91Ó°ÊÓ

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. $\stackrel{Áåž}{p}$represents the proportion of those workers who leave their jobs in order to retire.

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

The mean of the sampling distribution is equal to the true binomial proportion. i.e.$E\left(\stackrel{Áåž}{P}\right)=P$ and$\stackrel{Áåž}{p}$ 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 ${\mathrm{μ}}_{\hat{p}}=p=0.03$and

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

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

The probability is given by,

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

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.

The probability is given by,

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

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.