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
[tex]\bar{x} =\frac{ 147+123+119+161+136}{5}=137.2dB[/tex]
Variance of noise level
[tex]\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[/tex]
Standard deviation,
[tex]\sigma =\sqrt{164.16}=12.81dB[/tex]
a) Confidence interval is given by
[tex]\bar{x}-Z\times \frac{\sigma}{\sqrt{n}}\leq \mu\leq \bar{x}+Z\times \frac{\sigma}{\sqrt{n}}[/tex]
For 99% confidence interval Z = 2.576,
Number of noises, n = 5
Substituting
[tex]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[/tex]
The 99% confidence interval for the mean noise level = [122.44, 151.96]
b) Sample standard deviation
[tex]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[/tex]
Sample standard deviation, s = 17.3dB
