A simple random sample of 25 observations was selected from a normal population. The mean and standard deviation of this sample are 20 and 5, respectively.

a. Test ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{\mu }\mathbf{=}\mathbf{22}$against ${\mathit{H}}_{\mathbf{a}}\mathbf{:}\mathit{\mu }\mathbf{\ne }\mathbf{22}$at the 10% significance level.

b. Test ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{\mu }\mathbf{\ge }\mathbf{22}$against ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{\mu }\mathbf{\le }\mathbf{22}$at the 1% significance level.

The sample size is, n=25.

The mean of the sample, $\overline{x}=20$.

The sample standard deviation, $s=5$.

The null hypothesis is a common statistical theory that asserts that no statistical link or significance exists between two sets of observable data and measurable events. The null hypothesis is helpful since it can be tested to see whether or not there is a link between two measurable occurrences. It can tell the user whether the results produced are a result of chance and manipulation of a phenomena.

The null and alternative Hypothesis are given as follows:

$$\begin{gathered} H_0:\mu =22 \\ {H}_a:\mu \ne 22 \end{gathered}$$

The significance level is 10%.

The test statistic is computed as:

$$\begin{gathered} t=\frac{\overline{x}-\mu }{s/\sqrt{n}} \\ =\frac{20-22}{5/\sqrt{25}} \\ =\frac{-2}{1} \\ =-2 \end{gathered}$$

This is a two-tailed test. So, the critical value of t corresponding to the degrees of freedom (n-1)=24 obtained from the t-table is 1.711.

Here, $\left|t\right|>t_{\frac{a}{2}}\Rightarrow \left|-2\right|>1.711$. So, we reject the null hypothesis.

Hence, the t-statistic is falls in to the rejection region. So, we reject the null hypothesis.