Question

Hospital length of stay. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS in the United States is 4.5 days (Healthcare Cost and Utilization Project Statistical Brief, October 2014). A random sample of 20 hospitals in one state had a mean LOS of 3.8 days and a standard deviation of 1.2 days.

a. Use a 90% confidence interval to estimate the population's mean LOS for the state’s hospitals.

b. Interpret the interval in terms of this application.

c. What is meant by the phrase “90% confidence interval”?

Short answer

A confidence interval represents the likelihood that a parameter will drop among two possible values close to the mean. The level of uncertainty, as well as confidence in a sample process, is measured by the confidence level.

Step-by-step solution

 Step 1: (a) The population estimate mean LOS for the state's hospitals

A hospital sample is chosen at random from one state

$$\begin{gathered} \text{n=}20 \\ \begin{array}{c}\text{Mean}\left(\stackrel{-}{\chi }\right)=3.8\text{days}\\ \sigma =1.2\text{days}\end{array} \end{gathered}$$

The calculation is given below:

$$\begin{gathered} \text{Thedegreesoffreedom=n}-\text{1} \\ \text{=20}-\text{1} \\ \text{=19} \end{gathered}$$

90% confidence interval using.

Hence, the confidence interval is 0.90

$\begin{array}{c}\text{1}-\propto \text{=0.90}\\ \propto \text{=0.1}\\ \frac{\propto }{2}\text{=0.05}\end{array}$

The value ${\text{t}}_{\text{0.05,19}}$ is 1.729

localid="1651473954295" ${\stackrel{-}{\chi }}_{+\propto \text{12}}\left(\frac{\text{S}}{\sqrt{\text{n}}}\right)$

$$\begin{gathered} \stackrel{-}{\chi }=3.8 \\ {\text{t}}_{\propto \text{12}}=1.729 \\ \begin{array}{c}\text{S}=1,2\\ \text{n}=20\end{array} \end{gathered}$$

$\begin{array}{l}\text{=3.8}\pm \text{1.729}\left(\frac{\text{1.2}}{\sqrt{\text{20}}}\right)\\ =\text{3.8}\pm \text{1.729}\times 0.2683\\ =\text{3.8}\pm 0.4638\\ =(3.3362,4.2638)\end{array}$

The 90 percent Confident interval for the population implies that the LOS for the state's hospitals is$3.3362\text{and}4.2638.$

(b) Interpretation

In terms of this application, the interval.

Interpretation

90 percent confident that the populations mean length of remain (LOS) for the state’s hospitals$\left(\mu \right)$lies among $3.3362\text{and}4.2638$.

(c) Confidence Interval 90%

The term 90% confidence interval. Equally generated intervals will include the genuine population mean score$\left(\mu \right)$ is repeated sampling. A confidence interval expresses the likelihood that a number will fall within the upper as well as lower boundaries of a probability distribution.