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$\begin{array}{l}\stackrel{炉}{d}=\frac{{\displaystyle 鈭�d}}{n}\\ =\frac{12}{6}\\ =2\end{array}$

$\begin{array}{l}{s_d}^2=\frac{{\displaystyle 鈭�{(d鈭�\stackrel{炉}{d})}^2}}{n鈭�1}\\ =\frac{10}{5}\\ =2\end{array}$

Therefore, $\stackrel{炉}{d}=2$ and ${s_d}^2=2$.

${渭}_1$And ${渭}_2$ are the means of population 1 and 2.

So, ${渭}_d={渭}_1鈭�{渭}_2$

Here, $n=4$

So, the degree of freedom will be $=n鈭�1=5$

$伪=0.05$

From the t-table, the critical value at $5\%$a level of significance with a degree of freedom $5$ for a two-tailed test is $卤2.571$

The margin of error is,

$\begin{array}{l}ME=t_{伪/2}\left(\sqrt{\frac{{s_d}^2}{n}}\right)\\ =\left(2.571\right)\left(\sqrt{\frac{2}{6}}\right)\\ =1.4844\end{array}$

The Confidence Interval is,

$\begin{array}{l}\mathrm{CI}=\stackrel{炉}{\mathrm{d}}卤\mathrm{ME}\\ =\left(2\right)卤(1.4844)\\ =(0.516,3.484)\end{array}$

Here, $n=4$

So, the degree of freedom will be $=n鈭�1=5$

$伪=0.05$

From the t-table, the critical value at $5\%$ a level of significance with a degree of freedom $5$ for a two-tailed test is $卤2.571$

$\begin{array}{l}t=\frac{\stackrel{炉}{d}鈭�D_0}{\sqrt{\frac{{s_d}^2}{n}}}\\ =\frac{2鈭�0}{\sqrt{\frac{2}{6}}}\\ =3.46\end{array}$

Since$3.46>卤2.571$ so null hypothesis will be rejected. Therefore, it can be concluded that there is a significant difference between ${渭}_1$ and ${渭}_2$.