A simple random sample of size $n=40$ is drawn from a population. The sample mean is found to be 105.2 , and the sample standard deviation is found to be 19.6 . Is the population mean greater than 100 at the $\alpha=0.10$ level of significance? Determine the null and alternative hypotheses. \[ \begin{array}{l} H_{0}: \mu=100 \\ H_{1}: \quad \mu>100 \end{array} \] Compute the test statistic. $=\square$ (Round to two decimal places as needed.)

Step 1 ：State the null and alternative hypotheses. The null hypothesis is that the population mean is equal to 100, and the alternative hypothesis is that the population mean is greater than 100. In mathematical terms, this can be written as: \[H_{0}: \mu=100\] \[H_{1}: \mu>100\]

Step 2 ：Calculate the test statistic. The test statistic for a hypothesis test about a population mean is a z-score, which is calculated using the formula: \[z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}\] where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(\sigma\) is the sample standard deviation, and \(n\) is the sample size.

Step 3 ：Substitute the given values into the formula. In this case, \(\bar{x} = 105.2\), \(\mu_0 = 100\), \(\sigma = 19.6\), and \(n = 40\).

Step 4 ：Calculate the z-score. The test statistic is approximately 1.68. This value represents how many standard deviations the sample mean is above the hypothesized population mean.

Step 5 ：Final Answer: The test statistic is approximately \(\boxed{1.68}\).