In an experiment using 30 mice, the sample proportion of the mice that gained weight after a drug injection is 0.65. What is the 99.7% confi

I'm not a statistician as such so a proper one, (a proper statistician that is ), might take issue, however, so far as I'm aware, the S.E. that's being calculated here is the S.E. for the parent population, not the sample.  We don't have a standard deviation for the sample.

BTW, this is the method used in opinion poles for elections and so on.

I would like to see this answered too please :)

The Standard Error for the proportion of weight gains in the whole population will be

$$\displaystyle \sqrt{\frac{0.65 \times 0.35}{30}}=0.087\quad \text{ 3 dp},$$

which means that the 99.7% confidence limits will be

$$\displaystyle 0.65 \pm 3\times 0.087 = 0.65 \pm 0.26.$$

Thanks Bertie,

So the Standard Error of  a Bernoulli distribution  is just     

$$SE=\sqrt{\dfrac{p*q}{n}}$$               

(edited a little)

Ok thanks Bertie,

I shall alter my answer a little :)