Question
2) The data in dodecanol.csv were collected in an interlaboratory comparison of a particular chemical technique to measure the hydroxyl value. The investigators prepared large
2) The data in dodecanol.csv were collected in an interlaboratory comparison of a particular chemical technique to measure the "hydroxyl" value. The investigators prepared large amounts of the chemical dodecanol and sent part of that large amount to each of 10 different labs. Each lab measured their sample four times. There are a total of 40 measurements. The purpose of an interlaboratory comparison is to quantify the variability between labs and the variability between measurements taken at the same lab.You are willing to assume that the within-lab variability is the same at each lab.
Aside, not necessary to answer my questions.Interlaboratory comparisons are commonly done, especially for hard-to-measure chemicals and environmental pollutants. One hopes that both the variability between labs and the variability between measurements at the same lab are small. If not, there are problems with the chemical procedure. Large variability between labs is especially problematic.
a) Estimate the variance components for labs and measurements.
b) Calculate the variance of a single observation, i.e. one measurement at one lab. c) Express the variance components as proportions (or percents) of the variability in a single observation. d) Estimate the mean hydroxyl value for dodecanol and its se.
e) Incorrectly analyze the data assuming that the 40 observations are independent. What is this estimated mean and se? f) All statistical models are approximations, because real data never satisfy the assumptions. E.g., real data never have equal variances and are never normally distributed. However, the effects of violating an assumption are often small. Looking at your results from parts, d) and e), does the assumption of independence matter here? In other words, do the analyses in parts d) and e) give you similar means? Do they give you similar s.e's? Where the results differ, which value is more appropriate? I.e., which one should you report? Briefly explain your choice of more appropriate value.
Please use JMP with the answer and complete explanation.
Data:
compound | lab | concentration |
dodecanol | 1 | 5 |
dodecanol | 1 | 7.6 |
dodecanol | 1 | 4.2 |
dodecanol | 1 | 6.4 |
dodecanol | 2 | 5.1 |
dodecanol | 2 | 1 |
dodecanol | 2 | 0.2 |
dodecanol | 2 | 0.2 |
dodecanol | 3 | 3.3 |
dodecanol | 3 | 4.1 |
dodecanol | 3 | 4.6 |
dodecanol | 3 | 2.2 |
dodecanol | 4 | 10.1 |
dodecanol | 4 | 9.9 |
dodecanol | 4 | 11.6 |
dodecanol | 4 | 14.4 |
dodecanol | 5 | 2.8 |
dodecanol | 5 | 1.7 |
dodecanol | 5 | 2.4 |
dodecanol | 5 | 2.6 |
dodecanol | 6 | 8.9 |
dodecanol | 6 | 7.9 |
dodecanol | 6 | 7.2 |
dodecanol | 6 | 6.5 |
dodecanol | 7 | 9.2 |
dodecanol | 7 | 9.7 |
dodecanol | 7 | 5.3 |
dodecanol | 7 | 7.8 |
dodecanol | 8 | 7.8 |
dodecanol | 8 | 8.8 |
dodecanol | 8 | 9.3 |
dodecanol | 8 | 7 |
dodecanol | 9 | 4.4 |
dodecanol | 9 | 5.2 |
dodecanol | 9 | 10.6 |
dodecanol | 9 | 6.4 |
dodecanol | 10 | 4.2 |
dodecanol | 10 | 2.9 |
dodecanol | 10 | 2.5 |
dodecanol | 10 | 3.6 |
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