Question

Let${\mathit{t}}_{\mathbf{0}}$ be a specific value of t. Use Table III in Appendix D to find${\mathit{t}}_{\mathbf{0}}$ values such that the following statements are true.

a. $\mathbf{{\rm P}}\left(t\ge t_0\right)\mathbf{=}\mathbf{0}\mathbf{.025}$where $\mathbf{\text{df}}\mathbf{=}\mathbf{11}$

b.$\mathbf{{\rm P}}\left(t\ge t_0\right)\mathbf{=}\mathbf{0}\mathbf{.01}$ where$\mathbf{\text{df}}\mathbf{=}\mathbf{9}$

c.$\mathit{{\rm P}}\left(t\ge t_0\right)\mathbf{=}\mathbf{0}\mathbf{.005}$ where$\mathbf{\text{df}}\mathbf{=}\mathbf{6}$

d.$\mathbf{{\rm P}}\left(t\ge t_0\right)\mathbf{=}\mathbf{0}\mathbf{.05}$ where$\mathbf{\text{df}}\mathbf{=}\mathbf{18}$

Short answer

1. .The value of$t_0$ for${\rm P}\left(t\ge t_0\right)=0.025$where$\text{df}=11$ is 2.201
2. The value of$t_0$ for${\rm P}\left(t\ge t_0\right)=0.01$ where$\text{df}=9$is 2.821
3. The value of$t_0$ for${\rm P}\left(t\ge t_0\right)=0.005$where$\text{df}=6$ is 3.707

4. The value of$t_0$ for${\rm P}\left(t\ge t_0\right)=0.05$ where$\text{df}=18$ is 1.734

Step-by-step solution

Calculating the value of t0 Such that Ρ(t≥t0)=0.025

Given$\mathrm{{\rm P}}\left(\mathrm{t}\ge {\mathrm{t}}_0\right)=0.025$where$\text{df}=11$

By using t- distribution table, we will see the row with 11 degrees of freedom and column corresponds to$\alpha =0.025$ in the one-tail.

Hence from the critical values we get $t_{0.025,11}=2.201$

Calculating the value t0 of Such that Ρ(t≥t0)=0.01

Given${\rm P}(\mathrm{t}\ge {\mathrm{t}}_0)=0.01$where$\text{df}=9$

By using t- distribution table, we will see the row with 9 degrees of freedom and column corresponds to$\alpha =0.01$ in the one-tail.

Hence from the critical values we get$t_{0.01,9}=2.821$

Calculating the value of t0 Such that Ρ(t≥t0)=0.005

Given$\mathrm{{\rm P}}\left(\mathrm{t}\ge {\mathrm{t}}_0\right)=0.005$where$\text{df}=6$

By using t- distribution table, we will see the row with 6 degrees of freedom and column corresponds to$\alpha =0.005$ in the one-tail.

Hence from the critical values we get$t_{0.005,6}=3.707$

Calculating the value of t0 Such that Ρ(t≥t0)=0.05

Given$\mathrm{{\rm P}}\left(\mathrm{t}\ge {\mathrm{t}}_0\right)=0.05$ where$\text{df}=18$

By using t- distribution table, we will see the row with 18 degrees of freedom and column corresponds to$\alpha =0.05$ in the one-tail.

Hence from the critical values we get $t_{0.05,18}=1.734$.