A research firm conducted a survey to determine the mean amount steady smokers s

Question

A research firm conducted a survey to determine the mean amount steady smokers spend on cigarettes during a week. They found the distribution of amounts spent per week followed the normal distribution with a population standard deviation of $5. A sample of 49 steady smokers revealed that formula168.mml = $20.

What is the point estimate of the population mean?

Using the 95% level of confidence, determine the confidence interval for .

Using the 95% level of confidence, determine the confidence interval for .

Explanation / Answer

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    20          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    5          
n = sample size =    49          
              
Thus,              
Margin of Error E =    1.399974275          
Lower bound =    18.60002573          
Upper bound =    21.39997427          
              
Thus, the confidence interval is              
              
(   18.60002573   ,   21.39997427   ) [ANSWER]

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