.03*0.29 / 6 = 18.22 ± 0.48 4 Statistical comparison with standard Example : A method of analysis was tested for accuracy by performing it on a standard material, which contains 18.45% nickel N = 6; Mean = 18.22% Ni; S.D. ( s ) = 0.29 – Find the difference between the mean and the standard result • 18.22% - 18.45% = -0.23% – Compare the confidence limit with the above difference: – Then no significant difference has been demonstrated at this confidence level • Conclusions for this set? No significant difference 5 N t.s. ) - X ( IF μ Statistical comparison with standard • Suppose the results were: N = 12; Mean = 18.52% Ni; s = 0.18 – Has a significant error been demonstrated, with 90% confidence? t.s. X IF | x – μ | = 18.52% - 18.45% = 0.07% t * s / N = 1.80*0.18 / 3.4641 = 0.09% – No significant difference 6 N ) - (
3 Statistical comparison with standard Example 7-5 on page 153 : A new procedure for the rapid determination of the percentage of sulfur in kerosenes was tested on a sample known from its method of preparation to contain 0.123% ( μ 0 = 0.123%) S. The results were %S: 0.112, 0.118, 0.115, 0.119. Do the data indicate that there is a bias in the method at the 95% confidence level?

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This note was uploaded on 04/12/2010 for the course ENGINEERIN 1ac3 taught by Professor Xxx during the Spring '10 term at McMaster University.

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9 - Statistical comparison tests Comparisons between...

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