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Homework Assignment 02 Class Monday Sept 20, 2004. __________________________________________. Homework Assignment 02 ...is from Chapter 3. Problems assigned are: 5,11,12,15,21,22 This assignment is due at class time Monday, Sept 20

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  1. Homework Assignment 02 Class Monday Sept 20, 2004 __________________________________________ Homework Assignment 02...is from Chapter 3. Problems assigned are: 5,11,12,15,21,22This assignment is due at class time Monday, Sept 20 Prepare on regularly sized paper, one side only with multiple pages stapled. Excel sheets

  2. Volumetric Chloride Lab __________________________________________ Volumetric Chloride Lab

  3. Chapter 4 – Statistics Comparison of t-Test and F-Test __________________________________________ The t-test allowed us to test for significant differences in the mean values obtained by the two scientists. Since the difference in the 2 mean values was less than the tolerance quantity, there is no significant difference between the mean values of the two scientists at the 95% confidence level. The F-test allowed us to test for differences in the precision of the two scientists. Since the calculated value of F cal < Ftable, there is no significant difference between the precision of the two scientists at the 95% confidence level.

  4. Rejection of Suspect Data - The Q-test __________________________________________ Occasionally in a set of data there is one value that appears to not belong with the rest of the set. If the experimenter is aware of some mistake or malfunction, she/he do not need to employ one of these tests to reject that result. If no known error has occurred (so that the suspect result appears to be random), the analyst is then faced with whether to retain or reject this suspect value. He/she needs some sound basis for their decision, not just ‘eyeballing’ it. Your textbook describes one such test, the Q-test. After I have discussed the Q-test, I will then discuss two additional less rigorous, but useful tests for rejection of suspect data.

  5. Rejection of Suspect Data - The Q-test A Problem __________________________________________ Problem – Given the following set of data for the determination of % Acidic Substance in a Cleansing Agent. May the suspect result be rejected, or must it be retained by the criteria of the Q-test?  % Acid 10.19% 10.08% 10.52% 10.13% (suspect) Calculate the mean values both retaining and rejecting the suspect value. xmean (retaining) = 10.23% xmean (rejecting) = 10.13%

  6. Rejection of Suspect Data - The Q-test A Problem __________________________________________ Clearly the suspect value has a significant effect on the mean value. To employ the Q-test we need the range and the difference between the suspect value and the value nearest it. Range = (10.52 – 10.08) = 0.44 Difference of Suspect and its nearest value = (10.52 – 10.19) = 0.33 Qcal = (xsuspect – xnearest) / (Range) = 0.33 / 0.44 = 0.75 Since Qcal < Qtable (0.75 < 0.76) we must retainthe suspect value at the 90% confidence level.

  7. Rejection of Suspect Data - The Q-test __________________________________________ Referring to Table 4-4, textbook page 79 Qt = 0.76 for n = 4 at the 90% Confidence Level. Thus we must retain the suspect value by this criterion. (Not in your textbook, but Qtable at the 96% confidence level has a value of 0.85 for n = 4 a; by this criterion, the suspect value of 10.52% would also be retained.) a Skoog and West, “Fundamentals of Analytical Chemistry, 4e, c1982, CBS College Publishing, p62.

  8. Rejection of Suspect Data The 2.5- and 4-d Tests __________________________________________ Although less rigorous, these tests may also be used to decide whether to retain or reject a suspect. In order to use it, one needs to calculate the average deviation which is defined as average deviation = i |(x i – xmean)| / n

  9. Rejection of Suspect Data The 2.5- and 4-d Tests __________________________________________

  10. Rejection of Suspect Data The 2.5- and 4-d Tests __________________________________________ Since 4 x avg d < di (0.147 < 0.39) for the suspect value from the mean, we could reject the suspect value by the 4d rule. The 2.5d is done identically except the multiplier is 2.5 instead of 4; Since 2.5d < di (0.093 < 0.39) Clearly the 2.5d rule allows easier rejection than the 4d rule. The suspect value of 10.52 with di = 0.39 may be rejected by both of these criteria.

  11. End Monday, Sept 20, 2004 __________________________________________

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