Answer:
[tex] P(t_{16}>2.583)[/tex]
And for this case we can use the complement rule and we got:
[tex]P(t_{16}>2.583)= 1- P(t_{16} <2.583)[/tex]
And if we use the t- table with df =16, we got:
[tex]P(t_{16}>2.583)= 1-0.990= 0.01[/tex]
Step-by-step explanation:
For this case we know that the degrees of freedom are given:
[tex] df = n-1= 16[/tex]
And we want to find the following probability:
[tex] P(t_{16}>2.583)[/tex]
And for this case we can use the complement rule and we got:
[tex]P(t_{16}>2.583)= 1-P(t_{16} <2.583)[/tex]
And if we use the t-table with df =16, we got:
[tex]P(t_{16}>2.583)= 1-0.990= 0.01[/tex]
