1. The problem is to test the hypothesis about a population proportion $p$ with null hypothesis $H_0: p = 0.49$ and alternative hypothesis $H_1: p \neq 0.49$. 2. The test statistic given is $z = -3.1825$ and the p-value is $0.0015$. 3. For hypothesis testing of a proportion, the test statistic formula is: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$$ where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized population proportion, and $n$ is the sample size. 4. The p-value represents the probability of observing a test statistic as extreme as $-3.1825$ under the null hypothesis. 5. Since the p-value $0.0015$ is less than common significance levels (e.g., 0.05), we reject the null hypothesis $H_0$. 6. Conclusion: There is sufficient evidence to conclude that the population proportion $p$ is not equal to $0.49$.