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 6.38 (p. 350). Recall that 16 water specimens were randomly selected from various locations in a reservoir on the floor of a mine and that the amount of dioxide (milligrams/liter)—a measure of biodegradation—as well as presence of oil were determined for each specimen. These data are reproduced in the accompanying table.

a. Conduct a test to determine if the true mean amount of dioxide present in water specimens that contained oil was less than 3 milligrams/liter. Use\(\alpha = .10\)

The null and alternative hypothesis are given by

\(\begin{aligned}{H_0}:\mu = 3\\{H_a}:\mu < 3\end{aligned}\)

a) The mean and standard deviation is calculated as

\(\begin{aligned}\bar x &= \frac{{0.5 + 1.3 + 0.4 + 0.2 + 0.5 + 0.2}}{6}\\ &= \frac{{3.1}}{6}\\ &= 0.51\end{aligned}\)

\(\begin{aligned}sd &= \sqrt {\frac{{{{\left( { - 0.01} \right)}^2} + {{\left( {0.79} \right)}^2} + {{\left( { - 0.11} \right)}^2} + {{\left( { - 0.31} \right)}^2} + {{\left( { - 0.01} \right)}^2} + {{\left( { - 0.31} \right)}^2}}}{5}} \\ &= \sqrt {\frac{{0.0001 + .6241 + 0.0121 + 0.0961 + 0.0001 + 0.0961}}{5}} \\ &= \sqrt {\frac{{0.9375}}{5}} \\ &= 0.433\end{aligned}\)

Therefore, the mean and standard deviation are 0.51 and 0.433.

The test statistic is calculated as

\(\begin{aligned}t &= \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\\ &= \frac{{0.51 - 3}}{{\frac{{0.433}}{{\sqrt 6 }}}}\\ &= \frac{{ - 2.49}}{{0.176}}\\ &= - 14.14\end{aligned}\)

Therefore, the test statistic is -14.14.

Degrees of freedom are

\(\begin{aligned}df &= n - 1\\ &= 6 - 1\\ &= 5\end{aligned}\)

For \(\alpha = .10\,and\,df = 5\)

The tabulated value is -1.47.

The calculated value is less than the tabulated value.

Therefore, we reject the null hypothesis.