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-value method. Use the normal distribution as an approximation to the binomial distribution. Let \( p \) denote the population proportion of all teens aged 13 to 17 who have made new friends online. Identify the null and alternative hypotheses \( H_{0} \cdot \mathrm{P}=0.5 \) \( H_{1}: \mathrm{P} \neq 0.5 \) (Type integers or decimals. Do not round.) Identify the test statistic. \( z=\square \) (Round to two decimal places as needed)

To carry out the hypothesis test, we first need to compute the test statistic using the sample proportion. The sample size \( n \) is 1168, and the sample proportion \( \hat{p} \) is \( 0.53 \) (since \( 53\% \) of the teens made new friends online). The null hypothesis is stated as: \( H_{0}: p = 0.5 \) The alternative hypothesis is stated as: \( H_{1}: p \neq 0.5 \) Now, let's calculate the test statistic \( z \). The formula for \( z \) is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] Where: - \( \hat{p} = 0.53 \) - \( p_0 = 0.5 \) - \( n = 1168 \) Calculating the denominator: \[ \sqrt{\frac{0.5(1 - 0.5)}{1168}} = \sqrt{\frac{0.5 \times 0.5}{1168}} = \sqrt{\frac{0.25}{1168}} \approx \sqrt{0.0002149} \approx 0.0147 \] Now substituting into the \( z \) formula: \[ z = \frac{0.53 - 0.5}{0.0147} \approx \frac{0.03}{0.0147} \approx 2.04 \] The test statistic \( z \) is approximately \( 2.04 \) (rounded to two decimal places). So, the test statistic is: \( z = 2.04 \)