Step 1
State the null and alternative hypothesis
[tex]\begin{gathered} H_o=7.89 \\ H_a>7.89 \end{gathered}[/tex]Step 2
State the p-value of the significance level.
[tex]\begin{gathered} \alpha=0.01 \\ p=\frac{\alpha}{2}=\frac{0.01}{2}=0.005 \end{gathered}[/tex]Step 3
Calculate the statistical test
[tex]\begin{gathered} n=15 \\ \mu(\operatorname{mean})=11.09 \\ \sigma(s\tan dard\text{ deviation)=4.86} \end{gathered}[/tex]The t-test formula is given as
[tex]t=\text{ }\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}}[/tex]Where
[tex]\begin{gathered} \bar{x}=\operatorname{mean} \\ \mu=theoretical\text{ }value \\ \end{gathered}[/tex][tex]\begin{gathered} t=\frac{7.89-11.09}{\frac{4.86}{\sqrt[]{15}}} \\ t=\frac{7.89-11.09}{1.254846604} \\ t=\frac{-3.2}{1.254846604} \\ t=-2.550112492 \end{gathered}[/tex]Step 4
Find the p-value from the t-test.
[tex]\text{The p-value from the t-test is 0.01}209[/tex]Step 5
Conclusion
The result is not significant at p<0.01. Therefore, the null hypothesis is rejected. It cannot be concluded that the mean in this area is higher than the national average because the p-value is greater than 0.01t
