Seminars: Numerical Problems
91
At the 95% confidence level (P = 0.05), t = 2.12. This allows the limits of confidence to
be calculated as follows:
_
s
0.241
Xl ± t c 3.87 ± 2.12 r,; = 3.87 ± 0.13%
vn
v16
Then, Dixon's test is used to reject potential outliers (which both extreme data in the
set are).
The implicit range for the first dubious datum excludes the second:
4.60 - 4.02 0.58
Qcal =
-
= 0.66 at = 15, Qt = 0.32 (P = 0.01)
4.60 - 3.73 0.87
Since Qt < Qcab then the result should be rejected.
The implicit range for the second dubious datum excludes the first:
3.73 - 3.41 0.32
Qcal =
-
= 0.52 at = 15, Qt = 0.32 (P = 0.01)
4.02 - 3.41 0.61
Since Qt < Qcab this result should also be rejected.
In addition, both potential outliers fall beyond the limits of confidence (4.00 and
3.74), which confirms that the decision to reject them was correct. More rigorous
application of Dixon's test would lead to identifying 4.02 and 3.76 as two further
outliers.
At this point, the remaining 14 results are evaluated (including both extremes in
the whole set). The new results are
X2 = 3.85% s = 0.095% n = 14
At the 95% confidence level (P = 0.05), t = 2.14 and the confidence limits are
_
s
0.095
X2 ± t c 3.85 ± 2.14 r.-:; = 3.85 ± 0.05%
vn
v14
As can be seen, the precision is much better (s and the uncertainty are much lower)
after the outliers are discarded.
As can also be seen, the means Xl and X2 are quite similar as a result of an overestimated and an underestimated result being discarded, and the effects on mean Xl
being mutually countered. Should one or more results differing from the mean with
the same sign have been rejected, the effect on the mean would have been rather different.
(3) The precision (uncertainty) of a fast immunoassay method (A) for the
determination of phenol in water relative to that of the slower, classical
chromatographic method (B) must be determined. For this purpose, two
experiment series are conducted in order to apply both methods to
aliquots of the same sample. The results obtained (in llg/L) are as follows:
Method A: 1.7,1.8,2.1,2.2,1.7,1.9,1.5,1.3,1.9,1.7,1.8,1.4
Method B: 1.60,1.82,1.73,1.81,1.70,1.73
The problem involves evaluating the dispersion of the two data sets, which consist of
a different number of results owing to the slowness of method B. Inputing the
previous data into a calculator provides the following statistical parameters:
Method A
MethodB
(n = 12)
(n = 6)
XA = 1.75 p.g/L
XB = 1.73 p.glL
s = 0.26 p.g/L
s = 0.08 p.g/L
91
At the 95% confidence level (P = 0.05), t = 2.12. This allows the limits of confidence to
be calculated as follows:
_
s
0.241
Xl ± t c 3.87 ± 2.12 r,; = 3.87 ± 0.13%
vn
v16
Then, Dixon's test is used to reject potential outliers (which both extreme data in the
set are).
The implicit range for the first dubious datum excludes the second:
4.60 - 4.02 0.58
Qcal =
-
= 0.66 at = 15, Qt = 0.32 (P = 0.01)
4.60 - 3.73 0.87
Since Qt < Qcab then the result should be rejected.
The implicit range for the second dubious datum excludes the first:
3.73 - 3.41 0.32
Qcal =
-
= 0.52 at = 15, Qt = 0.32 (P = 0.01)
4.02 - 3.41 0.61
Since Qt < Qcab this result should also be rejected.
In addition, both potential outliers fall beyond the limits of confidence (4.00 and
3.74), which confirms that the decision to reject them was correct. More rigorous
application of Dixon's test would lead to identifying 4.02 and 3.76 as two further
outliers.
At this point, the remaining 14 results are evaluated (including both extremes in
the whole set). The new results are
X2 = 3.85% s = 0.095% n = 14
At the 95% confidence level (P = 0.05), t = 2.14 and the confidence limits are
_
s
0.095
X2 ± t c 3.85 ± 2.14 r.-:; = 3.85 ± 0.05%
vn
v14
As can be seen, the precision is much better (s and the uncertainty are much lower)
after the outliers are discarded.
As can also be seen, the means Xl and X2 are quite similar as a result of an overestimated and an underestimated result being discarded, and the effects on mean Xl
being mutually countered. Should one or more results differing from the mean with
the same sign have been rejected, the effect on the mean would have been rather different.
(3) The precision (uncertainty) of a fast immunoassay method (A) for the
determination of phenol in water relative to that of the slower, classical
chromatographic method (B) must be determined. For this purpose, two
experiment series are conducted in order to apply both methods to
aliquots of the same sample. The results obtained (in llg/L) are as follows:
Method A: 1.7,1.8,2.1,2.2,1.7,1.9,1.5,1.3,1.9,1.7,1.8,1.4
Method B: 1.60,1.82,1.73,1.81,1.70,1.73
The problem involves evaluating the dispersion of the two data sets, which consist of
a different number of results owing to the slowness of method B. Inputing the
previous data into a calculator provides the following statistical parameters:
Method A
MethodB
(n = 12)
(n = 6)
XA = 1.75 p.g/L
XB = 1.73 p.glL
s = 0.26 p.g/L
s = 0.08 p.g/L