Answer:
a) The 90% confidence interval for the mean noise level at such locations
(112.46 , 163.54)
b) The critical value that should be used in constructing the confidence interval
Z₀.₁₀ = 1.645
Step-by-step explanation:
Step(i) :-
Noise levels at 3 airports were measured in decibels yielding the following data
108 146 160
Mean of given data
[tex]x^{-} = \frac{108+146+160}{3} = 138[/tex]
x : 108 146 160
x-x⁻ : -30 8 22
(x-x⁻ )² : 900 64 484
Variance σ ² = ∑(x-x⁻ )²/ n-1
= [tex]\frac{900+64+484}{3-1}= 724[/tex]
Standard deviation
σ = √724 = 26.90
Step(ii):-
The 90% confidence interval for the mean noise level at such locations
[tex](x^{-} - Z_{0.90} \frac{S.D}{\sqrt{n} } , (x^{-} + Z_{0.90} \frac{S.D}{\sqrt{n} } )[/tex]
The critical value that should be used in constructing the confidence interval
Z₀.₁₀ = 1.645
[tex](138 - 1.645 \frac{26.90}{\sqrt{3} } , (138 + 1.645 \frac{26.90}{\sqrt{3} } )[/tex]
( 138 - 25.54 , 138 + 25.54 )
(112.46 , 163.54)