Questions: Noise levels at 3 construction sites were measured in decibels yielding the following data: 153, 163, 181 Construct the 95% confidence interval for the mean noise level at such locations. Assume the population is approximately normal. Step 4 of 4: Construct the 95% confidence interval. Round your answer to one decimal place.

Solution Steps

The mean noise level is calculated using the formula:

\[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{153 + 163 + 181}{3} = \frac{497}{3} \approx 165.7 \]

The sample variance is calculated as follows:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = \frac{(153 - 165.7)^2 + (163 - 165.7)^2 + (181 - 165.7)^2}{3-1} = \frac{201.33}{2} \approx 100.665 \]

The sample standard deviation is then:

\[ s = \sqrt{201.33} \approx 14.19 \]

For a 95% confidence interval, we use the formula:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 165.7\)
• \(t\) is the t-value for \(n-1 = 2\) degrees of freedom at 95% confidence, which is approximately \(4.3\).
• \(s = 14.19\)
• \(n = 3\)

Calculating the margin of error:

\[ \text{Margin of Error} = t \frac{s}{\sqrt{n}} = 4.3 \cdot \frac{14.19}{\sqrt{3}} \approx 4.3 \cdot 8.19 \approx 35.2 \]

Thus, the confidence interval is:

\[ (165.7 - 35.2, 165.7 + 35.2) = (130.5, 200.9) \]

Final Answer