Chapter 10.3, Problem 40E

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.

1. 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 0.01.
2. 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.

To determine

Check whether the sample data provide convincing evidence that the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.6.

Answer to Problem 40E

No, the sample data do not provide the evidence that the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.6.

Explanation of Solution

Calculation:

In a sample of 304 businesses of Country U, 201 businesses were found to answer that they monitor employees’ web site visits. It is assumed that the sample of business is the representative of all the businesses of Country U.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of Country U’s businesses that monitor employees’ web site visits.

Step 2:

Null hypothesis: $H_0:p=0.6$

That is, the proportion of Country U’s businesses that monitor employees’ web site visits is 0.6.

Step 3:

Alternative hypothesis: $H_a:p>0.6$

That is, the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.6.

Step 4:

Significance level $\alpha$:

It is given that the significance level $\alpha =0.01$.

Step 5:

Test statistic z:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$,

Where, $\widehat{p}$ is the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion $p=0.6$ in the test statistic.

$z=\frac{\widehat{p}-0.6}{\sqrt{\frac{0.6\left(1-0.6\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is not known.

Step 6:

Assumptions:

• Let $\widehat{p}$ be the sample proportion from a random sample.
• The large sample z test can be used if the sample size n satisfies the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.
• The sample size should not be greater than 10% of the population size.

Requirement check:

• It is assumed that the sample is from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =304\left(0.6\right)\\ =182.4\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =304\left(1-0.6\right)\\ =304\left(0.4\right)\\ =121.6\\ >10\end{array}$

Since $n\left(\text{hypothesized value}\right)$ and $n\left(1-\text{hypothesized value}\right)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

• Although, the population size is not known, it is reasonable to assume that the sample size of 304 acts as the representative for all the businesses of Country U. It is also definite that the sample size is less than 10% of population of the business of Country U.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-0.6}{\sqrt{\frac{0.6\left(1-0.6\right)}{n}}}$

The sample proportion $\widehat{p}$ is obtained as follows:

$\widehat{p}=\frac{x}{n}$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion:

$\begin{array}{c}\widehat{p}=\frac{201}{304}\\ =0.66\end{array}$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$\begin{array}{c}z=\frac{0.66-0.6}{\sqrt{\frac{0.6\left(1-0.6\right)}{304}}}\\ =\frac{0.06}{0.0281}\\ \approx 2.14\end{array}$

Thus, the value of test statistic is 2.14.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is a right-tailed test. Therefore, the P-value is the area under the z curve and to the right of the calculated z value.

The P-value for the test statistic value of 2.14 is obtained as follows:

$\begin{array}{c}P\text{-value}=\text{Area to the Right of }2.14\\ =P\left(z>2.14\right)\\ =1-P\left(z<2.14\right)\end{array}$

Use Table A in Appendix A: Standard Normal Cumulative Probabilities to find the z-value.

Procedure:

For z at 2.14:

• Locate 2.1 in the left column of the table.
• Obtain the value in the corresponding row below .04.

That is, $P\left(z<2.14\right)=0.9838$.

The probability of the event is as given below:

$\begin{array}{c}P\left(z>2.14\right)=1-P\left(z<2.14\right)\\ =1-0.9838\\ =0.0162\end{array}$

Thus, the P-value for the test statistic of 2.14 is 0.00162.

Step 9:

Decision rule:

If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ of 0.0162 is greater than the significance level 0.01.

That is, $P\text{-value}\left(=0.0162\right)>\text{Significance level}\left(=0.01\right)$.

The decision is that the null hypothesis is not rejected.

Conclusion:

Hence, the sample data do not provide the evidence that the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.6.

To determine

Check whether the sample data provide convincing evidence that the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.5.

Answer to Problem 40E

No, the sample data do not provide the evidence that the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.5.

Explanation of Solution

Calculation:

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes the population proportion of Country U’s businesses that monitor employees’ web site visits.

Step 2:

Null hypothesis: $H_0:p=0.5$

That is, the proportion of Country U’s businesses that monitor employees’ web site visits is 0.5.

Step 3:

Alternative hypothesis: $H_a:p>0.5$

That is, the proportion of Country U’s businesses that monitor employees’ web site visits is greater than 0.5.

Step 4:

Significance level $\alpha$:

It is given that the significance level $\alpha =0.01$.

Step 5:

Test statistic z:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$

Where, $\widehat{p}$ is the sample proportion, p is the hypothesized proportion, and n is the sample size.

Substitute the hypothesized proportion $p=0.5$ in the test statistic.

$z=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is not known.

Step 6:

Assumptions:

• Let $\widehat{p}$ be the sample proportion from a random sample.
• The large sample z test can be used if the sample size n satisfies the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.
• The sample size should not be greater than 10% of the population size.

Requirement check:

• It is assumed that the sample is from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =304\left(0.5\right)\\ =152\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =304\left(1-0.5\right)\\ =304\left(0.5\right)\\ =152\\ >10\end{array}$

Since $n\left(\text{hypothesized value}\right)$ and $n\left(1-\text{hypothesized value}\right)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

• Although, the population size is not known, it is reasonable to assume that the sample size of 304 acts as the representative for all the businesses of Country U. It is also definite that the sample size is less than 10% of the population of the business of Country U.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{n}}}$

The sample proportion $\widehat{p}$ is obtained as follows:

$\widehat{p}=\frac{x}{n}$

Where, x is the number of observations of interest and n is the sample size.

Substitute the corresponding values to get the sample proportion:

$\begin{array}{c}\widehat{p}=\frac{201}{304}\\ =0.66\end{array}$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$\begin{array}{c}z=\frac{0.66-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{304}}}\\ =\frac{0.16}{0.0287}\\ \approx 5.58\end{array}$

Thus, the value of test statistic is 5.58.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is a right-tailed test. Therefore, the P-value is the area under the z curve and to the right of the calculated z value.

The P-value for the test statistic value of 5.58 is obtained as follows:

$\begin{array}{c}P\text{-value}=\text{Area to the Right of 5}\text{.58}\\ =P\left(z>5.58\right)\\ =1-P\left(z<5.58\right)\end{array}$

Use Table A in Appendix A: Standard Normal Cumulative Probabilities to find the z-value.

However, the Standard Normal Cumulative Probabilities has the value up to $P\left(z<3.8\right)=1.0000$.

Therefore, the $P\left(z<5.58\right)=1.0000$.

The probability of the event is as given below:

$\begin{array}{c}P\left(z>5.58\right)=1-P\left(z<5.58\right)\\ =1-1\\ =0\end{array}$

Thus, the P-value for the test statistic of 5.58 is 0.

Step 9:

Decision rule:

If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ of 0 is less than the significance level 0.01.

That is, $P\text{-value}\left(=0\right)<\text{Significance level}\left(=0.01\right)$.

The decision is that the null hypothesis is rejected.

Conclusion:

Hence, the sample data provides the evidence that the proportion of businesses of Country U’s businesses that monitor employees’ web site visits is greater than 0.5.