Question

A random sample of the price of gasoline from 40 gas stations in a region gives an average price of $3.69 and sample standard deviation 0.25$. Construct a 95% Confidence Interval for the mean price of regular gasoline in that region A. (3.61, 3.77) B. (3.09, 4.29) C. (2.79, 4.59) D. (3.39, 3.99)

Answer 1

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

$t_{39,0.05} = 2.0227$

Step(ii):-

95% Confidence Interval for the mean price of regular gasoline in that region

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , (x^{-} + t_{\alpha } \frac{S}{\sqrt{n} })$

$(3.69 - 2.0227 \frac{0.25}{\sqrt{40} } , 3.69 + 2.0227 \frac{0.25}{\sqrt{40} })$

(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)