Question

The Eks Survey Company employs 2000 people to conduct telephone surveys. Because many people don't like to answer such surveys, many "hang-ups" (whereby the person hangs up without completing the survey) occur. The owner of Eks wants to determine the mean number of "hang-ups" per employee on a particular day, using 95% confidence level. He samples 36 employees, and finds that the mean number of "hang-ups" on that day was 38.6. Suppose that the standard deviation of the number of "hang-ups" for all employees is 20.4 What is the value of the margin of error? (round to four decimal places)

Answer 1

Use the next formula to find the margin or error:

$MOE=z*\frac{\sigma}{\sqrt{n}}$

z is the critical value (For a 95% confidence level the z-score is 1.95998)

σ is the standard deviation

n is the sample size

$\begin{gathered} MOE=1.95998*\frac{20.4}{\sqrt{36}} \\ \\ MOE\approx6.6639 \end{gathered}$

Then the margin of error is: 6.6639

$\begin{gathered} \bar{x}\pm MOE \\ \\ 38.6\pm6.6639 \end{gathered}$