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ENCH 213
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Diane Beauchemin
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Quiz

Description

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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