Use a t-test to test the claim about the population mean $\mu$ at the given level of significance $\alpha$ using the given sample statistics. Assume the Claim: $\mu=52,100 ; \alpha=0.01$ Sample statistics: $\bar{x}=51,039, s=2300, n=19$ Click the icon to view the t-distribution table. What are the null and alternative hypotheses? Choose the correct answer below. A. \[ \begin{array}{l} H_{0}: \mu \leq 52,100 \\ H_{a}: \mu>52,100 \end{array} \] C. \[ \begin{array}{l} H_{0}: \mu \neq 52,100 \\ H_{a}: \mu=52,100 \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu \geq 52,100 \\ H_{a}: \mu<52,100 \end{array} \] D. \[ \begin{array}{l} H_{0}: \mu=52,100 \\ H_{a} \cdot \mu \neq 52,100 \end{array} \] What is the value of the standardized test statistic? The standardized test statistic is -2.01 . (Round to two decimal places as needed.) What is(are) the critical value(s)? The critical value(s) is(are) $\square$. (Round to three decimal places as needed. Use a comma to separate answers as needed.)
Solution
Step 1 ：The null and alternative hypotheses are: \n \[ H_{0}: \mu=52,100 \] \n \[ H_{a}: \mu \neq 52,100 \]
Step 2 ：The value of the standardized test statistic is calculated using the formula for the t-statistic, which is (sample mean - population mean) / (sample standard deviation / sqrt(sample size)). Plugging in the given values, we get \[ t = \frac{51039 - 52100}{2300 / \sqrt{19}} = -2.01 \]
Step 3 ：The critical values are found using a t-distribution table and the given level of significance, which is 0.01. Since we are doing a two-tailed test, we need to divide the level of significance by 2 to get the critical value for each tail of the distribution. The critical values are \[ \pm 2.878 \]
Step 4 ：Final Answer: The null and alternative hypotheses are: \n \[ H_{0}: \mu=52,100 \] \n \[ H_{a}: \mu \neq 52,100 \] \n The value of the standardized test statistic is \[ \boxed{-2.01} \] \n The critical values are \[ \boxed{-2.878}, \boxed{2.878} \]
