A random sample of n 30 shoppers yields that the mean amoun

A random sample of n = 30 shoppers yields that the mean amount spent on shopping is \\bar{x}x = $63.00 with a population standard deviation of \\sigma= $20.00. Find a 90% confidence interval for the true mean amount spent by shoppers.

Solution

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 =    0.05          
X = sample mean =    63          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    20          
n = sample size =    30          
              
Thus,              
Margin of Error E =    6.006156235          
Lower bound =    56.99384376          
Upper bound =    69.00615624          
              
Thus, the confidence interval is              
              
(   56.99384376   ,   69.00615624   ) [ANSWER]

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In case you still use t distribution even though n = 30 here, then this is the alternative:

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 =    63          
t(alpha/2) = critical t for the confidence interval =    1.699127027          
s = sample standard deviation =    20          
n = sample size =    30          
df = n - 1 =    29          
Thus,              
Margin of Error E =    6.20433467          
Lower bound =    56.79566533          
Upper bound =    69.20433467          
              
Thus, the confidence interval is              
              
(   56.79566533   ,   69.20433467   ) [ANSWER]