On the basis of extensive tests, the yield point of a particular type of mild st

Question

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with a = 100. The composition of the bar has been slightly modified, but the modification is not believed to have affected either the normality or the value of sigma. Assuming this to be the case, if a sample of 81 modified bars resulted in a sample average yield point of 8495 lb, compute a 90% Cl for the true average yield point of the modified bar. (Round your answers to one decimal place.) How would you modify the interval in part (a) to obtain a confidence level of 98%? (Round your answer to two decimal places.)

Explanation / Answer

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 =    8495          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    100          
n = sample size =    81          
              
Thus,              
Margin of Error E =    18.27615141          
Lower bound =    8476.723849          
Upper bound =    8513.276151          
              
Thus, the confidence interval is              
              
(   8476.723849   ,   8513.276151   ) [ANSWER]

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b)

If c = 0.98, alpha/2 = (1-c)/2 = (1-0.98)/2 = 0.01.

Hence, we should change the z value to

zcrit =   2.326347874 = 2.33 [ANSWER]

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