When Brett Kavanaugh was nominated to be a Supreme Court justice, a survey of 1037 Americans showed that \( 50.6 \% \) of them disapproved of Kavanaugh. A newspaper published an article with this headline: "Majority of Americans Disapprove of Kavanaugh." Use a 0.10 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. \( H_{0} \cdot \mathrm{p}=0.5 \) \( H_{1} \cdot \mathrm{p}>0.5 \) (Type integers or decimals. Do not round.) Identify the test statistic. \( z=\square \) (Round to two decimal places as needed.)

To conduct the hypothesis test, we first need to identify the test statistic. The sample proportion (\( \hat{p} \)) of Americans disapproving of Kavanaugh is \( 0.506 \) based on the survey of 1037 respondents. The null hypothesis states that \( p = 0.5 \), while the alternative hypothesis states \( p > 0.5 \). The test statistic for a proportion can be calculated using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Where: - \( \hat{p} = 0.506 \) (sample proportion) - \( p_0 = 0.5 \) (proportion under the null hypothesis) - \( n = 1037 \) (sample size) Substituting the values: 1. Calculate the standard error: \[ SE = \sqrt{\frac{0.5(1 - 0.5)}{1037}} = \sqrt{\frac{0.5 \cdot 0.5}{1037}} = \sqrt{\frac{0.25}{1037}} \approx 0.0154 \] 2. Now, calculate the z-value: \[ z = \frac{0.506 - 0.5}{0.0154} \approx \frac{0.006}{0.0154} \approx 0.39 \] Thus, rounding to two decimal places, the test statistic is: \[ z \approx 0.39 \] So, the test statistic is: \[ z = 0.39 \]