A machine produces parts with lengths that are normally dist

A machine produces parts with lengths that are normally distributed with = 0.53. A sample of 18 parts has a mean length of 76.61.

(a) Give a point estimate for . (Give your answer correct to two decimal places.)


(b) Find the 90% confidence maximum error of estimate for . (Give your answer correct to three decimal places.)


(c) Find the 90% confidence interval for . (Give your answer correct to three decimal places.)

Solution

a)

It is the sample mean,

X = 76.61 [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 =    76.61          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    0.53          
n = sample size =    18          
              
Thus,              
Margin of Error E =    0.205   [ANSWER]      

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C)

Also,

Lower bound =    76.405          
Upper bound =    76.815 [ANSWER]