Question

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 150 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

The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.37, 0.472)

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`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:

`n = 356, \pi = (150)/(356) = 0.421`

95% confidence level

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

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.421 - 1.96\sqrt{(0.421*0.579)/(150)} = 0.37`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.421 + 1.96\sqrt{(0.421*0.579)/(150)} = 0.472`

The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.37, 0.472)