Answer:
[tex]t_{critical} \text{ at 0.01 level of significance, 100 degree of freedom} = -2.364[/tex]
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 20 ounces
Sample size, n = 101
Alpha, α = 0.01
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu = 20\text{ ounces}\\H_A: \mu < 20\text{ ounces}[/tex]
We use one-tailed t test to perform this hypothesis.
[tex]t_{stat} = -2.89[/tex]
Now,
[tex]t_{critical} \text{ at 0.01 level of significance, 100 degree of freedom} = -2.364[/tex]
Conclusion:
Since the calculated test statistic is less than the critical value we fail to accept the null hypothesis and reject it.
We accept the alternate hypothesis.
Thus, there is enough evidence to support the claim that the mean amount of beverage in a medium drink sold at this college's cafeteria is less than 20 ounces