Question

$H_2:{\mu }_1>{\mu }_2$

Short answer

Since the value of the test statistic is fall in the accepted region. Thus, the null hypothesis is not 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: Step 2

Let's compute the sample mean

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

$=\frac{1}{8}$

$=0.125$

The value of the standard deviation is:

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

$=\sqrt{\frac{113-\frac{(1{)}^2}{8}}{8-1}}$

$=4.0156$

The test statistics are,

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

Replace the above value in the given equation

$z=\frac{0.125}{\frac{4.0156}{\sqrt{8}}}$

$=0.09$

The value of degree of freedom is,

$df=n-1$

$$\begin{gathered} =8-1 \\ =7 \end{gathered}$$

The critical value for the level of significance of $0.1$is $-1.44.$