The confidence interval for the mean noise at locations is (127.09,148.51).
Given noise levels at 5 airports=152,154,139,124,120
We have to find the confidence interval for the mean noise and sample mean of the data. We know that mean is the sum divided by the number of items taken.
Mean = sum of X values/n
=137.8 (calculated in figure)
s=[tex]\sqrt{(x-x bar)^{2} /n-1}[/tex]
=[tex]\sqrt{972.8/4}[/tex]
=[tex]\sqrt{243.2}[/tex]
=15.59
t value at 80% confidence level with degree of freedom 4 (5-1) is 1.533.
standard error=t value*s/[tex]\sqrt{n}[/tex]
=1.533*15.59/[tex]\sqrt{5}[/tex]
=10.71
Upper level= Mean + standard error
=137.8+10.71
=148.51
Lower level=Mean - standard error
=137.8-10.71
=127.09
Hence the confidence interval for the true mean noise is (127.09,148.51) and sample mean is 137.8.
Learn more about confidence interval at https://brainly.com/question/15712887
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