STA437 Winter Assignment 3 - Solution

Transcription

1 STA437 Winter 7 - Assignment 3 - Solution Question 1 The MLE for μ is x = ( 4,6)`. By the invariance rincile of MLE, we know that the MLE for ρ can be calculated by 1 ˆ n ˆ ρ = V ΣV = V S nv where V = diag( s11,..., s 44 ) and S = S = n / The calculation yields ˆρ = =. 3 / n. Question 1 (a) Use SAS to calculate the test statistic = n( X μ )' S ( X μ ) T we get T = n 1 57, n α 3,17 =. n 17 = 9.74 < We cannot reject H μ '= 4,5,1. The critical value is given by ( ) F ( ) = F (.5) Since T : ( ) The P-value = and since it is greater then α =. 5 we can not reject H. (b) The simultaneous confidence interval for μi is given by X i ( n 1) ( n ) ± F, n s ( α ) n Here, X = ( 4.64, 45.4, 9.965)' and s = , s = and s = Also, ( n 1) 57 F, n ( α ) = F3,17 (.5) = Substituting these values in the above n 17 formula, we get the following 95% simultaneous CI s: ( 3.398, 5.88) μ ( 35.5, ) and ( 8.571, ) μ. 3 μ, 1 ii. 1

2 α sii (c) The Bonferroni confidence interval for μ i is given by X i ± tn 1. m n α.5 In our case m = 3 and t n 1 = t19 =. 65. Substituting these values in the above m 6 formula, we get the following 95% simultaneous CI s: ( 3.644, 5.636) μ ( 37.13, ) and ( 8.847, 11.83) μ. 3 μ, The Bonferroni method yiels narrower CI s with the same overall error rate of.5 and therefore, Bonferroni s method is better for this roblem. 1 Question (a) We have that n = 4, =, X = ( 1, )' and the samle covariance S = The Hotelling T for testing H : μ = ( 14, )' is T = n( X μ ) S ( X μ ) = ' (b) The distribution of T is ( n 1) n F, n = 3F,. (c) The critical value for this test is (.5) 57 3, = F. Since T = < 57. We cannot reject H : μ '= ( 14, ). Also, the P-value = > α =. 5 so we can not reject H.

3 Question 4 (a) The Q-Q lots for each variable are given below x Nor mal Quant i l es As we can see, the distribution of and uer values of this variable. X 1 aears to be normal with a few outlier in the lower x Nor mal Quant i l es X Again, the distribution of aears to be normal with a few outlier in the lower and uer values of this variable. 3

4 1 1 8 x Nor mal Quant i l es Once again, the distribution of and uer values of this variable. X 3 aears to be normal with a few outlier in the lower (b) The matrix of scatterlots is given below 9.91 x x x As we see the Scatter lot matrix shows ellitical shaes suggesting bivariate normality of comonents. 4

5 Aendix SAS code for Question 1 Proc iml; title 'Assignment 3 - Question 1'; x = {3 6, 4 4, 5 7, 4 7}; n=4; x_bar =(1/n)* X`*J(n,1,1); rint x_bar; S = (1/(n-1))*x`*(I(n)-(1/n)*J(n,1,1)*(J(n,1,1))`)*x; Sn = S*(n-1)/n; V = root(diag(sn)); rho_hat = inv(v)*sn*inv(v); rint rho_hat; SAS code for Question 3 Proc iml; title 'Assignment 3 - Question 3'; x = {4 4, 16 18, 1 18, 16 }; n=4; =; x_bar =(1/n)* X`*J(n,1,1); S = (1/(n-1))*x`*(I(n)-(1/n)*J(n,1,1)*(J(n,1,1))`)*x; rint x_bar s; /* calculating the Hotelling T^ statistic */ mu = {14,}; diff = x_bar-mu; S_inv = inv(s); T = n*(diff)`*(s_inv)*(diff); rint T; /* Finding the critical value from the F distribution */ F = finv(.95,,); /* Find the 95th ercintile of the F distribution */ Critical_value = (*(n-1)/(n-))*f; rint Critical_value; /* Finding the P-value */ F_statistics = ((n-)/(*(n-1)))*t; P_value = 1- robf(f_statistics,,); rint P_value; 5

