Question

Wait times A hospital claims that $75\%$ of people who come to its emergency room are seen by a doctor within $30$ minutes of checking in. To verify this claim, an auditor inspects the medical records of $55$ randomly selected patients who checked into the emergency room during the last year. Only $32$ $(58.2\%)$ of these patients were seen by a doctor within $30$ minutes of checking in.

a. If the wait time is less than $30$ minutes for $75\%$ of all patients in the emergency room, what is 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?

b. Based on your answer to part (a), is there convincing evidence that less than $75\%$ of all patients in the emergency room wait less than $30$ minutes? Explain your reasoning.

Short answer

Part (a)$0.20\%$

Part (b) Yes.

Step-by-step solution

Part (a) Step 1: Given information

$$\begin{gathered} p=75\%=0.75 \\ \hat{p}=44\%=0.582 \\ n=55 \end{gathered}$$

Part (a) Step 2: Concept

$$\begin{gathered} {\sigma }_{\hat{p}}=\frac{p(1-p)}{n} \\ z=\frac{x-\mu }{\sigma } \end{gathered}$$

Part (a) Step 3: Calculation

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

${\mu }_{\hat{p}}=p=0.75$

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

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

The $z-$score is

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

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$.08$

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

Part (b) Step 1: Calculation

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

${\mu }_{\hat{p}}=p=0.75$

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

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

The $z-$score is

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

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.