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: $\bar{x}=1243.48, n=300$ \[ H_{a}: \mu>1220 \] C. \[ \begin{array}{l} H_{0}: \mu \leq 1243.48 \\ H_{a}: \mu>1243.48 \end{array} \] E. \[ \begin{array}{l} H_{0}: \mu \geq 1220 \\ H_{a}: \mu<1220 \end{array} \] \[ H_{a} \cdot \mu<1243.48 \] D. \[ \begin{array}{l} H_{0}: \mu>1220 \\ H_{a}: \mu \leq 1220 \end{array} \] F. \[ \begin{array}{l} H_{0}: \mu>1243.48 \\ H_{a}: \mu \leq 1243.48 \end{array} \] Calculate the standardized test statistic. The standardized test statistic is (Round to two decimal places as needed.)

Step 1 ：Given the claim is that \(\mu>1220\). The null hypothesis, \(H_{0}\), is the statement that the mean is less than or equal to 1220, and the alternative hypothesis, \(H_{a}\), is the statement that the mean is greater than 1220.

Step 2 ：We are given the sample mean, \(\bar{x}=1243.48\), the sample size, \(n=300\), and the population standard deviation, \(\sigma=195.94\).

Step 3 ：We can use these values to calculate the standardized test statistic, which is a z-score. The formula for the z-score is \(z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\)

Step 4 ：Substitute the given values into the formula to find the z-score: \(\mu = 1220\), \(\bar{x} = 1243.48\), \(n = 300\), \(\sigma = 195.94\)

Step 5 ：The calculated z-score is approximately 2.08. This is the standardized test statistic that we were asked to find.

Step 6 ：Final Answer: The standardized test statistic is approximately \(\boxed{2.08}\)