ERT 207
ANALYTICAL CHEMISTRY
Lecture 4
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
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.
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
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.
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.
Various uses of t-test:
(a) True or accepted value and confidence limit
known.
The calculated t value (ItI) is then
compared to the t value from the table
(Table 3.1).
 If t calculated > ttable, then there is a
significant difference in the results of the
two methods used at a certain level of
confidence.
 If tcalculated < ttable, then there is no
significant difference in the results of the
two methods used at a certain level of
confidence.


Example 1: The true value (μ) 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?

Solution:

For the three analysis or N-1 =2(Table
3.1)
 t calculated
> ttable for 90 % confidence
levels, and
 t calculated
< ttable for 95% confidence
level and more.

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 %.
• 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?
• Solution:
• t = 2.9
• 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.
(b) Comparing mean for two sets of data.
• See Gary D.Christian (page 95)
The pooled standard deviation, sp
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)
• 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?
Degree of freedom (for paired t-test)=N1+N2-2
• t calculated > ttable 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.
• 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 Fe203. The results,
reported as % Fe for each analysis, were as
follows:
• Is there a significant difference between the
two methods?
• 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.
c) T test with multiple samples
• Average difference D is calculated and
individual deviations of each from D are used
to compute a standard deviation, sd
• t value is calculated from :
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.
Sample
Your Method
(mg/dL) ,x
Standard Method
(mg/dL) ,y
A
10.2
10.5
B
12.7
11.9
C
8.6
8.7
D
7.5
16.9
E
11.2
10.9
F
11.5
11.1
Solution :
Sample
Your
Method,
mg/dL
Standard
Method,
mg/dL
Di
Di-D
(Di-D)2
A
10.2
10.5
-0.3
-0.6
0.36
B
12.7
11.9
0.8
0.5
0.25
C
8.6
8.7
-0.1
-0.4
0.16
D
17.5
16.9
0.6
0.3
0.09
E
11.2
10.9
0.3
0.0
0.00
F
11.5
11.1
0.4
0.1
0.01
∑ 1.7
D = 0.28
∑ 0.87
• 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
• 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
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.
• 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.
• (iv) Compare this Qcalculated with the Q from the
Q Table (Table 3.3). The doubtful value is
removed with 90 % confidence level if the
Qcalculated is greater or equals to the Q value
obtained from the Q table.
• 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?
•
•
•
•
•
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 %
• Testing for 33.15%:
• The difference between the doubtful value
and the nearest one to it:
• = 33.60—33.15
• = 0.45 %
• 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.
• 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.
• 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%
• 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%
• 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
• Therefore, the value 8.75 should be accepted
at 90% confidence level (8.75 cannot be
removed).
Assignment 1
(5 questions)
1.
2.
3.
4.
5.