Crude oil biodegradation.Refer to the Journal of Petroleum Geology (April 2010) study of the environmental factors associated with biodegradation in crude oil reservoirs, Exercise 2.29 (p. 85). Recall that amount of dioxide (milligrams/liter) and presence/absence of crude oil were determined for each of 16 water specimens collected from a mine reservoir. The data are repeated in the accompanying table.

Dioxide Amount	Crude Oil Present
3.3	No
0.5	Yes
1.3	Yes
0.4	Yes
0.1	No
4.0	No
0.3	No
0.2	Yes
2.4	No
2.4	No
1.4	No
0.5	Yes
0.2	Yes
4.0	No
4.0	No
4.0	No

a.Find the mean dioxide level of the 16 water specimens. Interpret this value.

b.Find the median dioxide level of the 16 water specimens. Interpret this value.

c.Find the mode of the 16 dioxide levels. Interpret this value.

d.Find the median dioxide level of the 10 water specimens with no crude oil present.

e.Find the median dioxide level of the 6 water specimens with crude oil present.

f.Compare the results, parts d and e. Make a statement about the association between dioxide level and presence/absence of crude oil.

$$\begin{gathered} Mean=\frac{Sumofallobservation}{Numberofobservation} \\ Mean=\frac{3.3+0.5+1.3+0.4+0.1+4.0+0.3+0.2+2.4+2.4+1.4+0.5+0.2+4.0+4.0+4.0}{16} \end{gathered}$$

Therefore, the mean dioxide level in 16 specimens of water is 1.8125.On average, the water in the mine reservoirs contains 1.8125 milligrams/liter of dioxide.

First, we arrange the data in ascending order,

(0.1, 0.2, 0.2, 0.3, 0.4, 0.5, 0.5, 1.3, 1.4, 2.4, 2.4, 3.3, 4.0, 4.0, 4.0, 4.0)

As there are 16 observations,

$$\begin{gathered} Median=\frac{Sumof8thand9thvalue}{2} \\ =\frac{1.3+1.4}{2} \\ =\frac{2.7}{2} \\ =1.35 \end{gathered}$$

Therefore the median value is 1.35, i.e., 8 reservoirs have dioxide levels below 1.35, and 8 have above 1.35.

Mode dioxide level is 4.0, as observed in 4 specimens out of 16. 4.0 milligrams/liter is the dioxide level commonly found in the reservoirs.

Values for samples with no crude oil present are arranged in ascending order.

(0.1, 0.3, 1.4, 2.4, 2.4, 3.3, 4.0, 4.0, 4.0, 4.0)

As the number of samples is 10,

$$\begin{gathered} Median=\frac{Sumoftwomidvalues}{2} \\ =\frac{2.4+3.3}{2} \\ =\frac{5.7}{2} \\ =2.85 \end{gathered}$$

Therefore, the median amount of dioxide in water specimens with no crude oil present is 2.85 milligrams/liter.

Organizing the data of 6 samples with crude oil present from smaller to bigger values,

(0.2, 0.2, 0.4, 0.5, 0.5, 1.3)

Due to the sample being even-numbered,

$$\begin{gathered} Median=\frac{Sumoftwomidvalues}{2} \\ =\frac{0.4+0.5}{2} \\ =\frac{0.9}{2} \\ =0.45 \end{gathered}$$

Therefore, the median is 0.45 milligrams/liter.

Looking at the medians of the water specimen where crude oil is present and specimens where it is absent, we can say that the quantity of dioxide present in the water is more when crude oil is not present. Because the median here (crude oil absent) is 2.85 milligrams/liter, half values will be less than 2.85, and half will be more than 2.85.

When crude oil is present in the water, the amount of dioxide is low. As the median is 0.45milligrams/liter, half the samples will be less than that, and half will be more than that. But as the median is small, the amount of dioxide will always be below.