Let c 095 s 7 and n 23 Find the margin of error E Assume

Let c = 0.95, s = 7, and n = 23. Find the margin of error, E. Assume that the population is normally distributed.

Solution

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
      
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    7          
n = sample size =    23          
              
Thus,              

Margin of Error E =    2.860765193   [ANSWER]

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In case you use t distrbution for n < 30, this is the alternative:

Note that              

Margin of Error E = t(alpha/2) * s / sqrt(n)              
      
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) = critical t for the confidence interval =    2.073873068          
s = sample standard deviation =    7          
n = sample size =    23          
df = n - 1 =    22          

Thus,              

Margin of Error E =    3.027026993   [ANSWER]