I am really confused about the confidence interval for n=100 tosses and k=52 heads. If I estimate the probability $\hat{p}$:
$\hat{p} = \frac{52}{100} = 0.52$
$Var(\hat{p}) = pq = 0.52 * 0.48 = 0.2496$, being a bernoulli process
$\sigma = \sqrt{Var(\hat{p})} = 0.5$
$SE = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{10} = 0.05 $
$CI = 0.52 ± Z_{0.90} * 0.05 = [0.44, 0.60] $
This seems correct to me. However, when I calculate the confidence interval for the actual number of heads I get:
${X} = 52$
$Var(X) = npq = 100 * 0.52 * 0.48 = 24.96$, being a binomial distribution
$\sigma = \sqrt{Var(X)} = 5$
$SE = \frac{\sigma}{\sqrt{n}} = \frac{5}{10} = 0.5 $
$CI = 52 ± Z_{0.90} * 0.5 = [51.17, 52.82] $. I would expect this to be $[44, 60]$ as in my previous result.
Why are these two confidence intervals different?
Thank you!