ERT 207 ANALYTICAL CHEMISTRY Lecture 4 Slide 2 TOPICS TO BE
COVERED: UTILIZATION OF STATISTICS IN DATA ANALYSIS: * SIGNIFICANT
TESTING *THE STUDENT T -TEST *THE Q TEST: REJECTION OF A RESULT
CALIBRATION CURVE * INTERPRET SLOPE, INTERCEPT AND COEFFICENT OF
DETERMINATION Slide 3 TESTS OF SIGNIFICANCE Is there a difference?
1. to compare data among friends with the intention of gaining some
confidence that the data observed could be accepted or rejected. 2.
to decide whether there is a difference between the results
obtained using two different methods. All these can be confirmed by
doing some significant tests. Slide 4 THE STUDENT T -TEST In this
method, comparison is made between two sets of replicate
measurements made by two different methods; one of them will be the
test method, and the other one will be an accepted method. T-test
is used to decide whether there is a significant difference in the
mean for two sets of data True or accepted value and confidence
limit is known Comparing mean for two sets of data Slide 5 The
calculated t value is compared with the t value from the table
(Table 3.1 from Gary D.Christian) for the sets of data and at a
required level of confidence. Slide 6 Select a confidence level
(95% is good) for the number of samples analyzed (= degrees of
freedom +1). Confidence limit = x ts/N. It depends on the
precision, s, and the confidence level you select. Select a
confidence level (95% is good) for the number of samples analyzed
(= degrees of freedom +1). Confidence limit = x ts/N. It depends on
the precision, s, and the confidence level you select. Slide 7 (a)
True or accepted value and confidence limit known. Various uses of
t-test: Slide 8 The calculated t value (ItI) is then compared to
the t value from the table (Table 3.1). If t calculated > t
table, then there is a significant difference in the results of the
two methods used at a certain level of confidence. If t calculated
< t table, then there is no significant difference in the
results of the two methods used at a certain level of confidence.
Slide 9 Example 1: The true value () for Cl - in a standard sample
is 34.63 %. After three analysis, it is found that the mean is
34.71 % and s equals to 0.04 %. From the result, is there any
determinate error that occurs in the method used? Slide 10
Solution: Slide 11 For the three analysis or N-1 =2(Table 3.1) t
calculated > t table for 90 % confidence levels, and t
calculated < t table for 95% confidence level and more. Slide 12
Therefore there is no significant difference for confidence level
greater than 95 % and we can conclude that there is no determinate
error for confidence level greater than 95 %. Slide 13 Example 2:
You are developing a procedure for determining traces of copper in
biological materials using a wet digestion followed by measurement
by atomic absorption spectrophotometry. In order to test the
validity of the method, you obtain an NIST orchard leaves standard
reference material and analyze this material. Five replicas are
sampled and analyzed, and the mean of the results is found to be
10.8 ppm with a standard deviation of 0.7 ppm. The listed value is
11.7 ppm. Does your method give a statistically correct value at
the 95% confidence level? Slide 14 Solution: t = 2.9 Slide 15 There
are five measurements, so there are 4 degrees of freedom (N 1).
From Table 3.1, we see that the tabulated value of t at the 95%
confidence level is 2.776. This is less than the calculated value,
so there is a determinate error in the new procedure. That is,
there is a 95% probability that the difference between the
reference value and the measured value is not due to chance. Slide
16 (b) Comparing mean for two sets of data. Slide 17 See Gary
D.Christian (page 95) Slide 18 The pooled standard deviation, s p
is given by: x => values in each set N => num. of
measurements x =>means of each of k sets of analyses N k =>
degree of freedom from (N1-1) + (N2-2) + + (Nk-1) Slide 19 Example
1: Two bottles of beer were analyzed for the alcohol content. Four
samples from the first bottle with a mean value of 12.61 % alcohol
and six samples from the second bottle with a mean value of 12.39 %
alcohol. The pooled standard deviation is 0.07 %. Is there a
significant difference between the content of the two bottles?
