91Ó°ÊÓ

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.

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

$\begin{array}{c}\stackrel{¯}{\mathrm{x}}=\frac{1}{\mathrm{n}}\underset{\mathrm{i}=1}{\overset{\mathrm{n}}{∑}}{\mathrm{x}}_{\mathrm{i}}\\ =\frac{1}{6}(0.5+1.3+0.4+0.2+0.5+0.2)\\ =\frac{3.1}{6}\\ =0.5167\end{array}$

Standard deviation calculation:

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

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

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

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 $\mathrm{Î\pm }=5\%$is ${\mathrm{t}}_{\frac{\mathrm{Î\pm }}{2}}=2.571$

Margin of error

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

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

95% upper bound $\stackrel{¯}{\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 $(0.0895,0.9439)$.

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

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

Mean

$\begin{array}{c}\stackrel{¯}{x}=\frac{{\displaystyle \underset{i=1}{\overset{10}{∑}}{x}_i}}{10}\\ =\frac{25.9}{10}\\ =2.59\end{array}$

Standard deviation

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

S be standard deviation then

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

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$\mathrm{Î\pm }=5\%$is${\mathrm{t}}_{\frac{\mathrm{Î\pm }}{2}}=2.262$

Margin of error

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

Confidence interval

95% lower bound $\stackrel{¯}{\mathrm{x}}−\mathrm{E}=2.59−1.1027=1.4873$

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

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

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.