Question

Study of why EMS workers leave the job. A study of fulltimeemergency medical service (EMS) workers publishedin the Journal of Allied Health(Fall 2011) found that onlyabout 3% leave their job in order to retire. (See Exercise3.45, p. 182.) Assume that the true proportion of all fulltime

EMS workers who leave their job in order to retire is p= .03. In a random sample of 1,000 full-time EMS workers, let represent the proportion who leave their job inorder to retire.

1. Describe the properties of the sampling distribution of$\hat{\mathbf{p}}$.
2. Compute $\mathbf{P}\left(\stackrel{⏜}{p}<0.05\right)$Interpret this result.
3. Compute$\mathbf{P}\mathbf{(}\stackrel{\mathbf{⏜}}{\mathbf{p}}\mathbf{>}\mathbf{0}\mathbf{.}\mathbf{025}\mathbf{)}$Interpret this result.

Short answer

a.$Mean=0.03,s.d=0.0054,Z=\frac{\hat{p}-0.03}{0.0054}$

b. The probability is approximately 0.99989. So, we can interpret that, there is approximately a 99% chance of observing a sample proportion of 0.05 or less if the true proportion of full-time workers who leave their jobs in order to retire.

c. The probability is approximately 0.82275. So, we can interpret that, there is approximately an 82% chance of observing a sample proportion of 0.025 or greater if the true proportion of full-time workers who leave their jobs in order to retire.

Step-by-step solution

Given information 

Referring to exercise 3.45 (page 182), the true proportion of all full-time EMS workers who leave their jobs in order to retire is p = 0.03. There is a random sample of 1000 full-time workers. So, n = 1000. $\overbrace{p}$represents the proportion of those workers who leave their jobs in order to retire.

Describe the properties

a.

We know from the Central Limit Theorem that the sampling distribution of$\overbrace{p}$ is normal.

The mean of the sampling distribution is equal to the true binomial proportion. i.e.$E\left(\overbrace{P}\right)=P$ and$\overbrace{p}$ is an unbiased estimator of p. The standard deviation of the sampling distribution is$\sqrt{\frac{p\left(1-p\right)}{n}}$

For this case the mean of the sampling distribution is ${\mu }_{\hat{p}}=p=0.03$and

the standard deviation is${\sigma }_{\hat{p}}=\sqrt{\frac{0.03\left(1-0.03\right)}{1000}}=0.0054$

The Z-score of the sampling distribution is $Z=\frac{\hat{p}-0.03}{0.0054}$

Compute the probability when Pp^<0.05

b.

The probability is given by,

$\begin{array}{rcl}P\left(\hat{p}<0.05\right)& =& P\left(Z<\frac{0.05-0.03}{0.0054}\right)\\ & =& P\left(Z<3.7037\right)\\ & \approx & 0.99989\\ & & \end{array}$

From the z-score table, we find the probability is approximately 0.99989.

So, we can interpret that, there is approximately 99% chance of observing a sample proportion of 0.05 or less if the true proportion of full-time workers who leave their jobs in order to retire.

Compute the probability when Pp^>0.025 

c.

The probability is given by,

$\begin{array}{rcl}P\left(\hat{p}>0.025\right)& =& P\left(Z>\frac{0.025-0.03}{0.0054}\right)\\ & =& P\left(Z>-0.9259\right)\\ & \approx & 0.82275\\ & & \end{array}$

From the z-score table, we find the probability is approximately 0.82275.

So, we can interpret that, there is approximately an 82% chance of observing a sample proportion of 0.025 or greater if the true proportion of full-time workers who leave their jobs in order to retire.