Refer to Exercises 6.17 and 8.14. The distribution

Chapter 8, Problem 132SE

(choose chapter or problem)

Refer to Exercises 6.17 and 8.14. The distribution function for a power family distribution is given by

                               \(F(y)=\left\{\begin{array}{ll}0, & y<0 \\\left(\frac{y}{\theta}\right)^{\alpha}, & 0 \leq y \leq \theta \\1, & y>\theta\end{array}\right.\)

        

where \(\alpha, \theta>0\) Assume that a sample of size \(n\) is taken from a population with a power family distribution and that \(\alpha=c\) where \(c>0\) is known, a Show that the distribution function of \(Y_{(n)}=\max \left\{Y_{1}, Y_{2, \ldots}, Y_{n}\right\}\) is given by

                              \(F_{Y_{(n)}}(y)=\left\{\begin{array}{ll}0, & y<0 \\\left(\frac{y}{\theta}\right)^{n c}, & 0 \leq y \leq \theta \\1, & y>\theta\end{array}\right.\)

where \(\theta>0\)

b Show that \(Y_{(n)} / \theta\) is a pivotal quantity and that for \(0<k<1\)

                                 

                                     \(P\left(k<\frac{Y_{(n)}}{\theta} \leq 1\right)=1-k^{c n}\)

c Suppose that \(n=5\) and \(\alpha=c=2.4\)

i Use the result from part (b) to find \(k\) so that

                                      \(P\left(k<\frac{Y_{(5)}}{\theta} \leq 1\right)=0.95\)

ii Give a 95% confidence interval for \(\theta\)

Equation Transcription:

{

 

{       

Text Transcription:

F(y)={ ^0, y<0 (y /theta) ^alpha,  0 leq y leq theta 1,y>0

alpha,theta>0

n

alpha=c

c>0

 Y_(n)=max{Y_1,Y_2,...,Y_n}

F_Y_(y)={ ^0, y<0 (y /theta) ^(nc),  0 leq y leq theta 1,y>0

theta>0

Y_(n)/theta

0<k<1

P(k<Y_(n)1=1)-k^cn

n=5

alpha=c=2.4

k

P(k<Y_(5)/1)=0.95

theta