Analytical Measurement and Uncertainty

This study material discusses analytical measurement and uncertainty. It covers topics such as linearity of the calibration function, residual plot, ANOVA, and more. The document does not mention a specific course code, course name, or college/university.

Analytical measurement and
uncertainty
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Question 4
Replicate
(ci/ mg L-1) 1 2 3 4 5
20 41 42 41 40 40
30 59 57 60 59 61
40 80 78 82 79 83
50 98 100 95 103 97
60 121 126 122 117 120
70 142 137 144 141 146
80 158 152 160 161 154
90 178 172 185 177 180
(a) Linearity of the calibration function can be calculated by plotting the graph in following way
-
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(b) & (c)
Linearity of Calibration plot can be examined via three steps –
1. Test for significance for correlation coefficient
2. Examination of residual point by performing run test
3. ANOVA at 95% confidence of interval
Solution 1. Correlation coefficient for n = 8 is 0.9906 then,
Null Hypothesis (H0) – Correlation coefficient is not significant statistically
Alternative Hypothesis (H1) – Correlation coefficient is significant statistically
Using t-test statistic as –
t = r √n – 2
√1 – r2
= 0.999 √8 – 2
√1 – (0.999)2
= 2.447/ 0.447 = 5.57
Critical Value – Based on two-sided t-table, with (n-2) or 6 degrees of freedom, critical value at
95% confidence level is 2.447
Decision – Since calculated t-value of 5.57 > tabulated t-value of 2.447, therefore, null
hypothesis is rejected.
Conclusion – It has been concluded that correlation coefficient is significant statistically at 95%
confidence level.
2. A residual plot can be constructed by calculating residuals in following way –
Equation of linear portion of calibration plot is
Fluorescence = 19.798 x concentration + 20.536
Concentration Expectedfluorescence Predictedfluorescence Residual
20 41 41.6 0.6
30 59 55.4 3.6
40 80 81.2 -1.2
50 98 101.04 -3.04
60 121 120.8 0.2
70 142 140.6 1.4
80 158 160.4 -2.4
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90 178 180.2 -2.2
Now, Residual plot can be constructed as –
The residual plot as per the given figure, appears to show the randomly distributed
residuals, therefore, supporting a linear relationship.
Now, after performing the run test, where signs are increasing as +, -, +, …
Therefore, curve is not fitted to set of data points
3. Performing ANOVA
Concentration Expectedfluorescence Predictedfluorescence Residual Residual2 (Predictedflu
– Mean)2
20 41 41.6 0.6 0.36 4627.4
30 59 55.4 3.6 12.96 2940.4
40 80 81.2 -1.2 1.44 807.9
50 98 101.04 -3.04 9.2416 73.7
60 121 120.8 0.2 0.04 124.8
70 142 140.6 1.4 1.96 959.4
80 158 160.4 -2.4 5.76 2578.1
90 178 180.2 -2.2 4.84 4952.64
Mean = 109.625 Sum = Sum =
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36.6016 17064.3
Sources of
variation
Degrees of
freedom
Sum of squares Mean Squares F
Regression 1 17064.3 17064.3
Residuals 8 – 2 = 6 36.6016 6.1 2797.37
Total 8 – 1 = 7
Null Hypothesis = There is not a linear relationship between Concentration and Fluorescence
Alternative Hypothesis = There is a linear relationship between Concentration and Fluorescence
Test statistic = 2797.37
Critical value – From a one-sided F-test with 1 and 6 degrees of freedom at 95% confidence
level, the critical value is 7.71
Decision – As Critical calculated value 2797.37 > 7.71, reject the null hypothesis
Conclusion – There is a linear relationship between Concentration and Fluorescence
(d)
