If the sample mean and standard deviation for fill weights o
Solution
a)
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 =    11.98          
 z(alpha/2) = critical z for the confidence interval =    1.959963985          
 s = sample standard deviation =    0.16          
 n = sample size =    81          
               
 Thus,              
 Margin of Error E =    0.034843804          
 Lower bound =    11.9451562          
 Upper bound =    12.0148438          
               
 Thus, the confidence interval is              
               
 (   11.9451562   ,   12.0148438   ) [ANSWER]
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b)
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 =    0.005          
 X = sample mean =    11.98          
 z(alpha/2) = critical z for the confidence interval =    2.575829304          
 s = sample standard deviation =    0.16          
 n = sample size =    81          
               
 Thus,              
 Margin of Error E =    0.045792521          
 Lower bound =    11.93420748          
 Upper bound =    12.02579252          
               
 Thus, the confidence interval is              
               
 (   11.93420748   ,   12.02579252   ) [ANSWER]
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c)
The 99% confidence interval is wider than the 95% confidence interval. This is because the critical z score increases as confidence level increases.
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d)
We are 95% confident that the true mean fill weight is between 11.9451562 and 12.0148438. [ANSWER]


