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One study reports that 34​% of newly hired MBAs are confronted with unethical business practices during their first year of employment. One business school dean wondered if her MBA graduates had similar experiences. She surveyed recent graduates from her​ school's MBA program to find that 28​% of the 116 graduates from the previous year claim to have encountered unethical business practices in the workplace. Can she conclude that her​ graduates' experiences are​ different?

User Propagated
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Answer:


z=\frac{0.28 -0.34}{\sqrt{(0.34(1-0.34))/(116)}}=-1.364


p_v =2*P(Z<-1.364)=0.173

The p value obtained was a very high value and using the significance level assumed
\alpha=0.05 we have
p_v>\alpha so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.

Explanation:

1) Data given and notation

n=116 represent the random sample taken

X represent the number graduates from the previous year claim to have encountered unethical business practices in the workplace


\hat p=0.28 estimated proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace


p_o=0.34 is the value that we want to test


\alpha=0.05 represent the significance level

z would represent the statistic (variable of interest)


p_v represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is 0.34 or no.:

Null hypothesis:
p=0.34

Alternative hypothesis:
p \\eq 0.34

When we conduct a proportion test we need to use the z statistic, and the is given by:


z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}} (1)

The One-Sample Proportion Test is used to assess whether a population proportion
\hat p is significantly different from a hypothesized value
p_o.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:


z=\frac{0.28 -0.34}{\sqrt{(0.34(1-0.34))/(116)}}=-1.364

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:


p_v =2*P(Z<-1.364)=0.173

The p value obtained was a very high value and using the significance level assumed
\alpha=0.05 we have
p_v>\alpha so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.

User Daniil Fajnberg
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