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An article reported that, in a study of a particular wafer inspection process, 356 dies were examined by an inspection probe and 223 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.)

User Rickerby
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Answer:


0.626 - 1.96\sqrt{(0.626(1-0.626))/(356)}=0.549


0.626 + 1.96\sqrt{(0.626(1-0.626))/(356)}=0.703

The 95% confidence interval would be given by (0.549;0.703)

Explanation:

Notation and definitions


X=223 number of dies that passed the inspection.


n=356 random sample taken


\hat p=(223)/(356)=0.626 estimated proportion of dies that passed the inspection


p true population proportion of dies that passed the inspection

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution


p \sim N(p,\sqrt{(p(1-p))/(n)})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
\alpha=1-0.95=0.05 and
\alpha/2 =0.025. And the critical value would be given by:


z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96

The confidence interval for the mean is given by the following formula:


\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}

If we replace the values obtained we got:


0.626 - 1.96\sqrt{(0.626(1-0.626))/(356)}=0.549


0.626 + 1.96\sqrt{(0.626(1-0.626))/(356)}=0.703

The 95% confidence interval would be given by (0.549;0.703)

User Zeshan Munir
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