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

Answer:

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

(0.4867 , 0.6453)

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 150

The sample proportion

                                $p = \frac{x}{n} = \frac{85}{150} = 0.566$

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.

$(p^{-} - Z_{0.05} \frac{\sqrt{p(1-p)} }{\sqrt{n} } , p^{-} + Z_{0.05} \frac{\sqrt{p(1-p)} }{\sqrt{n} })$

$(0.566 - 1.96\frac{\sqrt{0.566(1-0.566)} }{\sqrt{150} } , 0.566 + 1.96\frac{\sqrt{0.566(1-0.566)} }{\sqrt{150} })$

( 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)