Let's suppose I am surveying $n$ democratic citizens about whether they will vote for Candidate A or Candidate B. Let's suppose my survey yields $x$ for Candidate A and $n-x$ for Candidate B, $x \in \Bbb Z_+$. Let $p$ denote the fraction of voters who prefer Candidate A. Let's suppose I estimate $p$ by using $\hat{p} = \frac{x}{n}$. Suppose my variance is $\hat{p}(1-\hat{p})$. Suppose $H_0 : p = 0.5$ and $H_A : p \ne 0.5$.
For $n=400, x = 215$, the estimator $\hat{p} = \frac{43}{80}$ is and a 95% confidence interval of $$\frac{43}{80} - 1.96 \frac{\sqrt{1591}}{1000}<\hat{p} < \frac{43}{80} + 1.96 \frac{\sqrt{1591}}{1000}$$
$$0.48864 < \hat{p} < 0.58656$$
My Question
Does this confidence interval tell me anything about the validity of my null hypothesis? That is, can I decide whether to accept or reject my null hypothesis based on $0.48864 < \hat{p} < 0.58656$?