Slide 20 Degree of freedom (for paired t-test) =N 1 +N 2 -2 Slide
21 t calculated > t table for 80, 90, 95 and 99 % levels of
confidence (Table 3.1). This shows that the analysis gives a
significant difference in the alcohol content. Slide 22 Example 2:
A new gravimetric method is developed for iron (III) in which the
iron is precipitated in crystalline form with an organoboron cage
compound. The accuracy of the method is checked by analyzing the
iron in an ore sample and comparing with the results using the
standard precipitation with ammonia and weighing of Fe 2 0 3. The
results, reported as % Fe for each analysis, were as follows: Slide
23 Is there a significant difference between the two methods? Slide
24 Slide 25 Slide 26 The tabulated t for nine degrees of freedom
(N1 + N2 2) at the 95% confidence level is 2.262, so there is no
statistical difference in the results by the two methods. Slide 27
Average difference D is calculated and individual deviations of
each from D are used to compute a standard deviation, s d t value
is calculated from : c) T test with multiple samples Slide 28
Example You are developing a new analytical method for the
determination of blood urea nitrogen (BUN). You want to determine
whether you method differs significantly from a standard one for
analyzing a range of sample concentrations expected to be found in
the routine laboratory. It has been ascertained that the two
methods have comparable precisions. Following are two sets of
results for a number of individual samples. Slide 29 SampleYour
Method (mg/dL),x Standard Method (mg/dL),y A10.210.5 B12.711.9
C8.68.7 D7.516.9 E11.210.9 F11.511.1 Slide 30 SampleYour Method,
mg/dL Standard Method, mg/dL DiDi D i -D(Di-D) 2
A10.210.5-0.3-0.60.36 B12.711.90.80.50.25 C8.68.7-0.1-0.40.16
D17.516.90.60.30.09 E11.210.90.30.00.00 F11.511.10.40.10.01 1.7
0.87 D = 0.28 Solution : Slide 31 Sd = 0.42 t = 1.63 The tabulated
t value at 95% confidence level for 5 degrees of freedom (6-1=5) is
2.571. Therefore, t calc < t table So, there is no significant
difference between the 2 methods at 95% confidence level Slide 32
95% confidence level is considered significant 99% level is highly
significant Smaller the calculated t value, more confident that
there is no significant difference between the two method Slide 33
3.3 THE Q TEST: REJECTION OF A RESULT Not all data obtained from an
experiment can be used. There is a possibility that some data have
a great difference from the true value and therefore should be
removed. The rule for data rejection is known as Q test. Slide 34
The Q test is used for rejection of values that are further away
from the true value. The Q value can be obtained as follows: (i)
Arrange the values in an order. (ii) Calculate the differences
between the highest and lowest values. (iii) Calculate the
difference between the uncertain value and the nearest value to it.
Divide this value with the value calculated in (ii) to obtain the Q
calculated. Slide 35 (iv) Compare this Q calculated with the Q from
the Q Table (Table 3.3). The doubtful value is removed with 90 %
confidence level if the Q calculated is greater or equals to the Q
value obtained from the Q table. Slide 36 Slide 37 Example: An
analysis of iron ore gives the following results: 33.78 %, 33.84 %,
33.60 % and 33.15 %. Which of the values should be removed?
Example: An analysis of iron ore gives the following results: 33.78
%, 33.84 %, 33.60 % and 33.15 %. Which of the values should be
removed? Slide 38 Solution: Doubtful values are 33.15 % and 33.84 %
Arrange the values: 33.84, 33.78, 33.60, 33.15 The difference
between the highest and the lowest value = 33.84% 33.15% = 0.69 %
Slide 39 Testing for 33.15%: The difference between the doubtful
value and the nearest one to it: = 33.6033.15 = 0.45 % Slide 40
From the Q Table, Q table = 0.76 (for 4 observations). The highest
value (33.84 %) can also be tested using the same method as above.
Slide 41 Using the Q test for the following results, determine
whether the value 8.75 should be rejected as an outlier at 90%
confidence level. 8.20, 8.35, 8.64, 8.25, 8.75, 8.45. Slide 42
Answer: a) Arranging according to descending order: 8.75, 8.64,
8.45, 8.35, 8.25, 8.20 b) Calculate difference between the highest
and lowest values: 8.75-8.20 = 0.55% Slide 43 c) Calculate
difference between uncertain value and nearest value to it : 8.75-
8.64 = 0.11% d) Divide this value with the value calculated in (b)
to obtain Qcalculated Therefore, 0.11% = 0.2 0.55% Slide 44 e)
Compare Q calculated with the Qtable at 90% confidence level. N = 6
Qcalculated = 0.2 Qtable = 0.56 Qcalculated < Qtable 0.2 <
0.56 Slide 45 Therefore, the value 8.75 should be accepted at 90%
confidence level (8.75 cannot be removed). Slide 46 Assignment 1 (5
questions) Slide 47 1. Slide 48 2. Slide 49 3. Slide 50 4. 5.
