When Brett Kavanaugh was nominated to be a Supreme Court justice, a survey of 1036 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 0.05 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 ptoportion of all Americans who disapproved of Kavanaugh. Identify the null and alternative hypotheses.
(Type integers or decimals. Do not round.)
Identify the test statistic.
\[
z=
\]
(Round to two dechrial places as needed.)
Identify the P-value.
$\mathrm{P}$-value $=$
(Round to three decimal places as needed.)
Answer
State the final answer: The null and alternative hypotheses are \(H_0: p = 0.5\) and \(H_a: p > 0.5\). The test statistic is \(\boxed{z = 0.39}\). The P-value is \(\boxed{0.350}\).
Step 1 ：State the null hypothesis and the alternative hypothesis. The null hypothesis is that the population proportion is equal to 0.5 (i.e., 50%), which means that the majority of Americans do not disapprove of Kavanaugh. The alternative hypothesis is that the population proportion is greater than 0.5, which means that the majority of Americans disapprove of Kavanaugh. So, we have \(H_0: p = 0.5\) and \(H_a: p > 0.5\).
Step 2 ：Calculate the test statistic using the formula for a z-score in a proportion test, which is \((p̂ - p0) / \sqrt{(p0 * (1 - p0)) / n}\), where p̂ is the sample proportion, p0 is the population proportion under the null hypothesis, and n is the sample size. Given that n = 1036, p̂ = 0.506, and p0 = 0.5, we find that the test statistic is z = 0.3862434465463463.
Step 3 ：Calculate the P-value, which is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, under the null hypothesis. We can find this value using a standard normal distribution table or a calculator with a normal distribution function. The calculated P-value is 0.3496581899088608.
Step 4 ：Round the test statistic and the P-value to the required number of decimal places. The test statistic, rounded to two decimal places, is \(z = 0.39\). The P-value, rounded to three decimal places, is 0.350.
Step 5 ：State the final answer: The null and alternative hypotheses are \(H_0: p = 0.5\) and \(H_a: p > 0.5\). The test statistic is \(\boxed{z = 0.39}\). The P-value is \(\boxed{0.350}\).
