Question

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 163 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.)

Answer 1

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Answer 2

(0.4062, 0.5098)

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

Z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`

For this problem, we have that:

356 dies were examined by an inspection probe and 163 of these passed the probe. This means that
`n = 365` and
`\pi = (163)/(356) = 0.458`

Assuming a stable process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.

So
`\alpha` = 0.05, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975[tex], so [tex]z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.458 - 1.96\sqrt{(0.458*0.542)/(356)} = 0.4062`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.458 + 1.96\sqrt{(0.458*0.542)/(356)} = 0.5098`

The correct answer is

(0.4062, 0.5098)