Chapter 7.2, Problem 46E

(a)

To determine

To Calculate: the probability that the proportion of patients who wait less than 30 minutes is 0.582 or less in a random sample of 55 patients.

Answer to Problem 46E

0.20%

Explanation of Solution

Given:

  $\begin{array}{c}p=75\%=0.75\\ \widehat{p}=44\%=0.582\\ n=55\end{array}$

Formula used:

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ z=\frac{x-\mu }{\sigma }\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportions $\widehat{p}$ is

  ${\mu }_{\widehat{p}}=p=0.75$

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.75\left(1-0.75\right)}{55}}\\ =0.0584\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.582-0.75}{0.0584}\\ =-2.88\end{array}$

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

  $\begin{array}{c}P\left(\widehat{p}\le 0.20\right)=P\left(z<-2.88\right)\\ =0.0020\\ =0.20\%\end{array}$

(b)

To determine

To Explain: that there is convincing evidence that less than 75% of all patients in the emergency room wait less than 30 minutes.

Answer to Problem 46E

Yes

Explanation of Solution

Given:

  $\begin{array}{c}p=75\%=0.75\\ \widehat{p}=44\%=0.582\\ n=55\end{array}$

Formula used:

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ z=\frac{x-\mu }{\sigma }\end{array}$

Calculation:

The mean of the sampling distribution of the sample proportions $\widehat{p}$ is

  ${\mu }_{\widehat{p}}=p=0.75$

The standard deviation of the sampling distribution of the sample proportion $\widehat{p}$ is

  $\begin{array}{c}{\sigma }_{\widehat{p}}=\sqrt{\frac{p\left(1-p\right)}{n}}\\ =\sqrt{\frac{0.75\left(1-0.75\right)}{55}}\\ =0.0584\end{array}$

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{0.582-0.75}{0.0584}\\ =-2.88\end{array}$

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

  $\begin{array}{c}P\left(\widehat{p}\le 0.20\right)=P\left(z<-2.88\right)\\ =0.0020\\ =0.20\%\end{array}$

Since the probability is not small ( less than 0.05), the event of a sample proportion of at most $0.20$ is unlikely to get by chance and therefore there is convincing evidence that less than $75\%$ of all patients in the emergency room wait less than 30 minutes.