Question

An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 244 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

Answer:

The 95% confidence interval for the proportion of all dies that pass the probe is (0.637, 0.733).

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

356 dies were examined by an inspection probe and 244 of these passed the probe.

This means that $n = 356, \pi = \frac{244}{356} = 0.685$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.  

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.685 - 1.96\sqrt{\frac{0.685*0.315}{356}} = 0.637$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.685 + 1.96\sqrt{\frac{0.685*0.315}{356}} = 0.733$

The 95% confidence interval for the proportion of all dies that pass the probe is (0.637, 0.733).