Inference from sample proportion to population proportion I hope I'm posting in the right forum!  My daughter has something like the following problem in her college math class:  In a random sample of 100 registered voters, 75 people say they will vote for Clinton (the assignment was made up before the election).  Using statistical inference, what is the probability that Clinton will receive more than 60 percent of votes in the general election?
I have changed the numbers out of general principle.  I hope that it doesn't affect the problem materially.  I don't see how to get the answer to this kind of problem.  I understand confidence intervals, but I don't see how to get from that to an answer.  I hope that I correctly classified the type of problem in the title to the post.
Thanks,
Mark
 A: As the question is worded, it's hard to see a frequentist solution, because a frequentist would presumably treat the proportion who are voting for Clinton as a fixed, single value of the population. But it can be interpreted in a Bayesian fashion like this: if you get 75 successes from 100 independent Bernoulli trials with a true probability of success $θ$, then what is the posterior probability that $θ > .6$? Use a conjugate prior for $θ$ (which will be a beta distribution) and you can get a closed form for the posterior distribution of $θ$. The cumulative distribution function that corresponds to this posterior can tell you the final probability that $θ > .6$.
A: Kodiologist is right in its answer that frequentist solution does not give you an answer to the question as you worded it.
It can still give you some information about the sample proportion.
By calculating the standard error of the sample proportion:
$$ \text{SE}_p = \sqrt{\dfrac{p(1-p)}{n}} $$
you can get the statistical significance of this result comparing it to the hypothesized population proportion ($p = 0.6$).
In r, you can do this with
n_sample <- 100
p_est <- 75/n_sample
pnorm(p_est, mean = 0.6, sd = sqrt(p_est*(1-p_est)/n_sample), lower.tail = FALSE)
#[1] 0.0002660028

Note that this result does not tell you that there is a 99.97 % chance that the candidate will get more than 60 % of the vote.
What it tells you is that in a random sample of 100 people from a population in which the candidate will get more than 60 % of the vote, you would only get this result (of 75 % success estimate) less than 0.03 % of the time.
In other words, you can claim your sample estimate is larger than 60 % with a statistical significance $p = 0.00027$ (or $p < 0.001$).
