Confidence Interval for Population Proportion. Clopper-Pearson Interval [closed]

Question

I am sincerely stuck on this problem. Please help

alpha is 0.05

Answer 1

A - Confidence interval with a Normal approximation:

$$\hat p\pm z_{1-\frac{\alpha}{2}}\ \sqrt{\frac{\hat p (1-\hat p)}{n}},\ \ \hat p=\frac{s}{n}$$

in which $s$ is the number of successes (defects in this case) and $n$ the number of boards. $z_{1-\frac{\alpha}{2}}$ is the quantile $1-\frac{\alpha}{2}$ of a standard Normal (e.g. close to 1.96 for $\alpha=0.05$, a 95% confidence probability).

Making the substitutions $s=2$, $n=25$ $\alpha=0.05$ leads to

$$\hat p\pm z_{1-\frac{\alpha}{2}}\ \sqrt{\frac{\hat p (1-\hat p)}{n}}=[-0.026,0.186]$$

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 & 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

$$[\text{QBeta}(\alpha/2, s,n-s+1), \text{QBeta}(1-\alpha/2, s+1,n-s)]$$

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 $s=2$, $n=25$, $\alpha=0.05$ and the Beta quantile function from R, that is qbeta:

$$[\text{qbeta}(0.025,2,24), \text{qbeta}(0.975,3,23)]=[0.0098,0.2603].$$

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).