91Ó°ÊÓ

T is given:

$$\begin{gathered} n=10 \\ c=0.95 \end{gathered}$$

The mean is:

$$\begin{gathered} \stackrel{¯}{x}=\frac{{\displaystyle \underset{i-1}{\overset{n}{∑}}{x}_i}}{n} \\ =\frac{0.015+0.028+0.177+0.121+0.102+0.107+0.019+0.066+0.058+0.111}{10} \\ =\frac{0.804}{10} \\ =0.0804 \end{gathered}$$

The sample variance is then as:

$$\begin{gathered} s^2=\frac{\underset{}{∑}{(x-\stackrel{¯}{x})}^2}{n-1} \\ =\frac{{(0.015-0.0804)}^2+(0.028-0.0804{)}^2+(0.177-0.0804{)}^2+(0.121-0.0804{)}^2+(0.102-0.0804{)}^2+(0.107-0.0804{)}^2+(0.019-0.0804{)}^2+(0.066-0.0804{)}^2+(0.058-0.0804{)}^2+(0.111-0.0804{)}^2}{10-1} \\ =0.0027 \end{gathered}$$

The sample standard deviation is then,

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{0.0027} \\ =0.0523 \end{gathered}$$

Now, the degree of freedom will be:

$$\begin{gathered} df=n-1 \\ =10-1 \\ =9 \end{gathered}$$

Now, the value of t will be:

$t_{\mathrm{Î\pm }/2}=2.262$

Now, the confidence interval will be:

$$\begin{gathered} \stackrel{¯}{x}-t_{\mathrm{σ}/2}×\frac{s}{\sqrt{n}} \\ =0.0804-2.262×\frac{0.0523}{\sqrt{10}} \\ =0.0430 \\ \stackrel{¯}{x}+t_{\mathrm{σ}/2}+\frac{s}{\sqrt{n}} \\ =0.0804+2.262×\frac{0.0523}{\sqrt{10}} \\ =0.1178 \end{gathered}$$

Thus we conclude that we are$95\%$ confident that the true mean zinc concentration at the bottom of the well is between$0.0430and0.1178$ higher than the true mean zinc concentration at the top of the well.