91Ó°ÊÓ

The sampling distribution of the sample proportions $\hat{p}$has a mean of

${\mathrm{μ}}_{\hat{p}}=p=0.75$

The sampling distribution of the sample proportion $\hat{p}$standard deviation is ${\mathrm{σ}}_{\hat{p}}=\frac{p(1−p)}{n}$

$$\begin{gathered} =\sqrt{\frac{0.75(1−0.75)}{55}} \\ =0.0584 \end{gathered}$$

The $z-$score is

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.582−0.75}{0.0584} \\ =−2.88 \end{gathered}$$

The associating probability using the normal probability table $P(Z<2.88)$ is given in the standard normal probability table in the row starting with $-2.8$ and the column starting with$.08$

$$\begin{gathered} P(\hat{p}≤0.20)=P(z<−2.88) \\ =0.0020 \\ =0.20\% \end{gathered}$$

The sampling distribution of the sample proportions p has a mean of

${\mathrm{μ}}_{\hat{p}}=p=0.75$

The sampling distribution of the sample proportion$p$ has a standard deviation of

$$\begin{gathered} {\mathrm{σ}}_{\hat{p}}=\sqrt{\frac{p(1−p)}{n}} \\ =\sqrt{\frac{0.75(1−0.75)}{55}} \\ =0.0584 \end{gathered}$$

The $z-$score is

$$\begin{gathered} z=\frac{x−\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0.582−0.75}{0.0584} \\ =−2.88 \end{gathered}$$

The associating probability using the normal probability table $P(Z<-2.88)$is given in the standard normal probability table in the row starting with $-2.8$and the column starting with $.08$

Because the likelihood is not modest (less than $0.05$), a sample proportion of at most $0.20$is unlikely to occur by chance, and there is solid evidence that less than $75\%$of all emergency room patients wait less than $30$minutes.