A poll of 1142 Americans showed that $46.8 \%$ of the respondents prefer to watch the news rather than read or listen to it. Use those results with a 0.05 significance level to test the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. 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 prefer to watch the news rather than read or listen to it. Identify the null and alternative hypotheses. \[ \begin{array}{l} H_{0}: p=5 \\ H_{1}: p<5 \end{array} \] (Type integers or decimals. Do not round.) Identify the test statistic. \[ z=-2.16 \] (Round to two decimal places as needed.) Identify the P-value. P-value $=.015$ (Round to three decimal places as needed.) State the conclusion about the null hypothesis, os well as the final conclusion that addresses the original claim. the null hypothesis. There sufficient evidence to the claim that fewer than half of Americans prefer to watch the news rather than read or
Solution
Step 1 ：Let's denote the population proportion of all Americans who prefer to watch the news rather than read or listen to it as \(p\). The null hypothesis \(H_{0}\) is that \(p = 0.5\) and the alternative hypothesis \(H_{1}\) is that \(p < 0.5\).
Step 2 ：The sample size is 1142 and 46.8% of them prefer to watch the news. So, \(n = 1142\) and \(x = 0.468 * n = 534.456\).
Step 3 ：We calculate the test statistic using the formula \(z = \frac{\hat{p} - p_{0}}{\sqrt{\frac{p_{0}(1 - p_{0})}{n}}}\), where \(\hat{p} = \frac{x}{n}\), \(p_{0} = 0.5\), and \(n = 1142\). Substituting these values, we get \(z = -2.16\).
Step 4 ：The P-value is the probability that we observe a test statistic as extreme as \(z = -2.16\) under the null hypothesis. Using the standard normal distribution, we find that the P-value is 0.015.
Step 5 ：Since the P-value (0.015) is less than the significance level of 0.05, we reject the null hypothesis \(H_{0}\).
Step 6 ：Therefore, there is sufficient evidence to support the claim that fewer than half of Americans prefer to watch the news rather than read or listen to it. The test statistic is \(\boxed{-2.16}\) and the P-value is \(\boxed{0.015}\).
