91Ó°ÊÓ

Let ${\mathrm{μ}}_1$and ${\mathrm{μ}}_2$be the means of populations 1 and 2, respectively, and ${\mathrm{μ}}_d$be the difference between the means of two populations.

Null Hypothesis $\left(H_0\right)$: There is no difference between the two population means.

$H_0:{\mathrm{μ}}_dâ‰\yen 0({\mathrm{μ}}_1−{\mathrm{μ}}_2â‰\yen 0)$

Alternate Hypothesis $\left(H_0\right)$: The mean of population 2 is larger than population 1.

$H_a:{\mathrm{μ}}_d<0({\mathrm{μ}}_1−{\mathrm{μ}}_2<0)$

$\begin{array}{l}\stackrel{¯}{d}=\frac{{\displaystyle ∑d}}{n}\\ =\frac{−37}{10}\\ =−3.7\end{array}$

$\begin{array}{l}{s}_d=\sqrt{\frac{{\displaystyle ∑d^2−\frac{{\left({\displaystyle ∑d}\right)}^2}{n}}}{n−1}}\\ =\sqrt{\frac{181−\frac{{(−37)}^2}{10}}{9}}\\ =\sqrt{\frac{44.1}{9}}\\ =2.2136\end{array}$

Here,$n=10$

So, the degree of freedom will be $=n−1=9$

$\mathrm{Î\pm }=0.10$

From the t-table, the critical value at $10\%$the level of significance with a degree of freedom 9 is $1.383$.

$\begin{array}{l}t=\frac{\stackrel{¯}{d}}{\frac{s_d}{\sqrt{n}}}\\ =\frac{−3.7}{\frac{2.2136}{\sqrt{10}}}\\ =−5.29\end{array}$

$\left|t\right|=5.29$

Since$5.29>1.383$ so null hypothesis will be rejected. Therefore, it can be concluded that the mean of population 2 is larger than population 1.

Here, $n=10$

So, the degree of freedom will be $=n−1=9$

$\mathrm{Î\pm }=0.10$

From the t-table, the critical value at $10\%$the level of significance with a degree of freedom $9$is role="math" localid="1652708493593" $1.383$.

The margin of error is,

$\begin{array}{l}ME=t_{\mathrm{Î\pm }/2}\left(\frac{s_d}{\sqrt{n}}\right)\\ =\left(1.833\right)\left(\frac{2.2136}{\sqrt{10}}\right)\\ =1.2831\end{array}$

The Confidence Interval is,

$\begin{array}{l}\mathrm{CI}=\stackrel{¯}{\mathrm{d}}Â\pm \mathrm{ME}\\ =(−3.7)Â\pm (1.2831)\\ =(−4.98,−2.42)\end{array}$