Sample size n = 23
Standard Deviation s = 16
Construct a 99% confidence interval
The expression for the confidence interval is :
[tex]\begin{gathered} \text{ Confidence Interval =}\bar{\text{ X}}\pm Z\frac{S}{\sqrt[]{n}}_{} \\ \text{ where X is the }mean,\text{ n is the sameple size, s= standard deviation} \\ Z=Chosen\text{ z value from the table} \end{gathered}[/tex]Substitute the value :
Z at the 99% level is Z = 2.576
[tex]\begin{gathered} \text{ Confidence Interval =}\bar{\text{ X}}\pm Z\frac{S}{\sqrt[]{n}}_{} \\ \text{ Confidence Interval=}\bar{\text{X}}\text{ }\pm(2.576)\frac{16}{\sqrt[]{23}} \\ \text{Confidence Interval=}\bar{\text{X}}\text{ }\pm8.594 \\ \text{Confidence Interval=(}\bar{\text{X}}\text{ +8.594),(}\bar{\text{X}}-8.594) \end{gathered}[/tex]
Answer :
[tex]\text{Confidence Interval=(}\bar{\text{X}}\text{ +8.594),(}\bar{\text{X}}-8.594)[/tex]
