# Solved:Suppose the number of beds filled per day in a medium sized hospital is normally distributed. A hospital administrator tells the board of

#### ByRaju Chaudhari

Sep 25, 2020

Suppose the number of beds filled per day in a medium sized hospital is normally distributed. A hospital administrator tells the board of directors that, on the average, at least 185 beds are filled on any given day. One of the board members believes that the average is less than 185 and she sets out to test to determine if she is correct. She secures a random sample of 16 days of data (shown below). Use $\alpha = .05$ and the sample data to test the board member's theory. Assume the number of filled beds per day is normally distributed in the population.

173, 149, 166, 180, 189, 170, 152, 194, 177, 169, 188, 160,
199, 175, 172, 187

#### Solution

Given that the sample size $n = 16$, sample mean $\overline{x}= 175$ and sample standard deviation $s = 14.283$.

#### Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : \mu \geq 185$ against $H_1 : \mu < 185$ ($\text{left-tailed}$)

#### Step 2 Test Statistic

The test statistic is

\begin{aligned} t& =\frac{\overline{x} -\mu}{s/\sqrt{n}} \end{aligned}
which follows $t$ distribution with $n-1$ degrees of freedom.

#### Step 3 Significance Level

The significance level is $\alpha = 0.05$.

#### Step 4 Critical Value(s)

As the alternative hypothesis is $\text{left-tailed}$, the critical value of $t$ $\text{is}$ $-1.753$.

The rejection region (i.e. critical region) is $\text{t < -1.753}$.

#### Step 5 Computation

The test statistic under the null hypothesis is
\begin{aligned} t&=\frac{ \overline{x} -\mu_0}{s/\sqrt{n}}\ &= \frac{175-185}{14.283/ \sqrt{16 }}\ &= -2.801 \end{aligned}

#### Step 6 Decision

The test statistic is $t =-2.801$ which falls $\text{inside}$ the critical region, we $\text{reject}$ the null hypothesis.

$p$-value Approach:

This is a $\text{left-tailed}$ test, so the p-value is the area to the left of the test statistic ($t=-2.801$) is p-value = $0.0067$.

The p-value is $0.0067$ which is $\text{less than}$ the significance level of $\alpha = 0.05$, we $\text{reject}$ the null hypothesis.

We conclude that the average number of beds filled on any given day is less than 185.