91影视

$\begin{array}{|rrr|}\hline \text{Sample 1}& \text{Sample 2}& \text{Difference D}\\ 0.43& 0.415& 0.015\\ 0.266& 0.238& 0.028\\ 0.567& 0.39& 0.177\\ 0.531& 0.41& 0.121\\ 0.707& 0.605& 0.102\\ 0.716& 0.609& 0.107\\ 0.651& 0.632& 0.019\\ 0.589& 0.523& 0.066\\ 0.469& 0.411& 0.058\\ 0.723& 0.612& 0.111\\ \hline\end{array}$

Mean $=0.0804$

Sd$=0.0523$

The mean is the sum of all values divided by the number of values:

$$\begin{gathered} \stackrel{炉}{x}=\frac{0.015+0.028+鈥�+0.058+0.111}{10} \\ 鈮�0.0804 \end{gathered}$$

$n$is the number of values in the data set.

The variance is the sum of squared deviations from the mean divided by $n-1$:

localid="1650087335628" $$\begin{gathered} s^2=\frac{(0.015-0.0804{)}^2+鈥�.+(0.111-0.0804{)}^2}{8-1} \\ 鈮�0.0025 \end{gathered}$$

The standard deviation is the square root of the variance:

localid="1650087345303" $$\begin{gathered} s=\sqrt{0.0025} \\ 鈮�0.0523 \end{gathered}$$

Determine the t-value by looking in the row starting with degrees of freedom $n-1=10-1=9$and in the row with $c=95\%$in table B:

$t^{*}=2.262$

The margin of error is then:

localid="1650087364022" $$\begin{gathered} E=t^{*}路\frac{s}{\sqrt{n}} \\ =2.262脳\frac{0.0523}{\sqrt{10}} \\ 鈮�0.0374 \end{gathered}$$

Then the confidence interval becomes:

localid="1650087403359" $$\begin{gathered} 0.0430=0.0804-0.0374 \\ =\stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E \\ =0.0804+0.0374 \\ =0.1178 \end{gathered}$$

$\begin{array}{|rrr|}\hline \text{Sample 1}& \text{Sample 2}& \text{Difference D}\\ 0.43& 0.415& 0.015\\ 0.266& 0.238& 0.028\\ 0.567& 0.39& 0.177\\ 0.531& 0.41& 0.121\\ 0.707& 0.605& 0.102\\ 0.716& 0.609& 0.107\\ 0.651& 0.632& 0.019\\ 0.589& 0.523& 0.066\\ 0.469& 0.411& 0.058\\ 0.723& 0.612& 0.111\\ \hline\end{array}$

Mean $=0.0803$

Sd$=0.0523$

The mean is the sum of all values divided by the number of values:

$$\begin{gathered} \stackrel{炉}{x}=\frac{0.015+0.028+鈥�+0.058+0.111}{10} \\ 鈮�0.0804 \end{gathered}$$

$n$is the number of values in the data set.

$n=10$

The variance is the sum of squared deviations from the mean divided by $n-1$:

localid="1650087429813" $$\begin{gathered} s^2=\frac{(0.015-0.0804{)}^2+鈥�..+(0.111-0.0804{)}^2}{8-1} \\ 鈮�0.0025 \end{gathered}$$

The standard deviation is the square root of the variance:

localid="1650087440721" $$\begin{gathered} s=\sqrt{0.0025} \\ 鈮�0.0523 \end{gathered}$$

Determine the t-value by looking in the row starting with degrees of freedom $n-1=10-1=9$and in the row with $c=95\%$in table B:

$t^{*}=2.262$

The margin of error is then:

localid="1650087455657" $$\begin{gathered} E=t^{*}路\frac{s}{\sqrt{n}} \\ =2.262脳\frac{0.0523}{\sqrt{10}} \\ 鈮�0.0374 \end{gathered}$$

Then the confidence interval becomes:

localid="1650087477978" $$\begin{gathered} 0.0430=0.0804-0.0374 \\ =\stackrel{炉}{x}-E<渭<\stackrel{炉}{x}+E \\ =0.0804+0.0374 \\ =0.1178 \end{gathered}$$

We are $95\%$confident that the true population mean difference is between $0.0430$and$0.1178$.

The confidence interval does not contain $0$, which indicates that there is a difference between the population means and thus there is sufficient evidence to support the claim of a difference in zinc concentrations at the top and bottom of wells in the region.