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6statisticaltests.pdf

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Department
Chemistry
Course Code
ENCH 213
Professor
Diane Beauchemin

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Statistical tests • give probabilities only ◦ you still have to interpret the results, decide on accepting/rejecting a value, especially for small data sets (i.e. apply good judgement) • may be helpful for questions: ◦ should an outlying value be rejected or retained in calculation of the mean? ◦ are 2 samples, analysed by the same method, significantly different in composition? ◦ does a difference in precision exist between 2 data sets from different workers/methods? Rejection/retention of outliers: Grubbs test 1. Compute the average, x , and standard deviation, s. 2. G exp= ‌ q – x ‌ /s (mean includes questionable value) → x =qquestionable result 3. Get G critrom Table 4-5 4. Apply test: reject if Gexp> G crit Example: Fe in large mocked soil sample %Pb =1.1, 3.4, 6.9, 3.9, 2.7, 2.8, 1.2, 2.3 • should the 6.9 value be rejected? • x = 3.0; s =1.9 • G exp 6.9 – 3.0 ‌ /1.9= 2.05 • G critor n=8 is 2.032 • Reject because G > exp crit • BLINDAPPLICATION OF TESTS IS DANGEROUS. USE GOOD COMMON SENSE. In case of outliers: • re-examine all data relating to the outlying value: properly kept lab notebok • if possible, estimate the precision expected from the procedure, to check if the outlying value is actually questionable. • repeat the analysis if sufficient sample and time are available. • otherwise, apply Grubbs test. • if retention is indicated, consider reporting the median rather than the mean Example Police have a hit-and-run case and need to identify the brand of red auto paint. The percentage of iron oxide, which gives paint its red color, found during analysis is as follows: 43.15, 43.81, 45.71, 43.23, 42.99, and 43.56%. What is the average percentage of the iron oxide in the paint sample? a. 43.3 b. 43.7 c. 43.6 Example answer: Grubbs test G calc G crit reject 45.71 42.99, 43.15, 43.23, 43.56, 43.81, 45.71 % Mean= 43.74 s=1.0 Recalculate mean= 43.3 (s=0.3) G calc (45.71-43.74)/(1.0) = 1.97 → decrease ins indicationofoutlier G crit822 Hypothesis testing • "null hypothesis" assumes that quantities compared are the same unless proven otherwise by testing at a given probability level: ◦ comparison of mean from analysis with µ certified value of SRM...) ◦ comparison of the means of 2 different sets ◦ comparison of standard deviations from 2 sets of data ◦ comparison of individual differences between two data sets Comparison of mean with an accepted value: Student’s t test • mean - μ = ± ts /√(n) • if |mean-μ| > ts/ √(n) ◦ null hypothesis wrong ◦ determinate error at this confidence level • another way is to calculate t calcnd compare it to t critrom Table 4-2 ◦ t calc= √n x |mean-μ|/s ◦ if t calc tcrit ▪ Null hypothesis wrong at this confidence level Example: New method for the determination of sulfur in kerosenes • data: % S = 0.112, 0.118, 0.115, 0.119 • true = µ = 0.123% S • is there a determinate error? • mean= 0.116% S, s=0.003% • tcalc √4 x |0.116-0.123|/0.003 = 4.7 • tcritor 3 degrees of freedom = 3.182 at 95% confidence level • tcalc tcrit systematic error indicated • At 99% level, t = 5.84: null hypothesis holds crit Q: A systematic error a) can be discovered and corrected. b) arises from the limitations on the ability to make a physical measurement. c) is also known as an indeterminate error. Example 2: Hg in mocked soil sample (small sample) • data: % Hg= 7.5 ± 6.6 • True = µ = 5.86% Hg • Is there a determinate error? • t = √8 |7.5-5.86|/6.6 = 0.548 calc • tcritor 7 degrees of freedom = 2.365 at 95% • confidence level • t < t : no systematic error indicated calc crit Choice of confidence level • if too severe (99.9%) ◦ significant effect may be missed • if too relaxed (50%) ◦ insignificant effect may be judged important • In general ◦ result at 95% confidence level: SIGNIFICANT ◦ result at 99% confidence level: HIGHLY SIGNIFICANT Comparison of the precision of two sets: F test • Fcalc= s1/s 22 2 ◦ s =1ariance of set 1 ◦ s =2ariance of set 2
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