Lecture 4
ERT 207
ANALYTICAL CHEMISTRY
13 JAN 2011
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
* SLOPE,INTERCEPT AND COEFFICENT OF
DETERMINATION
3.1 SIGNIFICANT TESTING
•
Why we do testing to our experimental data?
•
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.
3.2 THE STUDENT T -TEST
• 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 know
• 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.
(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?
• 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: 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) For special case, that is M = N.
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 he 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).
• 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 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
• Too low confidence level (e.g. 80%),
likely to conclude erroneously that
there is a significant difference
between two methods (type I error)
• On the other hand, too high confidence
level will require too large a difference
to detect (type II error)
• If a calculated t value is near tabular
value at the 95% confidence level, more
tests should be run to ascertain
definitely whether the two methods are
significantly different

Thank you for your attention