A survey of 25 randomly selected customers find the age is shown in years. The m

Question

A survey of 25 randomly selected customers find the age is shown in years. The mean is 31.72 years and the standard deviation is 9.45 years. A. construct a 98% confidence interval for the mean age of all customers assuming that the assumptions and conditions for the confidence interval have been met. B. how large is the margin of error? C. how would the confidence interval change if you had a see you at the standard deviation was known to be 10.0 years? 3343 az a146 30 35 a3 3 12 43 al 31 33 34 lo 7 13 3

Explanation / Answer

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.

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