Answer:
a. 0.025
b. 0.90
c. 0.05
d. 0.01
e. 0.95
f. 0.90
Step-by-step explanation:
From the T-distribution table, we have;
[tex]Pr(T_v > t_{\alpha;v }) = \alpha[/tex]
For a t distribution with 16 degrees of freedom, we have;
a. [tex]Pr(T_{16} > 2.120 }) = 0.025[/tex]
b. [tex]Pr(T_{16} < 1.337}) =1 - Pr(T_{16} > 1.337}) = 1 - 0.1 = 0.90[/tex]
c. [tex]Pr(T_{16} < -1.746}) =Pr(T_{16} > 1.746}) = 0.05[/tex]
d. [tex]Pr(T_{16} > 2.583}) = 0.01[/tex]
e. [tex]Pr(-2.120 < T_{16} < 2.120 }) = 1 - 2 \times Pr(T_{16} > 2.120 }) = 1 - 2 \times 0.025 = 0.95[/tex]
f. [tex]Pr(-1.746< T_{16} < 1.746}) = 1 - 2 \times Pr(T_{16} > 1.746}) = 1 - 2 \times 0.05 = 0.90[/tex]
The required probabilities using the t-distribution table will be a. 0.025, b. 0.90, c. 0.05, d, 0.01, e. 0.95, and f. 0.90.
Given:
A t-distribution with 16 degrees of freedom.
See the graph for the given distribution.
It is required to calculate the probability of:
Use the table of t-distribution to calculate the required probabilities.
[tex]P(T_\nu>t_{\alpha;\nu})=\alpha[/tex] where [tex]\nu[/tex] is the degree of freedom.
The required probabilities will be,
a) To the right of 2.120.
[tex]P(T_{16}>2.120)=0.025[/tex]
b) To the left of 1.337
[tex]P(T_{16}<1.337)=1-P(T_{16}>1.337)\\=1-0.1\\=0.9[/tex]
c) To the left of -1.746
[tex]P(T_{16}<-1.746)=P(T_{16}>1.746)\\=0.05[/tex]
d) To the right of 2.583.
[tex]P(T_{16}>2.583)=0.01[/tex]
e) Between -2.120 and 2.120.
[tex]P(-2.120<T_{16}<2.120)=1-2P(T_{16}>2.120)\\=1-2(0.025)=1-0.05\\=0.95[/tex]
f) Between -1.746 and 1.746.
[tex]P(-1.746<T_{16}<1.746)=1-2P(T_{16}>1.746)\\=1-2(0.05)=1-0.1\\=0.90[/tex]
Therefore, the required probabilities using the t-distribution table will be a. 0.025, b. 0.90, c. 0.05, d, 0.01, e. 0.95, and f. 0.90.
For more details, refer to the link:
https://brainly.com/question/7294680
