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Evgen [1.6K]
3 years ago
11

Noise levels at 5 airports were measured in decibels yielding the following data: 147,123,119,161,136 Construct the 99% confiden

ce interval for the mean noise level at such locations. Assume the population is approximately normal. Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.
Physics
1 answer:
lisov135 [29]3 years ago
4 0

Answer:

a) The 99% confidence interval for the mean noise level = [122.44, 151.96]

b) Sample standard deviation, s = 17.3dB

Explanation:

Noise levels at 5 airports = 147,123,119,161,136

Mean noise level

        \bar{x} =\frac{ 147+123+119+161+136}{5}=137.2dB

Variance of noise level

        \sigma^2 =\frac{ (137.2-147)^2+(137.2-123)^2+(137.2-119)^2+(137.2-161)^2+(137.2-136)^2}{5}\\\\\sigma^2=164.16

Standard deviation,

       \sigma =\sqrt{164.16}=12.81dB

a) Confidence interval  is given by

       \bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}

For 99% confidence interval Z = 2.576,

Number of noises, n = 5

Substituting

    137.2-2.576\times \frac{12.81}{\sqrt{5}}\leq \mu\leq 137.2+2.576\times \frac{12.81}{\sqrt{5}}\\\\122.44\leq \mu\leq 151.96

The 99% confidence interval for the mean noise level = [122.44, 151.96]

b) Sample standard deviation

        s=\sqrt{\frac{ (137.2-147)^2+(137.2-123)^2+(137.2-119)^2+(137.2-161)^2+(137.2-136)^2}{5-1}}\\\\s=17.3dB

   Sample standard deviation, s = 17.3dB

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