Question

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 728 law enforcement agencies. One question on the survey asked if the agency routinely reviewed applicant's social media activity during background checks. For purposes of this exercise, suppose that the 728 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.

Answer 1

a) The mean of the sampling proportion is p=0.25. The standard deviation is σ=0.016. The shape is like a bell.

b) I wouldn't be surprised as there is a 10% chance of getting this value p=0.27 or more for a sample proportion.

c) This sample proportion p=0.31 has a null probability of happening, so I would be surprised of getting that value.

Explanation:

The question is incomplete:

a) Describe the sampling distribution (shape, center and spread) of p for a sample size of n=728, if the null hypothesis is true (p=0.25).

b) Would you be surprised to observe a sample proportion p=0.27 for a sample size of n=728 if the null hypothesis (p=0.25) is true?

c) Would you be surprised to observe a sample proportion p=0.31 for a sample size of n=728 if the null hypothesis (p=0.25) is true?

a) The sampling distribution would be centered in the null hypothesis proportion (p-0.25). Then, the mean of the sampling proportion is p=0.25.

The standard deviation would be:

`\sigma_p=√(p(1-p)/n)=√(0.25*0.75/728)=√(0.000257555)\\\\\sigma_p=0.016`

As it would be a binomial-like distribution, the shape will be a bell-shape.

b) We can calculate the likelyhood of a value p=0.27 in this distribution calculating the z-value and looking for its probability in the standard normal distribution.

`z=(p-\pi)/\sigma_p=(0.27-0.25)/0.016=0.02/0.016=1.25\\\\P(z>1.25)=0.106`

I wouldn't be surprised as there is a 10% chance of getting this value p=0.27 or more for a sample proportion.

c) We repeat the calculation for p=0.31

`z=(p-\pi)/\sigma_p=(0.31-0.25)/0.016=0.06/0.016=5\\\\P(z>5)=0.000`

This sample proportion p=0.31 has a null probability of happening, so I would be surprised of getting that value.