Test the claim about the population mean $\mu$ at the level of significance $\alpha$. Assume the population is normally distributed. Claim: $\mu \geq 1479 ; \alpha=0.02 ; \sigma=30$. Sample statistics: $\bar{x}=1473, n=25$ Identify the null and alternative hypotheses. Choose the correct answer. A. \[ \begin{array}{l} \mathbf{H}_{0}: \mu \geq 1479 \\ H_{a}: \mu<1479 \end{array} \] c. \[ \begin{array}{l} H_{0}: \mu \geq 1473 \\ H_{a}: \mu<1473 \end{array} \] E. \[ \begin{array}{l} H_{0}: \mu>1473 \\ H_{a}: \mu \leq 1473 \end{array} \] Calculate the standardized test statistic. The standardized test statistic is $\square$. (Round to two decimal places as needed.) Determine the P-value. $P=\square$ (Round to three decimal places as needed.) Determine the outcome and conclusion of the test. $\mathrm{H}_{0}$. At the $2 \%$ significance level, there $\mathrm{H}_{0}$. At the $2 \%$ significance level, there B. \[ \begin{array}{l} H_{0}: \mu<1479 \\ H_{a}: \mu \geq 1479 \end{array} \] D. \[ \begin{array}{l} H_{0}: \mu>1479 \\ H_{a}: \mu \leq 1479 \end{array} \] F. \[ \begin{array}{l} H_{0}: \mu<1473 \\ H_{a}: \mu \geq 1473 \end{array} \]

Step 1 ：Identify the null and alternative hypotheses. The null hypothesis is usually a statement of no effect or no difference. The alternative hypothesis is what you might believe to be true or hope to prove true. In this case, the claim is that the population mean is greater than or equal to 1479. So, the null hypothesis would be the opposite of the claim, which is that the population mean is less than 1479. The alternative hypothesis would be the claim itself. Therefore, the correct answer is: \[ \begin{array}{l} H_{0}: \mu<1479 \ H_{a}: \mu \geq 1479 \end{array} \]

Step 2 ：Calculate the standardized test statistic. The formula for the test statistic in a one-sample z-test is: \[ Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} \] where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

Step 3 ：Find the P-value. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. If the P-value is less than the level of significance, we reject the null hypothesis.

Step 4 ：Determine the outcome and conclusion of the test. If we reject the null hypothesis, we conclude that there is enough evidence to support the claim. If we fail to reject the null hypothesis, we conclude that there is not enough evidence to support the claim.

Step 5 ：Calculate the test statistic and P-value. The calculated test statistic is -1.0 and the P-value is approximately 0.159. Since the P-value is greater than the level of significance (0.02), we fail to reject the null hypothesis.

Step 6 ：Therefore, at the 2% significance level, there is not enough evidence to support the claim that the population mean is greater than or equal to 1479.

Step 7 ：Final Answer: The null and alternative hypotheses are: \[ \begin{array}{l} H_{0}: \mu<1479 \ H_{a}: \mu \geq 1479 \end{array} \] The standardized test statistic is \(\boxed{-1.0}\). The P-value is \(\boxed{0.159}\). At the 2% significance level, there is not enough evidence to support the claim that the population mean is greater than or equal to 1479.