A random sample of the price of gasoline from 20 gas station

A random sample of the price of gasoline from 20 gas stations in a region gives the statistics below Complete parts a) through c). Find a 95% confidence interval for the mean price of regular gasoline in that region. Find the 90% confidence interval for the mean.

Solution

a)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    4.49          
t(alpha/2) = critical t for the confidence interval =    2.093024054          
s = sample standard deviation =    0.26          
n = sample size =    20          
df = n - 1 =    19          
Thus,              
Margin of Error E =    0.121683746          
Lower bound =    4.368316254          
Upper bound =    4.611683746          
              
Thus, the confidence interval is              
              
(   4.368316254   ,   4.611683746   )

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b)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    4.49          
t(alpha/2) = critical t for the confidence interval =    1.729132812          
s = sample standard deviation =    0.26          
n = sample size =    20          
df = n - 1 =    19          
Thus,              
Margin of Error E =    0.100527921          
Lower bound =    4.389472079          
Upper bound =    4.590527921          
              
Thus, the confidence interval is              
              
(   4.389472079   ,   4.590527921   ) [ANSWER]