Question

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotapingand Firing-Employees" (American Management Association, 2005) summarized a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet, and 131 had fired workers for e-mail misuse. Assume that the sample is representative of businesses in the United States. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of the Internet. b. What are two reasons why a \(90 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of e-mail would be narrower than the \(95 \%\) confidence interval calculated in Part (a)?

Short answer

a. The \(95 \%\) confidence interval for businesses firing their employees due to Internet misuse is \(0.22\) to \(0.30\), meaning there is \(95 \%\) confidence that the true proportion is within this range. b. A \(90 \%\) confidence interval would be narrower because it implies less certainty and uses a smaller Z-score, thereby resulting in a smaller margin of error.

Step-by-step solution

Compute the proportion and standard error

Firstly, find the proportion \(p\) of businesses that have fired workers for Internet misuse by dividing the number that have (137) by the total surveyed (526). This gives \(p = 137/526 = 0.26\). Now, calculate the standard error \(SE\) using the formula for the standard error of a proportion, \(SE = \sqrt{ p(1-p) / n }\), where \( n \) is the number of businesses. This results in \(SE = \sqrt{ 0.26 * (1 - 0.26) / 526 } = 0.02\).

Calculate the confidence interval

A \(95\%\) confidence interval for a proportion is given by \(p \pm 1.96 Se\), where \(1.96\) is the Z-score for \(95\%\) confidence level. Substituting the calculated values gives \(0.26 \pm 1.96 * 0.02\), yielding an interval of \(0.22\) to \(0.30\). This means there is \(95\%\) confidence that between \(22\%\) and \(30\%\) of all U.S. businesses have fired employees for Internet misuse.

Explain why a 90% confidence interval would be narrower

There are two main reasons why a \(90\%\) confidence interval for the proportion of businesses that have fired for e-mail misuse would be narrower than a \(95\%\) interval. Firstly, a lower confidence level implies less certainty about the range in which the true proportion lies, which leads to a narrower interval. Secondly, the Z-score corresponding to a \(90\%\) confidence level (\(1.645\) instead of \(1.96\)) is smaller, meaning that the margin of error (Z-score times the standard error) is also smaller.