Suppose that 53% of the population of voters were in favor of fighting the global warming. If we wanted to conduct a random sample of size $n$ of voters, how many should I survey if I want the probability that the sample proportion to be larger than $50\%$ with $95\%$ probability?
I know the we can normalize it using $$\frac{\hat p-p}{\sqrt{\frac{\hat p(1- \hat p)}{n}}}$$
but I couldn't figure out how to estimate it to get larger than 50%. Should this expression e equal to $X\sim (0,1)$ and $P(X>0.5)$?
I would appreciate explanation as well.
Technically, it depends on how the survey is conducted -- i.e., with or without replacement. In the following, I assume this is made with replacement (each voter is picked independently at random, with replacement -- i.e., the same voter could be picked several times).
Let $p=0.53$ be the probability to sample one voter in favor of fighting global warming, and $X_i$ be the random variable (indicator) that the $i$-th voter sampled is in favor of fighting global warming. $X_1,\dots,X_n$ are (a) independent and (b) identically distribututed according to a $\operatorname{Bernoulli}(p)$ distribution.
Let $$X\stackrel{\rm def}{=}\sum_{k=1}^nX_i$$ denote the (random variable) number of voters sampled that are in favor of fighting global warming. $X$ follows a $\operatorname{Binomial}(n,p)$ distribution (can you see why?). You are asked to find the smallest value of $n\geq 1$ for which $$ \mathbb{P}\left\{X < \frac{n}{2}\right\} < 0.05 $$ Can you use what you know about the tail of Binomial distribution (e.g.,properties of its CDF) to answer this?