Question

In exercise 10.117-10.122, the null hypothesis is $H_0:{\mu }_1={\mu }_2$and the alternative hypothesis is as specified. we have provided data from a simple random paired sample from the two population under consideration. in each case, use the paired $t$test to perform the required hypothesis test at the $10\%$significance level.

$H_{\alpha }:{\mu }_1>{\mu }_2$

Short answer

The null hypothesis is rejected.

Step-by-step solution

Given Information

Given in the question that,

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ {H}_{\alpha }:{\mu }_1>{\mu }_2 \end{gathered}$$

Level of significance is $0.1$we have to determine the paired t$t$-test for the given values.

Explanation

Let's compute the sample mean as follow:

$$\begin{gathered} \overline{d}=\frac{\sum d}{n} \\ =\frac{18}{10} \\ =1.8 \end{gathered}$$

Calculate the standard deviation :

$$\begin{gathered} s_d=\sqrt{\frac{\sum d_i^2-\frac{{\left(\sum d_i\right)}^2}{n}}{n-1}} \\ =\sqrt{\frac{138-\frac{(18{)}^2}{10}}{10-1}} \\ =3.4254 \end{gathered}$$

The formula of test statistics is

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

Substitute the given values

$$\begin{gathered} t=\frac{1.8}{\frac{3.4254}{\sqrt{10}}} \\ =1.66 \end{gathered}$$

The value of degree of freedom is:

$$\begin{gathered} dof=n-1 \\ =10-9 \\ =1 \end{gathered}$$

The critical value for the level of significance $0.1is1.383$

The value of test statistic is fall in the rejection region.

Therefore, the null hypothesis is rejected.