Question

Ask Your Teacher An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 183 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% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.462, 0.566).

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 interval $1-\alpha$, we have the following confidence interval of proportions.

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

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

356 dies were examined by an inspection probe and 183 of these passed the probe. This neabs that $n = 356, \pi = \frac{183}{356} = 0.514$

95% confidence interval

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.514 - 1.96\sqrt{\frac{0.514*0.486}{356}} = 0.462$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.514 + 1.96\sqrt{\frac{0.514*0.486}{356}}{119}} = 0.566$

The 95% (two-sided) confidence interval for the proportion of all dies that pass the probe is (0.462, 0.566).