Sewage and industrial pollutants dumped into a body of water can reduce the dissolved oxygen concentration and adversely affect aquatic species. In one study, weekly readings are taken from the same location in a river over a two-month period. The week 3 measurement in the data set is suspected of being an outlier. True or False: According to the Q test, the value at week 3 can be rejected at the 95% confidence level.
Week Dissolved O2 (ppm)
1 4.9
2 5.1
3 5.6
4 4.3
5 4.7
6 4.9
7 4.5
8 5.1

FALSE

Since 0.385 < 0.526, the value for week 3 is accepted.

Explanation:

Qexp = (|Xq - Xₙ₋₁|)/w

where Xq is the suspected outlier; Xₙ₋₁ is the next nearest data point; w is the range of data

First, the data are arranged in decreasing order, from highest to lowest:

3. 5.6

2. 5.1

8. 5.1

1. 4.9

6. 4.9

5. 4.7

7. 4.5

4. 4.3

Xq = 5.6; Xₙ₋₁ = 5.1; w = 5.6 - 4.3 = 1.3

Qexp = (|5.6 - 5.1|)/1.3 = 0.385

From tables, at 95% confidence level, for n = 8, Qcrit = 0.526

Since 0.385 < 0.526, the value for week 3 is accepted.

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )

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