5. The fraction of defective integrated circuits produced in a photolitho- graph

Question

5. The fraction of defective integrated circuits produced in a photolitho- graphy process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. (a) Calculate a 95% two-sided CI on the fraction of defective circuits (b) Calculate a 95% upper confidence bound on the fraction of defec- (c) Using the observed sample proportion, calculate the sample size produced by this particular tool. tive circuits. required to have a 99% confidence interval no wider than 0.02 units. (d) Repeat the calculation in (c), but using the most conservative estimate of sample size.

Explanation / Answer

Question A.

given that,

possibile chances (x)=13

sample size(n)=300

success rate ( p )= x/n = 0.0433

CI = confidence interval

confidence interval = [ 0.0433 ± 1.96 * Sqrt ( (0.0433*0.9567) /300) ) ]

= [0.0433 - 1.96 * Sqrt ( (0.0433*0.9567) /300) , 0.0433 + 1.96 * Sqrt ( (0.0433*0.9567) /300) ]

= [0.0203 , 0.0664]

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interpretations:

1. We are 95% sure that the interval [ 0.0203 , 0.0664] contains the true population proportion

2. If a large number of samples are collected, and a confidence interval is created

for each sample, 95% of these intervals will contains the true population proportion

Question B.

success rate ( p )= x/n = 0.0433

CI = confidence interval

confidence interval = [ 0.0433 ± 1.645 * Sqrt ( (0.0433*0.9567) /300) ) ]

= [0.0433 - 1.645 * Sqrt ( (0.0433*0.9567) /300) , 0.0433 + 1.645 * Sqrt ( (0.0433*0.9567) /300) ]

= [0.024 , 0.0627]

Question C.

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.01 is = 2.576

Sample Proportion = 0.04

ME = 0.02

n = ( 2.576 / 0.02 )^2 * 0.04*0.96

= 637.0345 ~ 638

Question D.

we’re using p = q = .50 for the most conservative estimate

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.01 is = 2.576

Sample Proportion = 0.5

ME = 0.02

n = ( 2.576 / 0.02 )^2 * 0.5*0.5

= 4147.36 ~ 4148

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