I dont understand this pls help
2 answers:
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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:
<u><em>Step(i)</em></u> :-
Noise levels at 3 airports were measured in decibels yielding the following data
108 146 160
Mean of given data
x : 108 146 160
x-x⁻ : -30 8 22
(x-x⁻ )² : 900 64 484
Variance σ ² = ∑(x-x⁻ )²/ n-1
=
Standard deviation
σ = √724 = 26.90
<u><em>Step(ii):</em></u>-
The 90% confidence interval for the mean noise level at such locations
The critical value that should be used in constructing the confidence interval
Z₀.₁₀ = 1.645
( 138 - 25.54 , 138 + 25.54 )
(112.46 , 163.54)