Question

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). One indicator of biodegradation is the level of dioxide in the water. Recall that 16 water specimens were randomly selected from various locations in a reservoir on the floor of a mine and the amount of dioxide (milligrams/liter) as well as presence of oil was determined for each specimen. These data are reproduced in the next table.

a. Estimate the true mean amount of dioxide present in water specimens that contain oil using a 95% confidence interval. Give a practical interpretation of the interval.

b. Repeat part a for water specimens that do not contain oil.

c. Based on the results, parts a and b, make an inference about biodegradation at the mine reservoir.

Short answer

a. The mean amount of dioxide present in water specimens that contain oil using a 95% confidence interval is $(0.0895,0.9439)$.One is 95% confident that true mean amount of dioxide present in water specimens is $(0.0895,0.9439)$.

b. The mean amount of dioxide present in water specimens that do not contain oil using a 95% confidence interval is $(1.4873,3.6927)$.

c. It is biodegradable.

Step-by-step solution

Given information

16 water specimens were randomly selected from various locations in a reservoir on the floor of a mine and the amount of dioxide (milligrams/liter) as well as presence of oil was determined for each specimen. These data are reproduced in the table.

Calculation of the true mean amount of dioxide present in water specimens that contain oil using a 95% confidence interval.

a)

Let x be the amount of dioxide present in water specimens.

Sample size 6

Confidence coefficient 95%=0.95

Mean Calculation: There 6 data values containing the dioxide present in the water-specimens.:0.5,1.3,0.4,0.2,0.5,0.2

The mean is

$\begin{array}{c}\overline{\mathrm{x}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\\ =\frac{1}{6}(0.5+1.3+0.4+0.2+0.5+0.2)\\ =\frac{3.1}{6}\\ =0.5167\end{array}$

Standard deviation calculation:

role="math" localid="1660736760291" ${\mathrm{s}}^2=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\frac{{\displaystyle \left(\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\right)}}{\mathrm{n}}}}{\mathrm{n}-1}$

$\begin{array}{c}\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2={0.5}^2+{1.3}^2+{0.4}^2+{0.2}^2+{0.5}^2+{0.2}^2\\ =2.43\end{array}$

$\begin{array}{c}{\mathrm{s}}^2=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}^2-\frac{{\displaystyle \left(\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\right)}}{\mathrm{n}}}}{\mathrm{n}-1}\\ =\frac{2.43-\frac{3{.41}^2}{6}}{6-1}\\ =0.1657\end{array}$

So sample standard deviation is

$\begin{array}{c}\mathrm{s}=\sqrt{{\mathrm{s}}^2}\\ =\sqrt{0.1657}\\ =0.4070\end{array}$

Confidence interval calculation

The critical value from t -table with degree of freedom 6-1=5 and $\alpha =5\%$is ${\mathrm{t}}_{\frac{\mathrm{\alpha }}{2}}=2.571$

Margin of error

$\begin{array}{c}\mathrm{E}={\mathrm{t}}_{\frac{\mathrm{\alpha }}{2}}\times \frac{\mathrm{s}}{\mathrm{n}}\\ =2.571\times \frac{0.4070}{\sqrt{6}}\\ =0.4272\end{array}$

95% lower bound $\overline{\mathrm{x}}-\mathrm{E}$= 0.5167-0.4272=0.0895

95% upper bound $\overline{\mathrm{x}}+\mathrm{E}=$0.5167+0.4272=0.9439

Hence, the mean amount of dioxide present in water specimens that contain oil using a 95% confidence interval is$(0.0895,0.9439)$ .One is 95% confident that true mean amount of dioxide present in water specimens is $(0.0895,0.9439)$.

Computation of part a for water specimens that do not contain oil.

b)

Let x be the amount of dioxide that do not present in water specimens.

Sample size 10

Confidence coefficient 95%=0.95

Mean:

Adding all data values we get

$\begin{array}{c}\sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}=3.3+0.1+4+0.3+2.4+2.4+1.4+4+4+4\\ =25.9\end{array}$

Mean

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}\\ =\frac{25.9}{10}\\ =2.59\end{array}$

Standard deviation

$\begin{array}{c}\sum _{\mathrm{i}=1}^{10}{\mathrm{x}}_{\mathrm{i}}^2=3{.3}^2+0{.1}^2+...+4^2\\ =88.47\end{array}$

S be standard deviation then

$\begin{array}{c}{\mathrm{s}}^2=\frac{{\displaystyle \sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}-\frac{\left(\sum _{\mathrm{i}=1}^{\mathrm{n}}{\mathrm{x}}_{\mathrm{i}}\right)}{\mathrm{n}}}}{\mathrm{n}-1}\\ =\frac{88.47-\frac{{25.9}^2}{10}}{10-1}\\ =2.3766\end{array}$

Hence the sample standard deviation is

$\begin{array}{c}\mathrm{s}=\sqrt{{\mathrm{s}}^2}\\ =\sqrt{2.3766}\\ =1.5416\end{array}$

Critical value:

The critical value from t -table with degree of freedom 10-1=9 and$\alpha =5\%$is${\mathrm{t}}_{\frac{\mathrm{\alpha }}{2}}=2.262$

Margin of error

role="math" localid="1660737368643" $\begin{array}{c}\mathrm{E}={\mathrm{t}}_{\frac{\mathrm{\alpha }}{2}}\times \frac{\mathrm{s}}{\sqrt{\mathrm{n}}}\\ =2.262\times \frac{1.1516}{\sqrt{10}}\\ =1.1027\end{array}$

Confidence interval

95% lower bound $\overline{\mathrm{x}}-\mathrm{E}=2.59-1.1027=1.4873$

95% upper bound $\overline{\mathrm{x}}+\mathrm{E}=2.59+1.1027=3.6927$

Hence 95%confidence interval that the amount of dioxide that do not present in water specimens is$(1.4873,3.6927)$

Inference about biodegradation at the mine reservoir.

c)

The mean amount of dioxide present in water specimens that contain oil using a 95% confidence interval is $(0.0895,0.9439)$.

95%confidence interval that the amount of dioxide that do not present in water specimens is$(1.4873,3.6927)$.

Value in the interval $(0.0895,0.9439)$are lower than the values in the interval$(0.0895,0.9439)$ .This implies true mean amount of dioxide present in water specimen containing oil is less than the true mean amount of dioxide present in the specimens not containing oil. Hence it is biodegradable.