hospital noise levels for a sample of 5 operating rooms taken in a hospital study, the mean noise level was 42.6 decibels and the standard deviation was 10.2. find the 99% confidence interval of the true mean of the noise levels in the operating rooms. assume the variable is normally distributed. round your answers to at least two decimal places.

Question

Accepted Answer

# Explanation:
## Step1: Determine the degrees of freedom and t - value
The degrees of freedom $df=n - 1$, where $n = 5$, so $df=5-1 = 4$.
For a $99\%$ confidence interval, $\alpha=1 - 0.99=0.01$, and $\frac{\alpha}{2}=0.005$.
Looking up the t - value in the t - distribution table (or using a calculator), $t_{\frac{\alpha}{2},df}=t_{0.005,4}=4.604$.

## Step2: Calculate the margin of error $E$
The formula for the margin of error when the population standard deviation $\sigma$ is unknown is $E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}$, where $s = 10.2$ and $n = 5$.
$$
\begin{align*}
E&=4.604\times\frac{10.2}{\sqrt{5}}\
&=4.604\times\frac{10.2}{2.236}\
&=4.604\times4.562\
&\approx21.00
\end{align*}
$$

## Step3: Calculate the confidence interval
The confidence interval formula is $\bar{x}-E<\mu<\bar{x}+E$, where $\bar{x}=42.6$
$$
\begin{align*}
\bar{x}-E&=42.6-21.00=21.60\
\bar{x}+E&=42.6 + 21.00=63.60
\end{align*}
$$

# Answer:
$21.60<\mu<63.60$

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QUESTION IMAGE

Question

Explanation:

Step1: Determine the degrees of freedom and t - value

Step2: Calculate the margin of error \(E\)

Step3: Calculate the confidence interval

Answer: