Answer:
95% Confidence Interval for the mean price of regular gasoline in that region
(3.61005, 3.76995)
Step-by-step explanation:
Step(i):-
Given sample size 'n' =40
mean of the sample x⁻ = 3.69 $
standard deviation of the sample "s' = 0.25 $
We will use students 't' distribution
degrees of freedom
ν = n-1 = 40-1 =39
[tex]t_{39,0.05} = 2.0227[/tex]
Step(ii):-
95% Confidence Interval for the mean price of regular gasoline in that region
[tex](x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , (x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })[/tex]
[tex](3.69 - 2.0227 \frac{0.25}{\sqrt{40} } , 3.69 + 2.0227 \frac{0.25}{\sqrt{40} })[/tex]
(3.69 - 0.07995 , 3.69 + 0.07995)
(3.61005, 3.76995)
Final answer:-
95% Confidence Interval for the mean price of regular gasoline in that region
(3.61005, 3.76995)
