A manufacturer of inexpensive brass weights weighs of a subs

A manufacturer of inexpensive brass weights, weighs of a subset of their 3 kg weights selected at random on a precision scale. The mean weight is found to be 2.9976 kg, with a standard deviation of 56.8 g. Assuming that the parent population of weights is normally distributed, what is the 90% confidence for a particular weight selected at random? a) If the sample size was 15 weights. b) If the sample size was 30 weights. c) If the sample size was 150 weights.

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 =    0.05          
X = sample mean =    2.9976          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    0.0568          
n = sample size =    15          
              
Thus,              
Margin of Error E =    0.024122925          
Lower bound =    2.973477075          
Upper bound =    3.021722925          
              
Thus, the confidence interval is              
              
(   2.973477075   ,   3.021722925   ) [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 =    0.05          
X = sample mean =    2.9976          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    0.0568          
n = sample size =    30          
              
Thus,              
Margin of Error E =    0.017057484          
Lower bound =    2.980542516          
Upper bound =    3.014657484          
              
Thus, the confidence interval is              
              
(   2.980542516   ,   3.014657484   ) [ANSWER]

******************

c)

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 =    2.9976          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    0.0568          
n = sample size =    150          
              
Thus,              
Margin of Error E =    0.007628339          
Lower bound =    2.989971661          
Upper bound =    3.005228339          
              
Thus, the confidence interval is              
              
(   2.989971661   ,   3.005228339   ) [ANSWER]