When Brett Kavanaugh was nominated to be a Supreme Court justice, a survey of 1037 Americans showed that \( 506 \% \) of them disapproved of Kavanaugh. A newspaper published an article with this headline: "Majority of Americans Disapprove of Kavanaugh. "Use a 010 significance level to test the claim made in that headline. Use the P-value method Use the normal distribution as an approximation to the binomial distribution. Let p denote the population proportion of all Americans who disapproved of Kavanaugh. Identify the null and alternative hypotheses. \( H_{0}: \mathrm{p}=0.5 \) \( H_{1}: \mathrm{p}>05 \) (Type integers or decimals. Do not round.) Identify the test statistic. \( \mathrm{z}=0.39 \) (Round to two decimal places as needed.) Identify the P-value P-value \( = \) (Round to three decimal places as needed.)

To solve this hypothesis test, we can follow the steps: 1. **Identify the sample proportion (\( \hat{p} \))**: The survey shows that 506 out of 1037 disapprove. Thus, \( \hat{p} = \frac{506}{1037} \approx 0.488 \). 2. **Calculate the test statistic (\( z \))**: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] where \( p_0 = 0.5 \) (the null hypothesis proportion) and \( n = 1037 \) (the sample size). \[ z = \frac{0.488 - 0.5}{\sqrt{\frac{0.5(1 - 0.5)}{1037}}} = \frac{-0.012}{\sqrt{\frac{0.25}{1037}}} \approx \frac{-0.012}{0.0154} \approx -0.779 \] 3. **Identify the P-value**: This is a one-tailed test to the right (since \( H_1: p > 0.5 \)). Using a standard normal distribution table or calculator: - For \( z = -0.779 \), we find the right-tail P-value: \[ P(z > -0.779) \approx 1 - 0.219 = 0.781 \] **Final Answers:** Identifying the P-value results in: P-value \( \approx 0.781 \) (rounded to three decimal places).