An article reported that, in a study of a particular wafer inspection process, 356 dies were...

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 154 of these passed the probe. Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe. (Round your answers to three decimal places.)

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Expert Solution

Solution :

Given that,

n = 356

x = 154

= x / n = 154 / 356 = 0.43

1 - = 1 - 0.43 = 0.57

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_{/ 2} * $\sqrt$ [( * (1 - )] / n)

= 1.96 * $\sqrt$ [(0.43 * 0.57)] / 356)

= 0.051

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.43 - 0.051 < p < 0.43 + 0.051

0.379 < p < 0.481

(0.379, 0.481)

orchestra answered 1 year ago

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