Since linear relationship is generally represented by a slope and an ordinate intercept, therefore,
using above yielded equation –
Fluorescence = 19.798 x concentration + 20.536
xi yi ypredicted (yi – yp)2 (xi -
mean)2 (xi)2
20 41 41.6 0.36 1225 400
30 59 55.4 12.96 626 900
40 80 81.2 1.44 225 1600
50 98 101.04 9.2416 25 2500
60 121 120.8 0.04 25 3600
70 142 140.6 1.96 225 4900
80 158 160.4 5.76 625 6400
90 178 180.2 4.84 1225 8100
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Mean =
55
Mean =
109.625
Total
= 36.6016
Total =
4201
Total =
28400
Now, Sy/x = √(yi – yp)2/ n-2
= √36.6016/6 = 2.47
Then, SDslope = Sy/x / √(xi - mean)2
= 2.47 / √4201
= 0.038
While
SDintercept = Sy/x √(xi)2/n(xi - mean)2
= 2.47 x √ 28400/ 8 x 4201
= 2.27
thus, at 95% confidence interval for slope = t(n-2) x SDslope
= 2.447 x 0.038
= 0.085 mL ng-1
and,
at 95% confidence interval for ordinate intercept = t(n-2) x SDintercept
= 2.447 x 2.27
= 5.55 arbitrary units
From equation –
Fluorescence = (20.536 ± 5.5) + (19.798 ± 0.1) x concentration
Therefore, when Fluorescence = 100, Concentration = 38.1
Fluorescence = 0, Concentration = 12.6
Now, estimate 95% confidence interval of these fluorescence in following way –
(SD)x = Sy/x √1 + 1/n + (y – mean)2
(slope) (slope)2 (x-mean)2
Thus, for Fluorescence = 100 arbitrary units,
SDx = 2.47 √1 + 1/8 + (100 – 109.625)2
(19.798) (19.798)2 x 4201
= 0.324
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95% confidence interval is 2.447 x 0.324 = 0.79 ng mL-1
which gives:- 38.1 ± 0.324 ng mL-1
similarly, for fluorescence = 0, will get same value
therefore, 12.6 ± 0.324 ng mL-1
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Question 7
Given table -
Temperature
% weight loss
1 hour 3 hour 15 hour
80 C⁰
7.57 6.61 8.02
6.74 7.32 8.05
100 C⁰
7.03 7.72 8.40
6.55 7.59 8.65
120 C⁰
9.04 8.48 10.99
9.60 9.20 11.76
Construct in proper way as –
Temperature % weight loss
1 hour 3 hour 15 hour Ti (Ti)2
80 C⁰ 14.31 13.93 16.07 44.31 1963.3761
100 C⁰ 13.58 15.31 17.05 45.94 2110.4836
120 C⁰ 18.64 17.68 22.75 59.07 3489.2649
Tj 46.53 46.92 55.87 ΣTi = 149.32 ΣTi2=
7563.1246
(Tj)2 2165.0409 2201.4864 3121.4569 ΣTj2 =
7487.984
ΣXijk2 =
1273.552
While, ΣTij2 = 2544.1754
Calculations:
Correction term = (ΣTi)2 = (149.32)2 = 1238.69
n x r x c 2 x 3 x 3
Between row-sum of squares = ( ΣTi2) – correction term = 7563.1246– 1238.69
n x c 2 x 3
= 21.83 with 1 degrees of freedom
Between column-sum of squares = ( ΣTj2) – correction term = (7487.984) – 1238.69
n x r 2 x 3
= 9.307 with 2 degrees of freedom
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Total sum of squares = (Σ(Xijk))2 – correction term
= 1273.552 – 1238.69
= 34.862 with 2 x 3 x 3 – 1 = 17 degrees of freedom
Residual sum of squares –
(Σ(Xijk))2 – ( ΣTij)2
2
= 1273.552 – 2544.1754
2
= 1.4643
Therefore, two-way ANOVA table can be set up in following way –
Source Sum of squares Degrees of freedom Mean square
Between row 21.83 1 21.83
Between column 9.307 2 4.6535
Interaction 2.2607 2 1.13035
Residual 1.4643 12 0.122
Total 34.862 17
(b) Testing if between-temperature, mean square is significant
Null Hypothesis – The between-temperature, mean square is not statistically significant
Alternative Hypothesis – The between-temperature, mean square is statistically significant
Test statistic (F) = Between-temperature mean square = 21.83 = 178.93
Residual mean square 0.122
Critical value – The critical value for a one-sided F-test, with 1 and 17 degrees of freedom, is
4.41 at 95% confidence level.