6 SAS code for Question title 'Assignment 3 - question '; /* Reading the data into SAS */ data sweat; infile 'C:\Documents and Settings\Hadas Moshonov\Deskto\STA437\Assignments\sweat.txt'; inut sweat_rate sodium otassium; /* Checking the content of the data file and rinting it*/ Proc contents data = sweat; roc rint data=sweat; roc iml; /* reading the data into a matrix X*/ use sweat; read all var {sweat_rate sodium otassium} into x; n = ; = 3; /* comuting x_bar and S */ x_bar =(1/n)* X`*J(n,1,1); S = (1/(n-1))*x`*(I(n)-(1/n)*J(n,1,1)*(J(n,1,1))`)*x; rint x_bar s; /* calculating the Hotelling T^ statistic */ mu = {4,5,1}; diff = x_bar-mu; S_inv = inv(s); T = n*(diff)`*(s_inv)*(diff); rint T; /* Finding the critical value from the F distribution */ F = finv(.95, 3,17); /* Find the 95th ercentile of the F distribution */ Critical_value = (*(n-1)/(n-))*f; rint Critical_value; /* Finding the P-value */ F_statistics = ((n-)/(*(n-1)))*t; P_value = 1- robf(f_statistics,3,17); rint P_value; 6

7 /* Finding simultaneous CI for mu_i using method 1 */ /* CI for mu_1 */ LCL = x_bar[1,]- root(critical_value*s[1,1]/n); UCL = x_bar[1,]+ root(critical_value*s[1,1]/n); rint 'CI for mu_1'; /* CI for mu_ */ LCL = x_bar[,]- root(critical_value*s[,]/n); UCL = x_bar[,]+ root(critical_value*s[,]/n); rint 'CI for mu_'; /* CI for mu_3 */ rint 'CI for mu_3'; LCL = x_bar[3,]- root(critical_value*s[3,3]/n); UCL = x_bar[3,]+ root(critical_value*s[3,3]/n); /* Finding Bonferroni's simultaneous CI for mu_i */ /* the Critical value t */ T_critic = tinv(1-.5/6,19); rint T_critic; /* CI for mu_1 */ rint ' Bonfferoni CI for mu_1'; LCL = x_bar[1,]- T_critic*root(s[1,1]/n); UCL = x_bar[1,]+ T_critic*root(s[1,1]/n); /* CI for mu_ */ rint ' Bonfferoni CI for mu_'; LCL = x_bar[,]- T_critic*root(s[,]/n); UCL = x_bar[,]+ T_critic*root(s[,]/n); /* CI for mu_3 */ rint ' Bonfferoni CI for mu_3'; LCL = x_bar[3,]- T_critic*root(s[3,3]/n); UCL = x_bar[3,]+ T_critic*root(s[3,3]/n); 7

8 SAS code for Question 4 title 'Assignment 3 - question 4'; /* Reading the data into SAS */ data data; infile 'C:\Documents and Settings\Hadas Moshonov\Deskto\STA437\Assignments\assign3data.txt' firstobs = ; inut obs x1 x x3; /* Checking the content of the data file and rinting it*/ Proc contents data = data; roc rint data=data; /* Part (a) - constraucting normal Q-Q lots */ roc univariate data = data; var x1 x x3; roc caability norint data = data; title 'Normal Probability Plots'; symbol1 v ='dot'; var x1 x x3; qqlot x1 x x3; /* Part (b) Producing matrix of air-wise scatterlots */ Proc insight data = data; title 'Scatterlot Matrix'; scatter x1 x x3 * x1 x x3; 8