A survey of 25 randomly selected customers find the age is s

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 =    0.01          
X = sample mean =    31.72          
t(alpha/2) = critical t for the confidence interval =    2.492159473          
s = sample standard deviation =    9.45          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
Margin of Error E =    4.710181404          
Lower bound =    27.0098186          
Upper bound =    36.4301814          
              
Thus, the confidence interval is              
              
(   27.0098186   ,   36.4301814   ) [ANSWER]

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B)
As we saw,

Margin of Error E =    4.710181404   [ANSWER]

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c)

it will probably become narrower, because critical z scores are smaller the t scores. However, the standard deviation is bigger, so we will see.

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 =    0.01          
X = sample mean =    31.72          
z(alpha/2) = critical z for the confidence interval =    2.326347874          
s = sample standard deviation =    10          
n = sample size =    25          
              
Thus,              
Margin of Error E =    4.652695748          
Lower bound =    27.06730425          
Upper bound =    36.37269575          
              
Thus, the confidence interval is              
              
(   27.06730425   ,   36.37269575   )
      
As we can see, there is less margin of error, so THIS IS NARROWER.