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: $\bar{x}=50,353, s=1600, n=20$
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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)
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Check answer
Answer
\(\boxed{\text{Final Answer:}}\) 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.
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.
