Answer:
95% Confidence Interval for the mean price of regular gasoline in that region
(3.61005, 3.76995)
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
<em>Given sample size 'n' =40</em>
<em>mean of the sample x⁻ = 3.69 $</em>
<em>standard deviation of the sample "s' = 0.25 $</em>
We will use students 't' distribution
degrees of freedom
ν = n-1 = 40-1 =39
<u><em>Step(ii):-</em></u>
95% Confidence Interval for the mean price of regular gasoline in that region
(3.69 - 0.07995 , 3.69 + 0.07995)
(3.61005, 3.76995)
<u>Final answer</u>:-
95% Confidence Interval for the mean price of regular gasoline in that region
(3.61005, 3.76995)