Pait 3 of 4 Points: 0 of 1 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 population is normally distributed. im: $\mu=51,400, \alpha=0.05$ Sample statistics: $\bar{x}=50,353, s=1600, n=20$ Click the icon to view the t-distribution table. hat are the null and alternative hypotheses? Choose the correct answer below. A. \[ \begin{array}{l} H_{0}-\mu=51,400 \\ H_{\text {a }} \mu \neq 51,400 \end{array} \] c. \[ \begin{array}{l} H_{0}-\mu \leq 51,400 \\ H_{a}-\mu>51,400 \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu \geq 51,400 \\ H_{a}: \mu<51,400 \end{array} \] \[ \text { D. } \begin{array}{l} H_{0}: \mu \neq 51,400 \\ H_{a}: \mu=51,400 \end{array} \] What is the value of the standardized test statistic? The standardized test statistic is -293 . (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) Clear all Check answer
Solution
Step 1 ：Given the population mean \(\mu = 51400\), the level of significance \(\alpha = 0.05\), the sample mean \(\bar{x} = 50353\), the sample standard deviation \(s = 1600\), and the sample size \(n = 20\).
Step 2 ：We first set up the null and alternative hypotheses. The null hypothesis \(H_0\) is that the population mean \(\mu\) is equal to 51400, and the alternative hypothesis \(H_a\) is that the population mean \(\mu\) is not equal to 51400. So, we have \(H_0: \mu = 51400\) and \(H_a: \mu \neq 51400\).
Step 3 ：We then calculate the standardized test statistic, which is given by the formula \((\bar{x} - \mu) / (s / \sqrt{n})\). Substituting the given values, we get a test statistic of approximately -2.93.
Step 4 ：We also need to find the critical value(s) for the t-distribution with \(n - 1 = 19\) degrees of freedom. For a two-tailed test with \(\alpha = 0.05\), the critical value(s) is(are) approximately 2.093.
Step 5 ：\(\boxed{\text{Final Answer:}}\) The null and alternative hypotheses are \(H_0: \mu = 51400\) and \(H_a: \mu \neq 51400\). The value of the standardized test statistic is -2.93. The critical value(s) is(are) 2.093.
