The data for a random sample of six paired observations are shown in the next table.

a. Calculate the difference between each pair of observations by subtracting observation two from observation 1. Use the differences to calculate $\overline{\mathbf{\text{d}}}{{\mathbf{\text{ands}}}_{\mathbf{\text{d}}}}^{\mathbf{\text{2}}}$.

b. If ${\mathbf{\text{μ}}}_{\mathbf{\text{1}}}{\mathbf{\text{andμ}}}_{\mathbf{\text{2}}}$are the means of populations 1 and 2, respectively, expressed ${\text{μ}}_{\text{d}}$in terms of ${\mathbf{\text{μ}}}_{\mathbf{\text{1}}}{\mathbf{\text{andμ}}}_{\mathbf{\text{2}}}$.

Pair	Sample from Population 1 (Observation 1)	Sample from Population 2(Observation 2)
$\begin{array}{l}\mathbf{\text{1}}\\ \mathbf{\text{2}}\\ \mathbf{\text{3}}\\ \mathbf{\text{4}}\\ \mathbf{\text{5}}\\ \mathbf{\text{6}}\end{array}$	$\begin{array}{l}\mathbf{\text{7}}\\ \mathbf{\text{3}}\\ \mathbf{\text{9}}\\ \mathbf{\text{6}}\\ \mathbf{\text{4}}\\ \mathbf{\text{8}}\end{array}$	$\begin{array}{l}\mathbf{\text{4}}\\ \mathbf{\text{1}}\\ \mathbf{\text{7}}\\ \mathbf{\text{2}}\\ \mathbf{\text{4}}\\ \mathbf{\text{7}}\end{array}$

c. Form a $\mathbf{95}\mathbf{\%}$ confidence interval for ${\mathbf{\text{μ}}}_{\mathbf{\text{d}}}$.

d. Test the null hypothesis ${\mathbf{\text{H}}}_{\mathbf{\text{0}}}{\mathbf{\text{:μ}}}_{\mathbf{\text{d}}}\mathbf{\text{=0}}$against the alternative hypothesis ${\mathbf{\text{H}}}_{\mathbf{\text{a}}}{\mathbf{\text{:μ}}}_{\mathbf{\text{d}}}\mathbf{\ne }\mathbf{\text{0}}$. Use$\mathbf{\text{α=.05}}$ .

$\begin{array}{l}\overline{d}=\frac{{\displaystyle \sum d}}{n}\\ =\frac{12}{6}\\ =2\end{array}$

$\begin{array}{l}{s_d}^2=\frac{{\displaystyle \sum {(d-\overline{d})}^2}}{n-1}\\ =\frac{10}{5}\\ =2\end{array}$

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

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

So, ${\mu }_d={\mu }_1-{\mu }_2$

Here, $n=4$

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

$\alpha =0.05$

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 $\pm 2.571$

The margin of error is,

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

The Confidence Interval is,

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

Here, $n=4$

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

$\alpha =0.05$

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 $\pm 2.571$

$\begin{array}{l}t=\frac{\overline{d}-D_0}{\sqrt{\frac{{s_d}^2}{n}}}\\ =\frac{2-0}{\sqrt{\frac{2}{6}}}\\ =3.46\end{array}$

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