A sample of 25 circuit boards yields 2

Thalia Ortiz

Thalia Ortiz

Answered question

2022-03-18

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?

Answer & Explanation

Brock Floyd

Brock Floyd

Beginner2022-03-19Added 6 answers

Step 1
A - Confidence interval with a Normal approximation:
p^±z1α2 p^(1p^)n,  p^=sn
in which s is the number of successes (defects in this case) and n the number of boards. z1α2 is the quantile 1α2 of a standard Normal (e.g. close to 1.96 for SPK\alpha=0.05ZSK, a 95% confidence probability).
Making the substitutions s=2, n=25α=0.05 leads to
p^±z1α2 p^(1p^)n=[0.026,0.186]
Problem: this is a crude approximation, as n=25, the lower limit for the "true" 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,ns+1), QBeta(1α2, s+1,ns)]
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).

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