Two analysts measure the percentage of ammonia in a chemical process over 9 days and find the following discrepancies between their results: Day 1 2 3 4 5 6 7 8 9 Analyst A 12.04 12.37 12.35 12.43 12.34 12.36 12.48 12.33 12.33 Analyst B 12.18 12.37 12.38 12.36 12.47 12.48 12.57 12.28 12.42 Investigate the mean discrepancy between their results and in particular give an interval in which you are 90% sure that it lies.

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Two analysts measure the percentage of ammonia in a chemical process over
9 days and find the following discrepancies between their results:
Day
1
2
3
4
5
6
7
8
9
Analyst A 12.04 12.37 12.35 12.43 12.34 12.36 12.48 12.33 12.33
Analyst B 12.18 12.37 12.38 12.36 12.47 12.48 12.57 12.28 12.42
Investigate the mean discrepancy between their results and in particular give
an interval in which you are 90% sure that it lies.
The answer I got is. Can you please interpret the answer?
1. Mean difference w 0.053, S = 0.0498, s = 0.0789. Assuming a standard
reference prior for and a variance known equal to 0.07892, the posterior distribution
of the effect of Analyst A over Analyst B is N(0.053, 0.07892/9) leading to a 90%
interval 0.053 ± 1.6449 × 0.0789/3, that is, (0.010, 0.097). If the variance not
assumed known then the normal distribution should be replaced by tg.
Transcribed Image Text:This is question. Two analysts measure the percentage of ammonia in a chemical process over 9 days and find the following discrepancies between their results: Day 1 2 3 4 5 6 7 8 9 Analyst A 12.04 12.37 12.35 12.43 12.34 12.36 12.48 12.33 12.33 Analyst B 12.18 12.37 12.38 12.36 12.47 12.48 12.57 12.28 12.42 Investigate the mean discrepancy between their results and in particular give an interval in which you are 90% sure that it lies. The answer I got is. Can you please interpret the answer? 1. Mean difference w 0.053, S = 0.0498, s = 0.0789. Assuming a standard reference prior for and a variance known equal to 0.07892, the posterior distribution of the effect of Analyst A over Analyst B is N(0.053, 0.07892/9) leading to a 90% interval 0.053 ± 1.6449 × 0.0789/3, that is, (0.010, 0.097). If the variance not assumed known then the normal distribution should be replaced by tg.
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