Questions: Hospital Noise Levels For a sample of 8 operating rooms taken in a hospital study, the mean noise level was 38.2 decibels and the standard deviation was 10.1. Find the 95% 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.

Solution Steps

To find the Z critical value for a 95% confidence level, we use the formula: \[ Z = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right) \] where \(\alpha = 0.05\). The Z critical value for this confidence level is found to be: \[ Z = 1.96 \]

The margin of error (ME) is calculated using the formula: \[ \text{ME} = Z \times \left(\frac{\sigma}{\sqrt{n}}\right) \] Substituting the values:

• \(Z = 1.96\)
• \(\sigma = 10.1\)
• \(n = 8\)

We find: \[ \text{ME} = 1.96 \times \left(\frac{10.1}{\sqrt{8}}\right) \approx 7.00 \]

The 95% confidence interval (CI) for the true mean is given by: \[ \text{CI} = \bar{x} \pm \text{ME} \] where \(\bar{x} = 38.2\). Thus, we calculate: \[ \text{Lower Bound} = 38.2 - 7.00 = 31.20 \] \[ \text{Upper Bound} = 38.2 + 7.00 = 45.20 \] Therefore, the 95% confidence interval is: \[ \text{CI} = (31.20, 45.20) \]

Final Answer