Test the claim about the population mean \( \mu \) at the level of significance \( \alpha \). Assume the population is normally distrbuted. Clairi. \( \mu<5215 ; \alpha=0.05 \) Sample statistics: \( \bar{x}=5317, s=5949, n=52 \) What are the null and altemative hypotheses? \( H_{0}: \mu \geq 5215 \) \( H_{a}: \mu<5215 \) (Type integers or decimals. Do not round.) Find the slandardized test statistic L . \( \mathbf{t}=0.12 \) (Round to two decimal places as needed.) Find the P-value. P= \( \square \) (Round to three decimal places as needed.)

Alright, I need to test the claim about the population mean μ at the level of significance α. The population is normally distributed. The claim is that μ is less than 5215, and the significance level α is 0.05. The sample statistics provided are the sample mean \( \bar{x} = 5317 \), the sample standard deviation \( s = 5949 \), and the sample size \( n = 52 \). First, I need to determine the null and alternative hypotheses. The null hypothesis \( H_{0} \) is that the population mean μ is greater than or equal to 5215, and the alternative hypothesis \( H_{a} \) is that μ is less than 5215. So, \( H_{0}: \mu \geq 5215 \) and \( H_{a}: \mu < 5215 \). Next, I need to find the standardized test statistic \( t \). The formula for the t-statistic when the population standard deviation is unknown is: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] Plugging in the values: \[ t = \frac{5317 - 5215}{5949 / \sqrt{52}} = \frac{102}{5949 / 7.211} = \frac{102}{826.5} \approx 0.123 \] Rounding this to two decimal places, \( t \approx 0.12 \). Now, I need to find the P-value. Since this is a one-tailed test (left-tailed because \( H_{a}: \mu < 5215 \)), I'll use the t-distribution table or a calculator to find the probability that \( t \) is less than 0.12 with \( n-1 = 51 \) degrees of freedom. Looking up \( t = 0.12 \) with 51 degrees of freedom, the P-value is approximately 0.456.