Question

The certain article reported that, in a study of a particular wafer inspection process, 150 dies were examined by an inspection probe and 85 of these passed the probe. Assuming a stable process, calculate a 95% confidence interval for the proportion of all dies that pass the probe.

Answer 1

95% of confidence intervals for the proportion of all dies that pass the probe.

(0.4867 , 0.6453)

Explanation:

Step(i):-

Given sample size 'n' = 150

The sample proportion

`p = (x)/(n) = (85)/(150) = 0.566`

Level of significance = 0.05

The critical value Z₀.₀₅ = 1.96

Step(ii):-

95% of confidence intervals for the proportion of all dies that pass the probe.

`(p^(-) - Z_(0.05) (√(p(1-p)) )/(√(n) ) , p^(-) + Z_(0.05) (√(p(1-p)) )/(√(n) ))`

`(0.566 - 1.96(√(0.566(1-0.566)) )/(√(150) ) , 0.566 + 1.96(√(0.566(1-0.566)) )/(√(150) ))`

( 0.566 - 0.0793 , 0.566 + 0.0793)

(0.4867 , 0.6453)

Conclusion:-

95% of confidence intervals for the proportion of all dies that pass the probe.

(0.4867 , 0.6453)