Respuesta :
SOLUTION:
Step 1 :
In this question, we are told that a marketing research company needs to estimate the average total compensation of CEOs in the service industry.
We also have that: Data were randomly collected from 38 CEOs and the 98% confidence interval was calculated to be ($2,181,260, $5,836,180).
Then, we are asked to find the margin error for the confidence interval.
Step 2:
We need to recall that:
[tex]\text{Higher Confidence Interval, CI}_{H\text{ = }}X\text{ + }\frac{Z\sigma}{\sqrt[]{n}}[/tex][tex]\text{Lower Confidence Interval , CI}_{L\text{ }}=\text{ X - }\frac{Z\sigma}{\sqrt[]{n}}[/tex]It means that:
[tex]\vec{}X\text{ = }\frac{CI_{H\text{ }}+CI_L}{2}[/tex][tex]\text{Margin of error, }\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{CI_{H\text{ - }}CI_L}{2}[/tex]where,
[tex]CI_H\text{ = }$5,836,180$\text{ }$$[/tex][tex]CI_{L\text{ }}=\text{ }2,181,260[/tex]putting the values into the equation for the margin of error, we have that:
[tex]\text{Margin of error,}\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{5,836,180\text{ - }2,181,260\text{ }}{2}[/tex][tex]\begin{gathered} =\text{ }\frac{3654920}{2} \\ =1,\text{ 827, 460} \end{gathered}[/tex]CONCLUSION:
The margin error for the confidence interval is 1, 827, 460
