According to a survey by Runzheimer International, the average cost of a fast-fo

Question

According to a survey by Runzheimer International, the average cost of a fast-food meal (quarter-pound cheeseburger, large fries, medium soft drink, excluding taxes) in Seattle is $4.82. Suppose this figure was based on a sample of 27 different establishments and the standard deviation was $0.37. Construct a 95% confidence interval for the population mean cost for all fast-food meals in Seattle. Assume the costs of a fast-food meal in Seattle are normally distributed. Using the interval as a guide, is it likely that the population mean is really $4.50? why or why not?

Explanation / Answer

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 =    4.82          
t(alpha/2) = critical t for the confidence interval =    2.055529439          
s = sample standard deviation =    0.37          
n = sample size =    27          
df = n - 1 =    26          
Thus,              
Margin of Error E =    0.146367125          
Lower bound =    4.673632875          
Upper bound =    4.966367125          
              
Thus, the confidence interval is              
              
(   4.673632875   ,   4.966367125   ) [ANSWER]

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Since 4.50 DOES NOT FALL INSIDE the interval, we are 95% confident that u IS GREATER THAN 4.50. [answer]

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