Statistics and Chemometrics Tutorial Answers Final PDF

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Statistics and Chemometrics Tutorial Answers 1. Confidence Limits The sodium ion content of a urine sample was determined by by using an ion-selective electrode. The following values were obtained: 102, 97, 99, 98, 101 and 106 mM. 1. What are the mean and standard deviation of these data? Mean = 100.5 mM, standard deviation = 3.27 mM 2. What are the 95% confidence limits for the sodium ion concentration? CL = 100.5  2.57 x 3.27/6 CL = 100.5  3.4 mM 3. What are the 99% confidence limits for the sodium ion concentration? CL = 100.5  4.03 x 3.27/6 CL = 100.5  5.4 mM 2. Paired T-Test Tablet 1 2 3 4 5 6 7 8 9 10

UV spectroscop y 84.63 84.38 84.08 84.41 83.82 83.55 83.92 83.69 84.06 84.03

IR reflectanc e 83.15 83.72 83.84 84.2 83.92 84.16 84.02 83.6 84.13 84.24

Difference, d 1.48 0.66 0.24 0.21 -0.1 -0.61 -0.1 0.09 -0.07 -0.21

_ mean value of d, d = 0.159 standard deviation of d, s = 0.570 tcalc = (0.159)2×10 / 0.570 = 0.882 t95% = 2.26 tcalc < t95% therefore the two methods are comparable; there is no significant difference between results. 3. T-Test

tcalc = (50.08−50.0)2/(0.02/8) = 11.3 t95% = 2.36 tcalc > t95% therefore there is a systematic error; method is unreliable. 4. Outliers Four gas chromatographic assays on a sample of pentachlorophenol were performed under the same conditions are gave the following results: 88.2%, 86.7%, 87.9% and 87.7%. 1. Calculate the mean and standard deviation for these data. Comment on the precision of the data. Mean = 87.63%, standard deviation = 0.65% Good precision; standard deviation less than 1% of measured value. 2. Calculate the mean and standard deviation for these data having discarded the outlier. Does this lead to an improvement in the measurement precision? Mean = 87.93%, standard deviation = 0.25% Improved precision 3. Calculate the Q value for this outlier and determine whether you can justify discarding it statistically. Qcalc = (86.7−87.7)2/(88.2−86.7) = 0.667 Q95% = 0.829 Qcalc < Q95% therefore do not discard 86.7%. 4. Subsequent measurements of 87.8%, 88.1% and 88.0% were obtained. Should the value of 86.7% be considered an outlier? Qcalc = (86.7−87.7)2/(88.2−86.7) = 0.667 Q95% = 0.568 Qcalc > Q95% therefore 86.7% can now classed as an outlier; discard value. 5. Theil’s Method y = 0.0204x + 0.004

For the full worked solution to this example see pages 173 to 174 in:

J N Miller and J C Miller, (2010) Statistics and Chemometrics for Analytical Chemistry, 6th Ed, Prentice Hall. For earlier editions look up Theil’s method in the index....