A sample of 25 circuit boards yields 2 defective boards after testing. What would be confidence interval for the true proportion of circuit boards in the lot from which the sample was drawn? Use the Clopper-Pearson Interval and also a normal approximation interval. In yur own words which do feel is better to use and why?

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Brock Floyd · Accepted Answer

Step 1A - Confidence interval with a Normal approximation:p^±z1−α2 p^(1−p^)n,  p^=snin which s is the number of successes (defects in this case) and n the number of boards. z1−α2 is the quantile 1−α2 of a standard Normal (e.g. close to 1.96 for SPKalpha=0.05ZSK, a 95% confidence probability).Making the substitutions s=2, n=25α=0.05 leads top^±z1−α2 p^(1−p^)n=[−0.026,0.186]Problem: this is a crude approximation, as n=25, the lower limit for the &quot;true&quot; frequency of defects is negative (-0.026), a non-sense.B - Confidence interval - Clopper Pearson (see for instance, Bilder vs. Loughin [2014] Categorical Data Analysis with R - CRC, page 15, for details)This interval is exact and usually defined by quantiles of a Beta distribution by[QBeta(α2,s,n−s+1), QBeta(1−α2, s+1,n−s)]in the expression, QBeta is the quantile function of a beta, and the parameters in each function represents, respectively, the probability for the quantile, the first parameter of the Beta, and the second parameter of the beta.Using s=2, n=25, α=0.05 and the Beta quantile function from R, that is qbeta:[qbeta(0.025,2,24),qbeta(0.975,3,23)]=[0.0098,0.2603].In this case the limits for the confidence interval, 0.0098 and 0.2603, are not negative. It is a large interval due to a low n (lots of uncertainty on the true frequency of defects).

A sample of 25 circuit boards yields 2

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