A sample of 25 pieces of laminate used in the manufacture of circuit boards was

Question

A sample of 25 pieces of laminate used in the manufacture of circuit boards was selected, and the amount of warpage (in.) under particular conditions was determined for each piece, resulting in a sample mean warpage of 0.0628 and a sample standard deviation of 0.0066.

(a) Calculate a prediction for the amount of warpage of a single piece of laminate in a way that provides information about precision and reliability. (Use a 95% prediction interval. Round your answers to four decimal places.)

(b) Calculate an interval for which you can have a high degree of confidence that at least 95% of all pieces of laminate result in amounts of warpage that are between the two limits of the interval. (Use a 99% tolerance interval. Round your answers to four decimal places.)

Explanation / Answer

Solution:- given n = 25 mean = 0.0628,sd = 0.0066

a)  95% confidence interval t = 2.064
df = n-1 = 25-1 = 24
= X ± 2.064 * (s(sqrt((n+1)/n)))
= 0.0628 ± 2.064 * (0.0066(sqrt((25+1)/25)))
= (0.0489 , 0.0767)

b)  => X ± critical value * s
= 0.0628 ± 3.457 * 0.0066

= (0.0400 , 0.0856)

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