Two machines produce DVDs, For quality control purposes, the rate of defective D

Question

Two machines produce DVDs, For quality control purposes, the rate of defective DVDs produced by each machine is analyzed, The observations revealed that of 700 DVDs produced by the first machine, I5 were defective; of 600 DVDs produced by the second machine, I2 were defective, Construct at 90% confidence interval on the true proportion of defective DVDs produced by machine 1, Construct a 90% confidence interval on the true proportion of defective DVDs produced by machine 2, Construct a 95% confidence interval on the difference between the true proportions of defective DVDs produced by the two machines.

Explanation / Answer

Machine 1: n1=700, p1=15/700= 0.021
Machine 2: n2=600, p2=12/600 = 0.02

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(a) Give a=0.1, Z(0.05)=1.645 (check standard normal table)
So 90% CI is
p1± Z*[p1*(1-p1)/n1]

--> 0.021 ± 1.645*sqrt(0.021*(1-0.021)/700)

--> ( 0.012, 0.03)

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(b) Give a=0.1, Z(0.05)=1.645 (check standard normal table)
So 90% CI is
p2± Z*[p2*(1-p2)/n2]

--> 0.02 ± 1.645*sqrt(0.02*(1-0.02)/600)

--> ( 0.011, 0.029)

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(c) Given a=0.05, Z(0.025)=1.96 (check standard normal table)

SO 95% CI is

(p1-p2) ± Z*[p1*(1-p1)/n1 + p2*(1-p2)/n2]

--> (0.021-0.02) ±1.96*sqrt(0.021*(1-0.02)/700 + 0.02*(1-0.02)/600)

--> ( -0.014, 0.016)

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