Use technology to help you test the claim about the population mean, $\mu$, at the given level of significance, $\alpha$, using the given sample statistics. Assume the population is normally distributed. Claim: $\mu>1220 ; \alpha=0.04 ; \sigma=195.94$. Sample statistics: $\bar{x}=1243.48, n=300$ \[ H_{a}: \mu>1 \angle 43.48 \] E. $H_{0}: \mu \geq 1220$ $H_{a}: \mu<1220$ \[ \begin{array}{l} H_{a}: \mu \leq 1 \angle L U \\ H_{0}: \mu>1243.48 \\ H_{a}: \mu \leq 1243.48 \end{array} \] Calculate the standardized test statistic. The standardized test statistic is 2.08 (Round to two decimal places as needed.) Determine the P-value. $\mathrm{P}=$ (Round to three decimal places as needed.)

Step 1 ：Given the sample mean (\(\bar{x}\)) is 1243.48, the population standard deviation (\(\sigma\)) is 195.94, the sample size (n) is 300, and the hypothesized population mean (\(\mu\)) is 1220.

Step 2 ：We can use these to calculate the z-score, which is the standardized test statistic. The formula for the z-score is: \[z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\]

Step 3 ：Substituting the given values into the formula, we get a z-score of approximately 2.08.

Step 4 ：After calculating the z-score, we can use a z-table or a statistical calculator to find the P-value. The P-value is the probability that a z-score is more extreme than the calculated z-score, given that the null hypothesis is true.

Step 5 ：The calculated P-value is approximately 0.019, which is less than the given level of significance, \(\alpha=0.04\). This means that we would reject the null hypothesis in favor of the alternative hypothesis.

Step 6 ：Final Answer: The standardized test statistic is approximately \(\boxed{2.08}\) and the P-value is approximately \(\boxed{0.019}\).