1. **State the problem:** We are testing the hypothesis about a population proportion $p$. Given: - Sample proportion $\hat{p} = 0.43$ - Null hypothesis $H_0: p = 0.5$ - Alternative hypothesis $H_1: p \neq 0.5$ - Decision rule: Reject $H_0$ if test statistic (T.S.) $< -1.96$ or $> 1.96$ 2. **Formula for test statistic for population proportion:** $$ T.S. = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} $$ where $p_0$ is the hypothesized population proportion and $n$ is the sample size. 3. **Important rules:** - The test statistic follows approximately a standard normal distribution under $H_0$. - The critical values $\pm 1.96$ correspond to a 5% significance level for a two-tailed test. 4. **Given test statistic:** $$ T.S. \approx -3.44 $$ 5. **Decision:** Since $-3.44 < -1.96$, the test statistic falls in the rejection region. 6. **Conclusion:** The correct conclusion should be to **reject** the null hypothesis $H_0$ because the test statistic is beyond the critical value. The statement "Fail to reject $H_0$" is incorrect based on the test statistic. **Summary:** - Test statistic $\approx -3.44$ - Critical values $\pm 1.96$ - Since $-3.44 < -1.96$, reject $H_0$ This means there is sufficient evidence to conclude that the population proportion $p$ is different from 0.5.