396,231 views
25 votes
25 votes
A marketing research company needs to estimate the average total compensation of CEOs in the service industry. Data were randomly collected from 38 CEOs and the 98% confidence interval was calculated to be ($2,181,260, $5,836,180). The margin error for the confidence interval is

User Ramesh Sivaraman
by
2.9k points

1 Answer

26 votes
26 votes

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:


\text{Higher Confidence Interval, CI}_{H\text{ = }}X\text{ + }\frac{Z\sigma}{\sqrt[]{n}}
\text{Lower Confidence Interval , CI}_{L\text{ }}=\text{ X - }\frac{Z\sigma}{\sqrt[]{n}}

It means that:


\vec{}X\text{ = }\frac{CI_{H\text{ }}+CI_L}{2}
\text{Margin of error, }\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{CI_{H\text{ - }}CI_L}{2}

where,


CI_H\text{ = }$$5,836,180$\text{ }$$$
CI_{L\text{ }}=\text{ }2,181,260

putting the values into the equation for the margin of error, we have that:


\text{Margin of error,}\frac{Z\sigma}{\sqrt[]{n}\text{ }}\text{ = }\frac{5,836,180\text{ - }2,181,260\text{ }}{2}
\begin{gathered} =\text{ }(3654920)/(2) \\ =1,\text{ 827, 460} \end{gathered}

CONCLUSION:

The margin error for the confidence interval is 1, 827, 460

User Mirushaki
by
2.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.