Use technology and 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
Claim: $\mu> 76 ; \alpha=0.05$ Sample statistics: $\bar{x}=76.4, s=3.2, n=26$
What are the null and alternative hypotheses? Choose the correct answer below.
A.
\[
\begin{array}{l}
H_{0} \mu=76 \\
H_{A} \mu \neq 76
\end{array}
\]
c.
\[
\begin{array}{l}
H_{0}: \mu \geq 76 \\
H_{A}: \mu< 76
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu \leq 76 \\
H_{A}: \mu> 76
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu \neq 76 \\
H_{A}: \mu=76
\end{array}
\]
What is the value of the standardized test statistic?
The standardized test statistic is $\square$. (Round to two decimal places as needed.)
Answer
Rounding to two decimal places, the value of the standardized test statistic is \(\boxed{0.64}\)
Step 1 ：The null and alternative hypotheses are given by: \[H_{0}: \mu \leq 76\] \[H_{A}: \mu>76\]
Step 2 ：The standardized test statistic is calculated as follows: \[(\text{sample mean} - \text{population mean}) / (\text{sample standard deviation} / \sqrt{\text{sample size}})\]
Step 3 ：Substituting the given values, we get: \[(76.4 - 76) / (3.2 / \sqrt{26}) = 0.6373774391991072\]
Step 4 ：Rounding to two decimal places, the value of the standardized test statistic is \(\boxed{0.64}\)
