Solution Found!
Applet Exercise ReferenceUse the structures of T and F
Chapter 7, Problem 32E(choose chapter or problem)
Applet Exercise
Find \(\mathrm{t}_{.05}\) for a t-distributed random variable with 5 df.Refer to part (a). What is \(\mathrm{P}\left(\mathrm{T}^{2}>t_{.05}^{2}\right)\)?Find \(\mathrm{F}_{.10}\) for an F-distributed random variable with 1 numerator degree of freedom and 5 denominator degrees of freedom.Compare the value of \(\mathrm{F}_{.10}\) found in part (c) with the value of \(t_{.05}^{2}\) from parts (a) and (b).In Exercise 7.33, you will show that if T has a t distribution with ν df, then \(\mathrm{U}=\mathrm{T}^{2}\) has an
F distribution with 1 numerator degree of freedom and ν denominator degrees of freedom. How does this explain the relationship between the values of \(\mathrm{F}_{.10}\) (1 num. df, 5 denom df) and \(t_{.05}^{2}\) (5 df) that you observed in part (d)?
Equation Transcription:
Text Transcription:
t.05
P(T2 >t.052)
F.10
F.10
t.052
U=T2
F.10
t.052
Questions & Answers
QUESTION:
Applet Exercise
Find \(\mathrm{t}_{.05}\) for a t-distributed random variable with 5 df.Refer to part (a). What is \(\mathrm{P}\left(\mathrm{T}^{2}>t_{.05}^{2}\right)\)?Find \(\mathrm{F}_{.10}\) for an F-distributed random variable with 1 numerator degree of freedom and 5 denominator degrees of freedom.Compare the value of \(\mathrm{F}_{.10}\) found in part (c) with the value of \(t_{.05}^{2}\) from parts (a) and (b).In Exercise 7.33, you will show that if T has a t distribution with ν df, then \(\mathrm{U}=\mathrm{T}^{2}\) has an
F distribution with 1 numerator degree of freedom and ν denominator degrees of freedom. How does this explain the relationship between the values of \(\mathrm{F}_{.10}\) (1 num. df, 5 denom df) and \(t_{.05}^{2}\) (5 df) that you observed in part (d)?
Equation Transcription:
Text Transcription:
t.05
P(T2 >t.052)
F.10
F.10
t.052
U=T2
F.10
t.052
ANSWER:
Step 1 of 6
a)
For the given student’s t-distribution, it is given that there are 5 degrees of freedom. The t-statistic at 0.05 level of significance can be obtained using the applet and the steps are as follows;