Respuesta :

Solution

Given that

sample size n = 11

The degree of freedom df = n - 1 = 11 - 1 = 10

At 99% confidence level,

[tex]\begin{gathered} \alpha=1-99\%=1-0.99=0.01 \\ \\ \frac{\alpha}{2}=0.005 \end{gathered}[/tex]

[tex]t_{\frac{\alpha}{2},df}=t_{0.005,10}=3.1693[/tex]

Checking the table,

Therefore, the Critical value t value = 3.169

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