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QUESTION A random sample of size $n$ is taken without replacement
from a very large sample of components and $r$ of the sample are
found to be defective.  Write down an approximate$99\%$ confidence
interval for the proportion of the population which are defective
stating clearly \underline{three} reasons who your interval is
only approximate.

If $n=400$ show that the length of the longest such interval is
about 0.13.


ANSWER
 99\% CI approximately
  $\frac{r}{n}\pm 2.58 \sqrt{\frac{\frac{r}{n}(1-\frac{r}{n})}{n}}$\\
  The distribution is really Hypergeometric but the batch is very
  large so the approximate distribution is Binomial n,p, n large
  hence we can use the normal to approximate. Variance
  =$p\frac{q}{n}$ but we use $\frac{r}{n}$ for p as an
  approximation.
  n=400, Length of interval $2\times2.58
  \sqrt{\frac{\frac{r}{n}(1-\frac{r}{n})}{n}}$. (p(1-p)is maximum
  when p$=\frac{1}{2}$)\\
  Hence maximum length $2\times 2.58 \sqrt
  {\frac{\frac{1}{2}\frac{1}{2}}{400}}=\frac{2.58}{20}\approx
  0.13.$

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