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Brrunno [24]
3 years ago
5

Applicants for temporary office work at Carter Temporary Help Agency who have successfully completed an administrative assistant

course are then placed in suitable positions by Nancy Dwyer and Darla Newberg. Employers who hire temporary help through the agency return a card indicating satisfaction or dissatisfaction with the work performance of those hired. From past experience it is known that 80% of the employees placed by Nancy are rated as satisfactory, and 65% of those placed by Darla are rated as satisfactory. Darla places 55% of the temporary office help at the agency, and Nancy places the remaining 45%. If a Carter office worker is rated unsatisfactory, what is the probability that he or she was placed by Darla? (Round your answer to three decimal places.)
Mathematics
1 answer:
dezoksy [38]3 years ago
5 0

Answer:

0.681

Step-by-step explanation:

Let's define the following events:

S: a Carter office worker is rated satisfactory

U : a Carter office worker is rated unsatisfactory

ND: a Carter office worker is placed by Nancy Dwyer

DN: a Carter office worker is placed by Darla Newberg

We have from the original text that

P(S | ND) = 0.8, this implies that P(U | ND) = 0.2.

P(S | DN) = 0.65, this implies that P(U | DN) = 0.35. Besides

P(DN) =  0.55 and P(ND) = 0.45, then we are looking for

P(DN | U), using the Bayes' formula we have

P(DN | U) = \frac{P(U | DN)P(DN)}{P(U | DN)P(DN) + P(U | ND)P(ND)} = \frac{(0.35)(0.55)}{(0.35)(0.55)+(0.2)(0.45)}=0.681

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