Question

The report "2007 Electronic Monitoring \& Surveillance Survey: Many Companies Monitoring. Recording, Videotaping-and Firing-Employees" (American Management Association, 2007 ) summarized the results of a survey of 304 U.S. businesses. Of these companies, 201 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 \(60 \%\) 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

Yes, there is sufficient evidence to conclude that more than 60% of US businesses, as well as the majority of US businesses, monitor employees' web site visits.

Step-by-step solution

State the Hypotheses

For part (a): The null hypothesis is that 60% businesses monitor their employees' website visits, represented as \(H_0: p = 0.60\). The alternative hypothesis is that more than 60% of the businesses monitor the websites, represented as \(H_1: p > 0.60\). For part (b): The null hypothesis is that majority of businesses (50%) monitor their employees' website visits, represented as \(H_0: p = 0.50\). The alternative hypothesis is that majority of businesses monitor their employees' website visits, represented as \(H_1: p > 0.50\).

Analyze sample data

The sample proportion of businesses that monitor their employees' websites is calculated by dividing the number of such companies by the total number of the companies surveyed. In this case, the sample proportion (\(\hat{p}\)) = 201/304 = 0.6618.

Perform the Hypothesis Test with Significance Level 0.01

Use the formula for hypothesis test: Z = ( \(\hat{p}\) - p) / sqrt [ p(1 - p) / n ]. For part (a): Using the values from step 2, and p = 0.60 from the null Hypothesis Z = (0.6618 - 0.60) / sqrt [ 0.60(1 - 0.60) / 304 ] = 4.07. For part (b): Using the values from step 2, and p = 0.50 from the null Hypothesis, Z = (0.6618 - 0.50) / sqrt [ 0.50(1 - 0.50) / 304 ] = 5.96.

Make a decision

Compare the calculated Z scores from step 3 with the critical Z score for the significance level of 0.01, which is 2.33. For part (a): Since the calculated Z score (4.07) is greater than the critical Z score, reject the null hypothesis and conclude that there is enough evidence to suggest that more than 60% businesses monitor their employees' web visits. For part (b): Since the calculated Z score (5.96) is greater than the critical Z score, reject the null hypothesis and conclude that there is enough evidence to suggest that majority of US businesses monitor their employees' web visits.

Key concepts

Null Hypothesis

The null hypothesis is a statement that suggests there is no effect or no difference in a situation and it serves as the starting point for hypothesis testing. It's essentially the default position that assumes no change or relationship exists. In the context of hypothesis testing, it's the hypothesis that we seek evidence against. In the exercise given, for part (a), the null hypothesis (\( H_0: p = 0.60 \)) expresses the belief that 60% of U.S. businesses monitor their employees' website visits.
For part (b), the null hypothesis (\( H_0: p = 0.50 \)) indicates that 50% of businesses monitor their employees' website visits, representing at least half or majority of such endeavors. When conducting a hypothesis test, we initially assume the null hypothesis is true until evidence suggests otherwise. This approach allows us to remain objective, as it relies on sampling evidence to make a decision.

Alternative Hypothesis

The alternative hypothesis represents what we want to prove in the hypothesis testing process. It suggests that there is an effect or a difference. For both parts of the exercise, the alternative hypotheses point to a belief that more than 60% or 50% of businesses monitor their employees' website activities. Hence, it goes against the null. For part (a), the alternative hypothesis (\( H_1: p > 0.60 \)) states that the proportion of businesses monitoring employees is greater than 60%.
For part (b), it asserts that more than 50% of businesses undertake monitoring (\( H_1: p > 0.50 \)). From a statistical standpoint, aligning with the alternative hypothesis means we have found substantial evidence to support that there is indeed a significant difference or effect in the population.

Significance Level

The significance level, often denoted as alpha (\( \alpha \)), is a threshold for determining when to reject the null hypothesis. It's the probability of making a Type I error, which is rejecting a true null hypothesis, so it represents how much risk we are willing to take. In this exercise, the significance level is set at 0.01 or 1%. A lower value, like 0.01, means that there is a stricter criterion to meet before concluding that the results are statistically significant.
This level indicates a 1% risk that the test wrongly refutes the null hypothesis. When the calculated test statistic exceeds the critical value corresponding to the significance level, in this case, the Z-score threshold set by 0.01, we reject the null hypothesis. Choosing a significance level before testing helps ensure the procedure remains objective and centered on probability rather than subjective judgment.

Z-score

The Z-score is a statistical measure that tells us how far away a data point is from the mean in terms of standard deviations. It's used in hypothesis testing to determine how far the sample proportion is from the hypothesized population proportion under the null hypothesis. In the exercise, the Z-score is computed by using the formula:\[ Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]where \( \hat{p} \) is the sample proportion, \( p \) is the assumed population proportion under the null hypothesis, and \( n \) is the sample size.
In part (a), the sample proportion of 0.6618 resulted in a Z-score of 4.07, which is far from the mean of 0.60, indicating a significant deviation. Similarly, for part (b), the Z-score calculated was 5.96. These scores are compared with a critical Z-value at the 0.01 significance level to determine if the deviations are statistically significant or not, leading to an informed decision about the hypotheses.

Proportion Testing

Proportion testing refers to statistical tests specifically designed to determine if there is a difference between observed proportions in a sample and expected proportions in the population. In this exercise, proportion tests are performed on the survey data to check if the proportion who monitor employees exceeds a certain threshold. This involves calculating the sample proportion by dividing the number of businesses monitoring web visits by the total number surveyed, resulting in \( \hat{p} = 0.6618 \).
The next step involves comparing this sample proportion to a hypothesized population proportion stated in the null hypothesis (0.60 for part a, and 0.50 for part b) using the Z-test formula. If the test statistic exceeds the critical value at the 1% significance level, we conclude that there is enough evidence to reject the null, suggesting that the actual proportions of businesses monitoring employees are indeed greater than what the null hypotheses state. This methodology is widely used in fields where comparing sample data to known or assumed population properties is essential.