Bertie, (or anyone with statistics knowledge) can you help please ?
Pretty please with sprinkles on top :)))

I do not know if this is correct.
I don't know what to do with the 0.04 either ![]()
I am copying Bertie's answer from here
$$\\SE=\sqrt{\dfrac{p*q}{n}}\\\\
p=\frac{62}{70},\qquad q=\frac{8}{70},\qquad and \qquad n=70\\\\
SE=\sqrt{\dfrac{p*q}{n}}\\\\
SE=\sqrt{\dfrac{\frac{62}{70}*\frac{8}{70}}{70}}\\\\
SE=0.0380\qquad $4 dec places)$$
$${\sqrt{{\frac{\left(\left({\frac{{\mathtt{62}}}{{\mathtt{70}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{70}}}}\right)\right)}{{\mathtt{70}}}}}} = {\mathtt{0.038\: \!027\: \!150\: \!037\: \!068\: \!1}}$$
$${\frac{{\mathtt{62}}}{{\mathtt{70}}}} = {\frac{{\mathtt{31}}}{{\mathtt{35}}}} = {\mathtt{0.885\: \!714\: \!285\: \!714\: \!285\: \!7}}$$
$$0.8857$$ is the mean of this distribution.
A 95% confidence interval is within 2 standard errors. Mmm
So the 95% conficence limits will be
$$\\\displaystyle 0.8857 \pm 2\times 0.0380 = 0.8857\pm 0.076\\\\
$So the lower limit is $0.8097 \\
$and the upper limit is $0.9617\\$$
Now I am guessing EVEN more. :)
NO I am not going to bother because I really have no idea what I am doing :((
Maybe Bertie can help :/