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Karo-lina-s [1.5K]
3 years ago
13

One study reports that 34​% of newly hired MBAs are confronted with unethical business practices during their first year of empl

oyment. 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?
Mathematics
1 answer:
MaRussiya [10]3 years ago
5 0

Answer:

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

p_v =2*P(Z  

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.  

Step-by-step 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 \neq 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{\frac{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{\frac{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  

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.  

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