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

Answer:

(0.4062, 0.5098)

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 163 of these passed the probe. This means that $n = 365$ and $\pi = \frac{163}{356} = 0.458$

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 $1 - \frac{0.05}{2} = 0.975[tex], so [tex]z = 1.96$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

The correct answer is

(0.4062, 0.5098)