Two analysts each made six determinations of the paracetamol content of the same batch of tablets. The results are shown below:
ANALYST A:
Paracetamol content % (m/m) = 84.32, 84.51, 84.63, 84.61, 84.64, 84.51
ANALYST B:
Paracetamol content % (m/m) = 84.24, 84.25, 84.41, 84.13, 84.00, 84.30
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.
So EACH analyst does a 7th replicate and obtains 93.48% and we want to know for EACH if that is an outlier; i.e., two answers?
To determine if the seventh analysis result (Trial 7) of 93.48 can be considered an outlier at 90% confidence, we first need to find the mean and standard deviation of the data set including Trials 1-6.
1. Calculate the mean:
- Add up all the values from the first six trials:
84.32 + 84.51 + 84.63 + 84.61 + 84.64 + 84.51 = 507.22
- Divide the sum by the number of observations (6):
507.22 / 6 = 84.54
2. Calculate the standard deviation:
- Calculate the squared difference for each trial by subtracting the mean from each value and squaring the result:
(84.32 - 84.54)^2, (84.51 - 84.54)^2, (84.63 - 84.54)^2, (84.61 - 84.54)^2, (84.64 - 84.54)^2, (84.51 - 84.54)^2
- Sum up all the squared differences:
0.0036 + 0.0009 + 0.00729 + 0.0049 + 0.00016 + 0.0009 = 0.01775
- Divide the sum by the number of observations minus 1 (since this is a sample):
0.01775 / (6-1) = 0.00355
- Take the square root of the result to obtain the standard deviation:
√0.00355 ≈ 0.05958
3. Calculate the maximum allowed difference from the mean at 90% confidence:
- Multiply the standard deviation by the critical value (Q) for seven observations, given as 0.507:
0.507 * 0.05958 ≈ 0.03025
4. Determine if Trial 7 (93.48) is an outlier:
- Compare the absolute difference between Trial 7 and the mean to the maximum allowed difference:
|93.48 - 84.54| ≈ 8.94
- If the absolute difference is greater than the maximum allowed difference, it is considered an outlier. In this case, 8.94 > 0.03025.
Conclusion:
Since the absolute difference between 93.48 and the mean (84.54) is greater than the maximum allowed difference at 90% confidence, we can consider it an outlier. Therefore, the correct answer is:
A) It is an outlier, reject, and do not include in the computation of the mean.