Question

hospital noise levels noise levels at various area urban hospitals were measured in decibels. the mean noise level in 163 ward areas was 50.2 decibels, and the population standard deviation is 5.2. find the 90% confidence interval of the true mean. round your answers to at least one decimal place. \\(\square < \mu < \square\\)

Explanation:

Step1: Identify given values

Sample mean $\bar{x}=50.2$, sample size $n=163$, population standard deviation $\sigma=5.2$, confidence level 90%.

Step2: Find z-score for 90% confidence

For 90% confidence, two-tailed z-score $z_{\alpha/2}=1.645$.

Step3: Calculate standard error

Standard error $SE=\frac{\sigma}{\sqrt{n}}=\frac{5.2}{\sqrt{163}}\approx\frac{5.2}{12.767}\approx0.407$.

Step4: Compute margin of error

Margin of error $E=z_{\alpha/2}\times SE=1.645\times0.407\approx0.670$.

Step5: Find confidence interval

Lower bound: $50.2 - 0.670\approx49.5$; Upper bound: $50.2 + 0.670\approx50.9$.

Answer:

49.5 < μ < 50.9