Suppose you carry out a significance test of $H_0:\mu =5$versus $H_0:\mu >5$based on a sample of size n =20 and obtain t = 1.81.

(a) Find the P-value for this test using (i) Table B and (ii) your calculator. What conclusion would you draw at the 5% significance level? At the 1% significance level?

(b) Redo part (a) using an alternative hypothesis of$H_0:\mu \ne 5$

A p-value estimates the probability of obtaining the noticed outcomes, it is consistent with expectations that the invalid hypothesis. The lower the p-esteem, the more prominent the statistical significance of the noticed difference

the sample size n = $20$

t= $1.81$

the hypotheses are,

$\begin{array}{r}{H}_0:\mu =5\\ H_a:\mu >5\end{array}$

calculating the degree of freedom we have,

$$\begin{gathered} df=n-1 \\ =20-1=19 \end{gathered}$$

The p-value is,

$$\begin{gathered} \begin{array}{c}=P(t>|t\left|\right)\\ =P(t>|1.81\left|\right)\end{array} \\ =0.0431 \end{gathered}$$

The p-value is greater than the significance level $\alpha =0.01$and less than the significance level $\alpha =0.05$

the sample size n = $20$

t= $1.81$

the null and alternative hypotheses are,

$\begin{array}{r}{H}_0:\mu =5\\ H_a:\mu \ne 5\end{array}$

The degree of freedom is $19$

calculating the p value for two tailed test,

$=2\times P(t>|t\left|\right)$

$$\begin{gathered} \begin{array}{c}=2\times P(t>|1.81\left|\right)\\ =2\times 0.0431\end{array} \\ =0.086 \end{gathered}$$

Since the p-value is greater than the significance levels the null hypothesis is not rejected.

Hence there is sufficient evidence.