Two analysts each made six determinations of the paracetamol content of the same batch of tablets. The results are shown below:
Analyst Paracetamol content % (m/m)
A 84.32 84.51 84.63 84.61 84.64 84.51
B 84.24 84.25 84.41 84.13 84.00 84.30
Questions:
We can measure of precision in terms of standard deviation. Which statement is TRUE for the precision of results of both analysts?
A) Analyst A is more precise than Analyst B
B) Analyst B is more precise than Analyst A
C) Analyst A and Analyst B have the same precision
D) Both analysts have incorrect analyses E) We need more information to conclude on this matter
If a seventh analysis (Trial 7) was made in the above paracetamol analysis and the result was 93.48, can this be considered as an outlier at 90% confidence? The critical value (Q) for seven (7) observations is 0.507.
A) It is an outlier, reject, and do not include in the computation of mean.
B) It is not an outlier, do not reject, and include in the computation of mean.
Answer
To determine the precision of the results for each analyst, we can calculate the standard deviation for each set of data.
For Analyst A:
Standard Deviation = √((Σ(x - x̄)^2) / n)
where x represents each individual measurement, x̄ represents the mean of the measurements, and n represents the number of measurements.
Using the given data for Analyst A:
84.32, 84.51, 84.63, 84.61, 84.64, 84.51
Calculating the mean:
x̄ = (84.32 + 84.51 + 84.63 + 84.61 + 84.64 + 84.51) / 6
= 507.22 / 6
= 84.537
Calculating the standard deviation:
Standard Deviation = √(( (84.32 - 84.537)^2 + (84.51 - 84.537)^2 + (84.63 - 84.537)^2 + (84.61 - 84.537)^2 + (84.64 - 84.537)^2 + (84.51 - 84.537)^2) / 6)
= √(((-0.217)^2 + (0.027^2 + 0.093^2 + 0.073^2 + 0.103^2 + 0.027^2)) / 6)
= √((0.047 + 0.001 + 0.009 + 0.005 + 0.013 + 0.001) / 6)
= √(0.076 / 6)
= √0.012667
≈ 0.1126
For Analyst B:
Using the same method as above, we can calculate the standard deviation for Analyst B using the given data:
84.24, 84.25, 84.41, 84.13, 84.00, 84.30
Calculating the mean:
x̄ = (84.24 + 84.25 + 84.41 + 84.13 + 84.00 + 84.30) / 6
= 505.33 / 6
≈ 84.2217
Calculating the standard deviation:
Standard Deviation = √(( (84.24 - 84.2217)^2 + (84.25 - 84.2217)^2 + (84.41 - 84.2217)^2 + (84.13 - 84.2217)^2 + (84.00 - 84.2217)^2 + (84.30 - 84.2217)^2) / 6)
= √(((-0.0217)^2 + (0.0283^2 + 0.1883^2 + 0.0917^2 + 0.2217^2 + 0.0783^2)) / 6)
= √((0.000471 + 0.000801 + 0.035368 + 0.008405 + 0.049192 + 0.006133) / 6)
= √(0.10037 / 6)
= √0.016762
≈ 0.1294
Comparing the standard deviations:
The standard deviation for Analyst A is approximately 0.1126, and the standard deviation for Analyst B is approximately 0.1294.
Therefore, the statement that is TRUE for the precision of the results of both analysts is: A) Analyst A is more precise than Analyst B.
Now, let's determine if the seventh analysis (Trial 7) with a result of 93.48 can be considered an outlier at 90% confidence, given the critical value (Q) for seven observations is 0.507.
To calculate the outlier boundary:
Outlier Boundary = Q * (standard deviation of the data)
where Q is the critical value and the standard deviation is calculated considering all seven observations.
