Answer:
-2.508
Explanation:
From the given information:
The null hypothesis and the alternative hypothesis can be computed as:
![H_o : \mu = 5](https://img.qammunity.org/2021/formulas/mathematics/college/wlyexi2wlv6xuk78hgyz9dtyn9lopf0dmb.png)
![H_1: \mu < 5](https://img.qammunity.org/2021/formulas/mathematics/college/785uilzti53avnnu5f9v71z4bqwzm18mxg.png)
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;
= -2.508 (since it is left tailed)