A poll of 1168 teens aged 13 to 17 showed that \( 53 \% \) of them have made new friends online. Use a 0.01 significance level to test the claim that half of all teens have made new friends online. Use the P-val method. Use the normal distribution as an approximation to the binomial distribution. 0 of 4 Let p denote the population proportion of all teens aged 13 to 17 who have made new friends online Identify the null and altemative hypotheses. \( \mathrm{H}_{0} \cdot \mathrm{p}=0.5 \) \( \mathrm{H}_{1} \mathrm{p} \neq 0.5 \) (Type integers or decimals. Do not round.) Identify the test statistic. \( \mathrm{z}=2.05 \) (Round to two decimal places as needed.) Identify the P-value. P-value \( =\square \) (Round to three decimal places as needed.) a

**Step 1. State the Hypotheses** We are testing the claim that half of all teens have made new friends online. - Null hypothesis: \( H_0: p = 0.5 \) - Alternative hypothesis: \( H_1: p \neq 0.5 \) --- **Step 2. Compute the Test Statistic** The test statistic for a proportion is given by \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}, \] where - \(\hat{p} = 0.53\) is the sample proportion, - \(p_0 = 0.5\) is the hypothesized proportion, and - \(n = 1168\) is the sample size. Calculate the standard error: \[ \text{SE} = \sqrt{\frac{0.5 \times 0.5}{1168}} = \sqrt{\frac{0.25}{1168}} \approx \sqrt{0.000214} \approx 0.01464. \] Now, compute \(z\): \[ z = \frac{0.53 - 0.5}{0.01464} \approx \frac{0.03}{0.01464} \approx 2.05. \] --- **Step 3. Find the P-value** Since \( H_1: p \neq 0.5 \) is a two-tailed test, the P-value is given by \[ \text{P-value} = 2 \Pr(Z > |2.05|). \] Looking up the normal probability for \(z = 2.05\), we have \[ \Pr(Z > 2.05) \approx 0.0202. \] Thus, the two-tailed P-value is \[ \text{P-value} = 2 \times 0.0202 \approx 0.0404. \] Rounded to three decimal places, the P-value is \[ 0.040. \] --- **Final Answers** - Null and alternative hypotheses: - \( H_0: p = 0.5 \) - \( H_1: p \neq 0.5 \) - Test statistic: \( z = 2.05 \) - P-value: \( 0.040 \)