Question

The report "2005 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring, Recording, Videotaping-and Firing-Employees" (American Management Association, 2005) summarized the results of a survey of 526 U.S. businesses. Four hundred of these companies indicated that they monitor employees' web site visits. For purposes of this exercise, 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 \(.01\). 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 \(.01\).

Short answer

To answer part a, if the calculated Z value for part a is greater than 2.33, there is sufficient evidence to conclude that more than 75% of U.S. businesses monitor employees' website visits at a 0.01 significance level. For part b, if the calculated Z value for part b is greater than 2.33, there is sufficient evidence to conclude that more than 50% of U.S. businesses monitor employees' website visits at a 0.01 significance level.

Step-by-step solution

Identify the Null and Alternative Hypotheses for part a

The null hypothesis \(H_0\) is that 75% (or 0.75) of U.S. businesses monitor employees' web site visits. The alternative hypothesis \(H_a\) is that more than 75% (or 0.75) of U.S. businesses monitor employees' web site visits. \nSo, \(H_0: p = 0.75\) and \(H_a: p > 0.75\)

Calculate the Test Statistics for part a

The test statistic is given by the formula \(Z = \frac{{\hat{p} - p}}{{\sqrt{\frac{{pq}}{{n}}}}}\), where \(\hat{p}\) is the sample proportion, \(p\) and \(q\) are the hypothesised population proportion and its complement respectively, and \(n\) is the sample size. Substituting the known values, \(Z = \frac{{0.76 - 0.75}}{{\sqrt{\frac{{0.75 \times 0.25}}{{526}}}}\)

Make a Decision for part a

Compare the calculated Z value with the critical Z value from the Z table for a 0.01 significance level (which is 2.33 for a one-tailed test). If the calculated Z value is greater than the critical Z value, reject the null hypothesis.

Identify the Null and Alternative Hypotheses for part b

The null hypothesis \(H_0\) is that 50% (or 0.5) of U.S. businesses monitor employees' web site visits. The alternative hypothesis \(H_a\) is that more than 50% (or 0.5) of U.S. businesses monitor employees' web site visits. \nSo, \(H_0: p = 0.5\) and \(H_a: p > 0.5\)

Calculate Test Statistics for part b

Calculate the test statistic using the same formula as in part a, but replacing p with 0.5 and q with 0.5 as per the null hypothesis for part b. The test statistic becomes \(Z = \frac{{0.76 - 0.5}}{{\sqrt{\frac{{0.5 \times 0.5}}{{526}}}}\)

Make a Decision for part b

Just like in part a, compare the calculated Z value with the critical Z value from the Z table for a 0.01 significance level (again, which is 2.33 for a one-tailed test). If the calculated Z value is greater than the critical Z value, reject the null hypothesis.