Using the data (including Trial 7):
84.32, 84.51, 84.63, 84.61, 84.64, 84.51, 93.48
Calculating the mean:
x̄ = (84.32 + 84.51 + 84.63 + 84.61 + 84.64 + 84.51 + 93.48) / 7
= 595.3 / 7
≈ 84.7571
Calculating the standard deviation:
Standard Deviation = √(( (84.32 - 84.7571)^2 + (84.51 - 84.7571)^2 + (84.63 - 84.7571)^2 + (84.61 - 84.7571)^2 + (84.64 - 84.7571)^2 + (84.51 - 84.7571)^2 + (93.48 - 84.7571)^2) / 7)
≈ √(( (0.4371^2 + 0.2471^2 + 0.1271^2 + 0.1471^2 + 0.1171^2 + 0.2471^2 + 8.7229^2) / 7)
≈ √(( 0.191 + 0.061 + 0.016 + 0.021 + 0.014 + 0.061 + 76.040) / 7)
≈ √(0.191 / 7)
≈ √0.02728
≈ 0.1653
Calculating the outlier boundary:
Outlier Boundary = 0.507 * 0.1653
≈ 0.0839
Since the value for Trial 7 (93.48) exceeds the outlier boundary (0.0839), it can be considered an outlier at 90% confidence.
Therefore, the statement: A) It is an outlier, reject, and do not include in the computation of mean, is correct.
To determine the precision of the results for both analysts, we need to calculate the standard deviation for each set of data. The standard deviation is a measure of how spread out the data points are from the mean.
For Analyst A:
- Calculate the mean of the paracetamol content for Analyst A: (84.32 + 84.51 + 84.63 + 84.61 + 84.64 + 84.51) / 6 = 84.5283
- Calculate the differences between each data point and the mean: (84.32 - 84.5283) = -0.2083, (84.51 - 84.5283) = -0.0183, (84.63 - 84.5283) = 0.1017, (84.61 - 84.5283) = 0.0817, (84.64 - 84.5283) = 0.1117, (84.51 - 84.5283) = -0.0183
- Square each difference: (-0.2083)^2 = 0.04329, (-0.0183)^2 = 0.000335, (0.1017)^2 = 0.01034, (0.0817)^2 = 0.006673, (0.1117)^2 = 0.01248, (-0.0183)^2 = 0.000335
- Calculate the sum of the squared differences: 0.04329 + 0.000335 + 0.01034 + 0.006673 + 0.01248 + 0.000335 = 0.073373
- Divide the sum by the total number of data points minus 1 (6-1): 0.073373 / 5 = 0.0146746
- Take the square root of the result to get the standard deviation: √(0.0146746) ≈ 0.121115
For Analyst B:
- Follow the same steps as above to calculate the standard deviation for Analyst B.
To compare the precision of the results of both analysts, we compare their standard deviations.
If the standard deviation of one analyst is smaller than the other, it suggests that their results are more consistent and, therefore, more precise.
Now, looking at the options for the first question:
A) Analyst A is more precise than Analyst B
B) Analyst B is more precise than Analyst A
C) Analyst A and Analyst B have the same precision
D) Both analysts have incorrect analyses
E) We need more information to conclude on this matter
To determine which statement is true, compare the standard deviations calculated for both analysts. If the standard deviation for Analyst A is smaller than the standard deviation for Analyst B, then the correct statement is A) Analyst A is more precise than Analyst B. If the standard deviations are equal, then the correct statement is C) Analyst A and Analyst B have the same precision.
Now, for the second question regarding the potential outlier (Trial 7) at a 90% confidence level:
To determine if Trial 7 is an outlier, we need to calculate the upper and lower values of the range where observations are not considered outliers.
For a 90% confidence level, we can use the critical value (Q) of 0.507. Since we have a total of 7 observations, we can calculate the range (R) where values are not considered outliers by multiplying the critical value by the standard deviation: R = Q * standard deviation.
If the value of Trial 7 falls within this range (mean ± R), it is not considered an outlier, and we can include it in the computation of the mean. Otherwise, if it falls outside of the range, it is considered an outlier, and we should exclude it.
So, calculate the mean for the 7 observations (including Trial 7) and calculate the range R by multiplying the critical value (Q) by the standard deviation. Then check if Trial 7 falls within the range (mean ± R) to determine if it is an outlier or not.
1. Calculate the standard deviation for each set and compare
2. Run the Dixon Q test for removing a single outlier and compare Qexpiment vs Q table value of 0.507. If Qexp > 0.507 the outlier may be removed.