Question

Suppose x_(1),x_(2),dots are i.i.d. from a distribution with pdf f_(x)(x;p)=(1-p)+2px,0<=x<=1; where 0<=p<=1. Furthermore, E[x]=0.5+(p)/(6) and Var[x]=(1)/(36)(3-p^(2)). a) Does the factorization theorem suggest that the sample mean, ar{x} , is sufficient for p ? b) By the central limit theorem, the following probability statement follows: P(-z_((1+gamma )/(2))<=sqrt(n)((ar{x} )-(0.5+(p)/(6)))/(sqrt((1)/(36)(3-p^(2))))<=z_((1+gamma )/(2)))gamma or equivalently P(-z_((1+gamma )/(2))<=sqrt(n)(6(ar{x} )-3-p)/(sqrt((3-p^(2))))<=z_((1+gamma )/(2)))gamma Generated a formula for an approximate gamma confidence interval for p from the above probability statement. Hint: provided a>0, the inequality at^(2)+bt+c<=0 has solution tin[(-b-sqrt(b^(2)-4ac))/(2a),(-b+sqrt(b^(2)-4ac))/(2a)] to c) Suppose a sample of size n=400 yields a sample mean of ar{x} =0.6. Calculate an approximate 95% confidence interval for p. Suppose X1, X2, ... are i.i.d. from a distribution with pdf fx(x;p)=(1-p)+2px 0x1; where 0p1 6 3pts a) Does the factorization theorem suggest that the sample mean, , is sufficient for p? 6pts b) By the central limit theorem, the following probability statement follows: 0.5+p 6 Z(1+y/2 ~ y or equivalently Generated a formula for an approximate y confidence interval for p from the above probability statement. Hint: provided a> 0, the inequality at+bt+co has solution -b-b2-4ac-b+b2-4ac 2a 2a to 3pts c) Suppose a sample of size n = 400 yields a sample mean of x = 0.6. Calculate an approximate 95% confidence interval for p. to

          Suppose x_(1),x_(2),dots are i.i.d. from a distribution with pdf
f_(x)(x;p)=(1-p)+2px,0<=x<=1; where 0<=p<=1.
Furthermore, E[x]=0.5+(p)/(6) and Var[x]=(1)/(36)(3-p^(2)).
a) Does the factorization theorem suggest that the sample mean, ar{x} , is
sufficient for p ?
b) By the central limit theorem, the following probability statement
follows:
P(-z_((1+gamma )/(2))<=sqrt(n)((ar{x} )-(0.5+(p)/(6)))/(sqrt((1)/(36)(3-p^(2))))<=z_((1+gamma )/(2)))~~gamma  or equivalently 
P(-z_((1+gamma )/(2))<=sqrt(n)(6(ar{x} )-3-p)/(sqrt((3-p^(2))))<=z_((1+gamma )/(2)))~~gamma 
Generated a formula for an approximate gamma  confidence interval for p
from the above probability statement.
Hint: provided a>0, the inequality at^(2)+bt+c<=0 has solution
tin[(-b-sqrt(b^(2)-4ac))/(2a),(-b+sqrt(b^(2)-4ac))/(2a)]
to
c) Suppose a sample of size n=400 yields a sample mean of ar{x} =0.6.
Calculate an approximate 95% confidence interval for p.
Suppose X1, X2, ... are i.i.d. from a distribution with pdf fx(x;p)=(1-p)+2px 0x1; where 0p1
6
3pts
a) Does the factorization theorem suggest that the sample mean,  , is sufficient for p?
6pts
b) By the central limit theorem, the following probability statement follows: 0.5+p 6 Z(1+y/2 ~ y or equivalently
Generated a formula for an approximate y confidence interval for p from the above probability statement. Hint: provided a> 0, the inequality at+bt+co has solution -b-b2-4ac-b+b2-4ac 2a 2a
to
3pts
c)
Suppose a sample of size n = 400 yields a sample mean of x = 0.6. Calculate an approximate 95% confidence interval for p.
to

suppose x1x2dots are iid from a distribution with pdf fxxp1 p2px0x1 where 0p1 furthermore ex05p6 and varx1363 p2 a does the factorization theorem suggest that the sample mean barx is suff 98554

Added by Alfredo G.

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