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zzz [600]
3 years ago
5

Ask Your Teacher An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an ins

pection 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.)
Mathematics
1 answer:
Arte-miy333 [17]3 years ago
6 0

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).

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