Question

$H_a*{\mu }_1<{\mu }_2$

Short answer

Since the value of the test statistic is fall in the rejection region. Thus, the null hypothesis is rejected.

Step-by-step solution

Given Information

The hypothesis is

$$\begin{gathered} H_0:{\mu }_1={\mu }_2, \\ {H}_a:{\mu }_1<{\mu }_2 \end{gathered}$$

and the level of significance is$0.1$.

Explanation

Let's compute the sample mean as follow

$\overline{d}=\frac{\sum d}{n}$

$=\frac{-21}{9}$

$=-2.333$.

Compute the value of standard deviation:

$s_d=\sqrt{\frac{\sum d_i^2-\frac{{\left(\sum d_i\right)}^2}{n}}{n-1}}$

$$\begin{gathered} =\sqrt{\frac{121-\frac{(-21{)}^2}{9}}{9-1}} \\ =3 \end{gathered}$$

The test statistics are,

$t=\frac{\overline{d}}{\frac{s_d}{\sqrt{n}}}$

Substitute the given values in the above equation.

$$\begin{gathered} t=\frac{-2.333}{\frac{9}{\sqrt{9}}} \\ =-2.333 \end{gathered}$$

The value of degree of freedom is,

$$\begin{gathered} dof=n-1 \\ =9-1 \\ =8 \end{gathered}$$

The critical value for the level of significance of $0.1$ is $-1.397.$