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If anyone can explain this to me that would be greatly appreciated. Thank you so much.

RECTTATON EXERCISE Error Analysis A CEM unknown was determined with an oxidation-reduction titration. The student didnt use the results from the first trial titration and the results of the next three were as follows: 2.014 g, 1.956 g, and 2.017 g. 262 student performed an experiment in which the mass of copper in an a. What is the mean and standard deviation of these three trials? Mean: ,916 2.ol4-1.9406-1996+(2 3 -1 Standard Deviation: 0344 b, with a 95% confidence level, what is the maximum error in this analysis? Assuming that all errors are random (no determinant or gross errors) what is the minimum grade that that student could expect on that lab (with a 95% confidence level)? See the histogram on the next page. c. d. Not satisfied with the possibility of this grade, the student chooses to do another titration and the result is 2.019 g. for the mass of the unknown. Its starting to look like the 1.956 g value is significantly different from the other three. Can the student reject this value with a confidence level of greater than 95%? e. What is the mean value and standard deviation that the student should report? Mean: Standard Deviation

f, with a 95% confidence level, what is the maximum error in this result? g. What is the minimum grade that the student can expect on this experiment, assuming no determinant errors? Bear in mind that this grade determination is a statistical estimate of the accuracy based upon the precision of the result. It assumes that the only errors are uncontrollable random errors. If there are determinant errors, this will not be a good predictor of the expected grade. CEM 262 Experiment5 Error and Grade Distribution 20 2 15 10 0 0.00 0.05 0.10 0.15 0.20 Errors

Table 4-2 Values of Students t Confidence level (%) Degrees of freedom 63.656 127.321 636.578 7.453 12.924 2.132 0.727 2.571 3.365 1.943 1.895 1.860 1.833 5.959 0.711 2.365 2.306 2.228 2.086 3.355 4.587 0.687 0.683 0.681 2.787 2.750 3.551 1.671 1.658 1.645 2.915 3.373 3.291 0.674 2.576 In calculating confidence intervals, ơ may he substituted for s in Equation 4-3 if you have a great deal of experience with a particular method and have therefore determined its True population standard deviation. If σ is used instead of 5, the value of, to use in Equa tion 4-3 comes from the botom row of this table. Table 4-4 Critical values of G for rejection of outlier Number of observations (95% confidence) 1.672 1.822 2.032 2.110 2.176 10 2.557 Gealculated|questionable value mean|/s. If Gcalculated> Gable the value in question can be rejected with 95% confidence. Values in this table are for a one-sided test, as recommended by ASTM SOURCE: ASTM E 178-02 Standard Practice for Dealing with Outlying Obsenations. F. E. Grubbs and G. Beck, Technometrics 1972, 14, 847

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