Answer:
-2.508
Step-by-step explanation:
From the given information:
The null hypothesis and the alternative hypothesis can be computed as:
[tex]H_o : \mu = 5[/tex]
[tex]H_1: \mu < 5[/tex]
This is a left-tailed test.
The sample size n = 23
Then, the degree of freedom df = n - 1
df = 23 - 1
df = 22
The level of significance ∝ = 0.01
Using the student t-table to determine the critical value;
[tex]t_{\alpha, df} = t_{0.01, 22}[/tex] = -2.508 (since it is left tailed)
