Noise levels at 7 concerts were measured in decibels yielding the following data:
\[
162,162,169,160,183,167,147
\]
Construct the $90 \%$ confidence interval for the mean noise level at such locations. Assume the population is approximately normal.

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Step 1 of 4: Calculate the sample mean for the given sample data. Round your answer to one decimal place.

Step 1 ：First, we calculate the sample mean (\(\bar{x}\)) of the given data. The sample mean is the sum of all the data points divided by the number of data points. In this case, we have 7 data points. So, \(\bar{x} = \frac{162+162+169+160+183+167+147}{7} = 165.7\).

Step 2 ：Next, we calculate the sample standard deviation (s). The sample standard deviation is the square root of the sum of the squared differences between each data point and the sample mean, divided by the number of data points minus 1. In this case, \(s = \sqrt{\frac{(162-165.7)^2+(162-165.7)^2+(169-165.7)^2+(160-165.7)^2+(183-165.7)^2+(167-165.7)^2+(147-165.7)^2}{7-1}} = 11.1\).

Step 3 ：Now, we can calculate the 90% confidence interval for the mean noise level. The formula for a confidence interval is \(\bar{x} \pm z\frac{s}{\sqrt{n}}\), where \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the sample standard deviation, and \(n\) is the number of data points. For a 90% confidence level, the z-score is approximately 1.645. So, the confidence interval is \(165.7 \pm 1.645\frac{11.1}{\sqrt{7}}\).

Step 4 ：Finally, we calculate the confidence interval. The lower bound of the confidence interval is \(165.7 - 1.645\frac{11.1}{\sqrt{7}} = 160.1\), and the upper bound of the confidence interval is \(165.7 + 1.645\frac{11.1}{\sqrt{7}} = 171.3\). So, the 90% confidence interval for the mean noise level at such locations is \(\boxed{[160.1, 171.3]}\).