A study, done as part of a yield improvement effort at a semiconductor manufactu

Question

A study, done as part of a yield improvement effort at a semiconductor manufacturing facility, provided defect data for a sample of 450 wafers. Of the 450, 116 wafers were judged to be of bad quality.

a) Calculate a 95% confidence interval for the proportion of wafers that were of bad quality.

b) Using the sample data, how many wafers would have to be sampled to reduce the error to within 0.03?

c) Now, assume the sample data had not yet been collected and you are trying to determine how many observations would be needed (at most) to estimate the proportion defective within 0.03. Determine the number of wafers that would need to be sampled.

Explanation / Answer

a)

95% confidence interval for the proportion of wafers that were of bad quality =0.2174 ; 0.2982

b)

c) for no estimate of proportion ; conservative estimated proportion =0.5

                                                                                           x                    =   116 sample size n                    = 450 sample proportion     p x/n= 0.2578 std error       =Se            =(p*(1-p)/n) = 0.0206 for 95 % CI value of z= 1.9600 margin of error E=z*std error                            = 0.0404 lower confidence bound=sample proportion-margin of error 0.2174 Upper confidence bound=sample proportion+margin of error 0.2982

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