Question

The report "2007 Electronic Monitoring and Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2007) summarized a survey of 304 U.S. businesses. Of these companies, 201 indicated that they monitor employees' web site visits. Assume that it is reasonable to regard this sample as representative of businesses in the United States. a. Is there sufficient evidence to conclude that more than \(75 \%\) of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of 0.01 . b. Is there sufficient evidence to conclude that a majority of U.S. businesses monitor employees' web site visits? Test the appropriate hypotheses using a significance level of 0.01 .

Short answer

There is not sufficient evidence to conclude that more than \(75\%\) of U.S. businesses monitor employees' web site visits. However, there is sufficient evidence to conclude that a majority of U.S. businesses do monitor employees' web site visits.

Step-by-step solution

Set Up the Hypotheses

In hypothesis testing, we always have two contradicting hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). For part (a) of our problem, we want to test if more than \(75\%\) of businesses control employees' website visits, so: \(H_0: p = 0.75\) and \(H_a: p > 0.75\). For part (b), we test if a majority of businesses, i.e., more than \(50\%\), control visits, so: \(H_0: p = 0.50\) and \(H_a: p > 0.50\). p refers to the proportion of all US businesses that monitor their employees' Internet usage.

Calculate Test Statistic and P-value

We determine the test statistic using the formula \(Z = (p - p_0) / \sqrt{(p_0 * (1 - p_0))/n}\). Where \(n\) is the sample size, \(p\) is the sample proportion, and \(p_0\) is the proportion under the null hypothesis. Using the sample data, for (a) we get \(Z1 = (201/304 - 0.75) / \sqrt{(0.75 * (1 - 0.75))/304} = -1.13\), and for (b), \(Z2 = (201/304 - 0.50) / \sqrt{(0.5 * (1 - 0.5))/304} = 10.26\). Then for each case, we calculate the p-value, which is the probability of getting a Z value as extreme as ours assuming the null hypothesis is true. p-value is given by \(P(Z > Z_{observed})\) for a right-tailed test.

Compare P-value to Significance Level

In each case, compare the calculated p-value to the provided significance level, \(\alpha\). If p-value is less than or equal to \(\alpha\), we reject the null hypothesis. If p-value is greater than \(\alpha\), we don't have enough evidence to reject the null hypothesis. Special tables (or computer software) give us the probability that Z is more than any given value.

Make Conclusion

Based on results from steps 2 and 3, make a conclusion for both parts. In part (a), as the p-value for Z1 would be larger than 0.01, we fail to reject the null hypothesis. We do not have enough evidence to support that more than \(75\%\) of U.S. businesses monitor employees' website visits. For part (b), since the p-value for Z2 will be less than 0.01, we reject the null hypothesis and conclude that the majority of U.S. businesses monitor their employees' web visits.