SOLUTION
The formula for finding the confidence interval is:
[tex]CI=\bar{x}\pm Z^{}(\frac{\sigma}{\sqrt[]{n}})[/tex]mean=21.3
S.D=2.3
Sample size (n) =150
From theory, the critical value (Z) for a 99% confidence interval is 2.58
So Z=2.58
Substituting all these parameters into the confidence interval formula,
we will obtain:
[tex]CI=21.3\pm(2.58\times(\frac{2.3}{\sqrt[]{150}}))[/tex][tex]\begin{gathered} =21.3\pm(2.58\times0.18779) \\ =21.3\pm(0.4844982) \end{gathered}[/tex][tex]\begin{gathered} =21.3+0.4844982\text{ to 21.3-0.4844982} \\ =21.7844982\text{ to 20.8155018} \\ =21.78\text{ to 20.82 (to the nearest hundredth)} \end{gathered}[/tex]The final answer is 21.78 to 20.82 confidence interval.