Decision – As 178.93 > 4.41, so Null hypothesis is rejected
Conclusion – It has been concluded that between-temperature, mean square is statistically
significant at 95% confidence level.
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Testing between- weight loss mean square and residual mean square
Null Hypothesis – The between-weight loss, mean square is not statistically significant
Alternative Hypothesis – The between-weight loss, mean square is statistically significant
Test statistic (F) = Between-weight loss mean square = 4.6535 = 37.172
Residual mean square 0.122
Critical value – The critical value for a one-sided F-test, with 1 and 17 degrees of freedom, is
4.41 at 95% confidence level.
Decision – As 37.172 > 4.41, so Null hypothesis is rejected
Conclusion – It has been concluded that between-weight loss, mean square is statistically
significant at 95% confidence level.
(c) Reproducibility = t30 √(2) [sw2 + si2 + sb2]
where, sw2 = residual mean square = 0.122
si2 = interaction mean square – residual mean square
number of determinants in each cell
= 1.13035 – 0.122 = 0.336
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sb2 = between column-mean square – residual mean square
number of determinants under each cell
= 4.6535 – 0.122 = 0.775
6
So, Reproducibility = 2.04 √2 [0.122+0.336+0.775]
= 2.04 x 1.57
= 3.2
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Question 11
Sample [Nitrite] / mg L-1 Sample [Nitrite] / mg L-1
1 12.28 11 12.34
2 12.37 12 12.17
3 12.00 13 12.34
4 12.23 14 12.35
5 12.38 15 11.89
6 12.18 16 12.12
7 12.01 17 12.18
8 12.23 18 12.20
9 12.33 19 12.09
10 12.38 20 12.15
From this table –
Mean = 12.21
Standard deviation (SD) = 0.140109%
so, ¯x ± 2 x SD/√n
= 12.21 ± 2 x 0.140109/√1
= 12.49 and 11.93
and,
= ¯x ± 3 x SD/√n
= 12.21 ± 3 x 0.140109/√1
= 12.63 and 11.79
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This chart shows that there is no point which lie above warning points
(b)
(1) Given
k = ½ x SD = ½ x 0.14 = 0.07%
(2) Given
h = ±4 SD
= ±4 x 0.14
= ±0.56%
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(3) Construct a CUSUM Chart –
Target = 12.2, s = 0.14 and k = 0.07
Target 12.2
SD 0.14
K 0.07
Sample Xi (SH)i (SL)i
1 12.28 0 0
2 12.37 0.01 0
3 12 0.11 0
4 12.23 0 -0.13
5 12.38 0 -0.03
6 12.18 0.11 0
7 12.01 0.02 0
8 12.23 0 -0.12
9 12.33 0 -0.02
10 12.38 0.06 0
11 12.34 0.17 0
12 12.17 0.24 0
13 12.34 0.14 0
14 12.35 0.21 0
15 11.89 0.29 0
16 12.12 0 -0.24
17 12.18 0 -0.25
18 12.2 0 -0.2
19 12.09 0 -0.13
20 12.15 0 -0.17
Use values of (SH)i and (SL)i to construct CUSUM plot
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Here, Blue line or (SH)i shows positive shift from mean,
while, orange line or (SH)i shows negative shift from mean
As in above CUSUM chart, the centre line is at zero and cumulative sums is positively deviated
on high side from zero at Sample 20. However, all the results are lie within assigned action limit
(h). Henceforth, outcomes in CUSUM chart are in lie agreement with those that are shown in
above Shewhart control chart.
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