Question

hospital noise levels noise levels at various area urban hospitals were measured in decibels. the mean noise level in 170 ward areas was 50.7 decibels, and the population standard deviation is 5.0. find the 99% confidence interval of the true mean. round your answers to at least one decimal place.

Explanation:

Step1: Find the z - value

For a 99% confidence interval, the significance level \(\alpha=1 - 0.99=0.01\). Then \(\frac{\alpha}{2}=\frac{0.01}{2}=0.005\).
The z - value \(z_{\frac{\alpha}{2}}\) is the value such that \(P(Z>z_{\frac{\alpha}{2}})=0.005\). Looking up in the standard normal table, \(z_{\frac{\alpha}{2}} = 2.576\).

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

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\frac{\alpha}{2}}\times\frac{\sigma}{\sqrt{n}}\).
Given \(n = 170\), \(\sigma=5.0\), and \(z_{\frac{\alpha}{2}}=2.576\).
\(E=2.576\times\frac{5.0}{\sqrt{170}}\)
\(\sqrt{170}\approx13.04\), \(E = 2.576\times\frac{5.0}{13.04}\)
\(E=2.576\times0.3835\approx0.99\)

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is given by \(\bar{x}-E<\mu<\bar{x} + E\).
Given \(\bar{x}=50.7\) and \(E\approx0.99\)
\(\bar{x}-E=50.7 - 0.99=49.71\approx49.7\)
\(\bar{x}+E=50.7+0.99 = 51.69\approx51.7\)

Answer:

\(49.7<\mu<51.7\)