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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鈥檚 company had implemented a 鈥渨histle-blower鈥 hotline鈥攖hat 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鈥 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. TestH0:p=.5 against Ha:p<.5using data from the 2015 survey and =.05Give a practical interpretation of the results.

b. Test H0:p=.5against Ha:p>.5using data from the 2013 survey and=.05 Give a practical interpretation of the results.

Short Answer

Expert verified

a. At =.05, 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%.

b.At =.05, 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%.

Step by step solution

01

Given information

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 isn1=1508

The sample proportion is

p^1=x1n1=6801508=0.451

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 isn2=681

The sample proportion is

p^2=x2n2=436681=0.640

02

Testing population proportion for the 2015 survey

We have to test

H0:p=.5

Against

Ha:p<.5

The test statistic for testing these hypotheses is

Z=p^1-pp1-pn1=0.451-0.500.501-0.501508=-0.0490.00016578=-3.81

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

Z=Z0.05=-1.645alternativehypothesisisleft-tailed

We can see that

Z<Z

Hence, we reject the null hypothesis H0.

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%.

03

Testing population proportion for the 2013 survey

We have to test

H0:p=.5

Against

Ha:p>.5

The test statistic for testing these hypotheses is

Z=p^2-pp1-pn2=0.640-0.500.501-0.50681=0.140.00036711=7.31

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

Z=Z0.05=1.645alternativehypothesisisright-tailed

We can see that

Z>Z

Hence, we reject the null hypothesis H0.

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%.

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Most popular questions from this chapter

A random sample of n observations is selected from a normal population to test the null hypothesis that 碌=10.Specify the rejection region for each of the following combinations of \(Ha,\alpha ,\) and n:

a.\(Ha:\)碌\( \ne 10;\alpha = .05.;n = 14\)

b.\(Ha:\)碌\( > 10;\alpha = .01;n = 24\)\(\)

c.\(Ha:\)碌\( > 10;\alpha = .10;n = 9\)

d.\(Ha:\)碌 <\(10:\alpha = .01;n = 12\)

e.\(Ha:\)碌\( \ne 10;\alpha = .10;n = 20\)

f. \(Ha:\)碌<\(10;\alpha = .05;n = 4\)

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d. Convert each of the b values you calculated in part a to the power of the test at the specified value of m. Plot the power on the vertical axis against m on the horizontal axis. Compare the graph of part b with the power curve of this part.

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