solutions_9.pdf

Probability and StatisticsSolutions 9Guanghua School of ManagementPeking UniversityDecember 7,20121/9Question 1The data is from a normal distribution and the variance isunknown,let =1 ,where is the confidencecoefficient,from the fact thatU=n(Xn)S tn1,whereXn=1nnXi=1Xi,S2=1n 1nXi=1(XiXn)2,n=8.Then the confidence interval will beXn tn1,/2S/n,Xn+tn1,/2S/n,2/9Question 1where tn1,/2is the upper/2 quantile of the t distribution withdegree of freedom of(n 1).When =0.9,=0.1,t7,0.05=1.8946.So the confidenceinterval with confidence coefficient 0.9 will be 2.7191,3.4058.Similarly,we can calculate the confidence intervalfor confidence coefficients 0.95 and 0.99 will be 2.634,3.491 and 2.4284,3.6966.3/9Question 2The data is from the normal distribution with knownvariance 2,then fromZ=n(Xn)N(0,1),So the confidence interval with confidence coefficient 0.95isXn z/2/n,Xn+z/2/n,(1)where =0.05 and z/2is the upper/2 quantile ofstandard normal distribution.4/9Question 2The length of confidence interval is2z/2n=2 1.96n 0.1.Then n 1537.5/9Question 3Let C denote the critical region of the test,for ,thenthe power function is(|)=Pr(X C|)=Pr(X 1|)=Z11exdx=e.The size of the test will be()=sup|(|)=sup1e=e1.6/9Question 4We firstly calculate the distribution of Yn.If y 0,thenFYn(y)=0.If y 0,FYn(y)=P(Yn y)=nYi=1Pr(Xi y)=(y)n.If y ,then FYn(y)=1.Then the power function of the test is(|)=Pr(Yn 1.5|)=FYn,(1.5)=1 1.5(1.5)n 1.57/9Question 4The size of the test is()=sup0(|)=sup2(|)=(34)n.8/9The power of the test is(|)=Pr(Y 1 or Y 7|p)=1 Pr(2 Y 6|p)=1 6Xk=2Ck20pk(1 p)(20k).So when p equals 0,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1,the power is 1,0.3941,0.1558,0.3996,0.7505,0.9423,0.9935,0.9997,0.9999,0.9999,1.The size of the test is=Pr(Y 1 or Y 7|p=0.2)=0.1558.9/9