Question

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 10 defectives. (a) Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. Round the answers to 4 decimal places.

Answer 1

The correct answer is

(0.0128, 0.0532)

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:

In a random sample of 300 circuits, 10 are defective. This means that
`n = 300` and
`\pi = (10)/(300) = 0.033`

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

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))/(300)} = 0.033 - 1.96\sqrt{(0.033*0.967)/(300)} = 0.0128`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(300)} = 0.033 + 1.96\sqrt{(0.033*0.967)/(300)} = 0.0532`

The correct answer is

(0.0128, 0.0532)