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an article summarizes a report of law enforcement agencies regarding the use of social media to screen applicants for employment. the report was based on a survey of 731 law enforcement agencies. one question on the survey asked if the agency routinely reviewed applicants' social media activity during background checks. for purposes of this exercise, suppose that the 731 agencies were selected at random, and that you want to use the survey data to decide if there is convincing evidence that more than 25% of law enforcement agencies review applicants' social media activity as part of routine background checks.

Respuesta :

a) The sampling proportion's mean value is 0.25. 0.016 is the standard deviation. The form resembles a bell.

b) There is a 10% possibility of getting this number, p=0.27 or more, given a sample proportion, so I wouldn't be surprised.

c) Since there is no chance that the sample fraction of 0.31 will occur, I would be shocked if it did.

Given,

a) The null hypothesis proportion would be the middle of the sampling distribution (p-0.25). So, p=0.25 is the sampling proportion's mean value.

This would be the standard deviation:

σp = √(p (1 - p) / n) = √(0.25 × 0.75/731) = 0.016

The distribution would resemble a binomial distribution, hence the form would be bell-shaped.

b) By calculating the z-value and checking for its probability in the standard normal distribution, we may determine the likelihood of a value p=0.27 in this distribution.

z = (p - π) / σp = (0.27 - 0.25) / 0.016 = 0.02/0.016  = 1.25

p(z > 1.25) = 0.106

There is a 10% probability of achieving this value, p=0.27 or more, for a sample proportion, so I wouldn't be surprised.

c) We repeat the calculation for p=0.31

z = (p - π) / σp = (0.31 - 0.25) / 0.016 = 0.06/0.016  = 5

p(z > 5) = 0.000

I would be astonished to see that value because the probability of this sample fraction occurring at p=0.31 is zero.

Learn more about probability here;

https://brainly.com/question/15727765

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