The opinions of employees at Asia-Pacific firms regarding fraud, bribery, and corruption in the workplace were elicited in the 2015 AsiaPacific (APAC) Fraud Survey. Interviews were conducted with a sample 1,508 employees of large APAC companies. One question concerned whether or not the employee’s company had implemented a “whistle-blower” hotline—that is, a phone number that an employee can call to report fraud or other types of misconduct without fear of retribution. The 2015 survey found that 680 of the 1,508 respondents’ companies had not implemented a whistle-blower hotline. In comparison, 436 of the 681 respondents in the 2013 survey reported their companies had not implemented a whistle-blower hotline.

a. Test${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{p}\mathbf{=}\mathbf{.}\mathbf{5}$ against ${\mathit{H}}_{\mathbf{a}}\mathbf{:}\mathit{p}\mathbf{<}\mathbf{.}\mathbf{5}$using data from the 2015 survey and $\mathit{\alpha }\mathbf{=}\mathbf{.}\mathbf{05}$Give a practical interpretation of the results.

b. Test ${\mathit{H}}_{\mathbf{0}}\mathbf{:}\mathit{p}\mathbf{=}\mathbf{.}\mathbf{5}$against ${\mathit{H}}_{\mathbf{a}}\mathbf{:}\mathit{p}\mathbf{>}\mathbf{.}\mathbf{5}$using data from the 2013 survey and$\mathit{\alpha }\mathbf{=}\mathbf{.}\mathbf{05}$ Give a practical interpretation of the results.

Asia Pacific (APAC) Fraud Survey, out of 1508 respondents surveyed in 2015, 680 said that their companies had not implemented a whistle-blower hotline

That is

The size of the sample is$n_1=1508$

The sample proportion is

$$\begin{gathered} {\hat{p}}_1=\frac{x_1}{n_1} \\ =\frac{680}{1508} \\ =0.451 \end{gathered}$$

And

Also,out of 681 respondents surveyed in 2013, 436 said that their companies had not implemented a whistle-blower hotline.

That is

The size of the sample is$n_2=681$

The sample proportion is

$$\begin{gathered} {\hat{p}}_2=\frac{x_2}{n_2} \\ =\frac{436}{681} \\ =0.640 \end{gathered}$$

We have to test

$H_0:p=.5$

Against

$H_a:p<.5$

The test statistic for testing these hypotheses is

$$\begin{gathered} Z=\frac{{\hat{p}}_1-p}{\sqrt{\frac{p\left(1-p\right)}{n_1}}} \\ =\frac{0.451-0.50}{\sqrt{\frac{0.50\left(1-0.50\right)}{1508}}} \\ =\frac{-0.049}{\sqrt{0.00016578}} \\ =-3.81 \end{gathered}$$

Now critical value at a 5% significance level using the standard normal table is

$$\begin{gathered} Z_{\alpha }=Z_{0.05} \\ =-1.645\left(alternativehypothesisisleft-tailed\right) \end{gathered}$$

We can see that

$Z<Z_{\alpha }$

Hence, we reject the null hypothesis $H_0$.

Interpretation:

At the 5% significance level, we have sufficient evidence to conclude that the true proportion of companies that had not implemented a whistle-blower hotline in 2015 is less than 50%.

We have to test

$H_0:p=.5$

Against

$H_a:p>.5$

The test statistic for testing these hypotheses is

$$\begin{gathered} Z=\frac{{\hat{p}}_2-p}{\sqrt{\frac{p\left(1-p\right)}{n_2}}} \\ =\frac{0.640-0.50}{\sqrt{\frac{0.50\left(1-0.50\right)}{681}}} \\ =\frac{0.14}{\sqrt{0.00036711}} \\ =7.31 \end{gathered}$$

Now critical value at a 5% significance level using the standard normal table is

$$\begin{gathered} Z_{\alpha }=Z_{0.05} \\ =1.645\left(alternativehypothesisisright-tailed\right) \end{gathered}$$

We can see that

$Z>Z_{\alpha }$

Hence, we reject the null hypothesis $H_0$.

Interpretation:

At the 5% significance level, we have sufficient evidence to conclude that the true proportion of companies that had not implemented a whistle-blower hotline in 2013 is greater than 50%.