Answer:
Last option
[tex]{\displaystyle {\overline {x}}} \± \frac{2.58*s}{\sqrt{n}}[/tex]
Step-by-step explanation:
[tex]{\displaystyle {\overline {x}}}[/tex]
If we extract a sample of size n from a normal population and we want to estimate the population mean then the confidence interval for the mean will be:
[tex]{\displaystyle {\overline {x}}} \± Z_{\alpha/ 2}*\frac{s}{\sqrt{n}}[/tex]
Where
Where [tex]{\displaystyle {\overline {x}}}[/tex] is an estimator for the mean μ.
n is the sample size and s is its standard deviation.
[tex]\alpha[/tex] is the level of significance
[tex]\alpha[/tex]= 1-Confidence Level
[tex]\alpha= 1-0.99[/tex]
[tex]\alpha = 0.01[/tex]
Then [tex]Z_{\frac{\alpha}{2}} = Z_{\frac{0.01}{2}} = Z_{0.005}=2.58[/tex]
Then the 99% confidence interval for the population mean is
[tex]{\displaystyle {\overline {x}}} \± \frac{2.58*s}{\sqrt{n}}[/tex]
