Applet Exercise Suppose that Y has an F distribution with

Chapter 7, Problem 28E

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Applet Exercise Suppose that Y has an F distribution with \(v_{1}=4\)numerator degrees of freedom and \(\mathrm{v}_{2}=6\) denominator degrees of freedom.

Use Table 7, Appendix 3, to find \(\mathrm{F}_{.025}\) Also find \(\mathrm{F}_{.025}\) using the applet F-Ratio Probabilities and Quantiles.Refer to part (a). What quantile of Y does \(\mathrm{F}_{.025}\)  correspond to? What percentile?Refer to parts (a) and (b). Use the applet F-Ratio Probabilities and Quantiles to find \(\mathrm{F}_{.975}\), the .025 quantile (2.5th percentile) of the distribution of Y.If U has an F distribution with \(\mathrm{v}_{1}=6\) numerator and \(\mathrm{v}_{2}=4\) denominator degrees of freedom, use Table 7, Appendix 3, or the F-Ratio Probabilities and Quantiles applet to find \(\mathrm{F}_{.025}\).In Exercise 7.29, you will show that if Y is a random variable that has an F distribution

with ν1 numerator and \(\mathrm{V}_{2}\)  denominator degrees of freedom, then U = 1/Y has an F distribution with \(\mathrm{v}_{2}\) numerator and \(\mathrm{v}_{1}\) denominator degrees of freedom. Does this result explain the relationship between \(\mathrm{F}_{.975}\) from part (c) (4 numerator and 6 denominator degrees of freedom) and \(\mathrm{F}_{.025}\) from part (d) (6 numerator and 4 denominator degrees of freedom)? What is this relationship?

Equation Transcription:

 

 

 

Text Transcription:

v1=4

 v2=6

F.025

F.025

F.025

F.975

v1=6

v2=4

F.025

v2

U=1/Y

v2

v1

F.975

F.025