Question

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

Answer 1

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{ }\frac{3654920}{2} \\ =1,\text{ 827, 460} \end{gathered}$

CONCLUSION:

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