Question

Use Table III, Appendix D to determine the${\mathbf{t}}_{\mathbf{0}}$ values foreach of the following probability statements and their respectivedegrees of freedom (df ).

a.$\mathbf{P}\left(\mathbf{t}⩽{\mathbf{t}}_{\mathbf{0}}\right)=.\mathbf{25}\mathbf{with}\mathbf{df}=\mathbf{15}$

b.$\mathbf{P}\left(\mathbf{t}⩾{\mathbf{t}}_{\mathbf{0}}\right)\mathbf{=}\mathbf{.}\mathbf{1}\mathbf{withdf}\mathbf{=}\mathbf{8}$

c.$\mathbf{P}\left(-{\mathbf{t}}_{\mathbf{0}}⩽\mathbf{t}⩽{\mathbf{t}}_{\mathbf{0}}\right)\mathbf{=}\mathbf{.}\mathbf{01}\mathbf{withdf}\mathbf{=}\mathbf{19}$

d.$\mathbf{P}\left(-{\mathbf{t}}_{\mathbf{0}}⩽\mathbf{t}⩽{\mathbf{t}}_{\mathbf{0}}\right)\mathbf{=}\mathbf{.}\mathbf{05}\mathbf{withdf}\mathbf{=}\mathbf{24}$

Short answer

a.-0.6912

b.1.39682

c.0.0127

d.0.06337

Step-by-step solution

Defining t-test assumptions

A t-test is a statistical test for verifying hypothesis testing. It comes from t-distribution which varies according to its degrees of freedom.It is also used to compare the means of two groups.

Solving subpart a.

By referring to the t-distribution table, we have to compute the probability,

$P\left(t⩽t_0\right)=.25withdf=15$

The value of that satisfies this equation is: -0.6912.

Solving subpart b.

By referring to the t-distribution table, we have to compute the probability,

$P\left(t⩾t_0\right)=.1withdf=8$

The value of that satisfies this equation is: 1.39682.

Solving subpart c.

By referring to the t-distribution table, we have to compute the probability,

$P\left(-t_0⩽t⩽t_0\right)=.01withdf=19$

The value of${\text{t}}_{\text{0}}$ that satisfies this equation is: 0.0127.

Solving subpart d.

By referring to the t-distribution table, we have to compute the probability,

$P\left(-t_0⩽t⩽t_0\right)=.05withdf=24$

The value of that satisfies this equation is: 0.06337.