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Missing Data Analysis. Complete Data: n=100 Sample means of X and Y 0.0479 10.1720 Sample variances and covariances of X Y 0.7543 3.1699 3.1699 17.2721.

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Presentation on theme: "Missing Data Analysis. Complete Data: n=100 Sample means of X and Y 0.0479 10.1720 Sample variances and covariances of X Y 0.7543 3.1699 3.1699 17.2721."— Presentation transcript:

1 Missing Data Analysis

2 Complete Data: n=100 Sample means of X and Y 0.0479 10.1720 Sample variances and covariances of X Y 0.7543 3.1699 3.1699 17.2721 Population mean of Y is 10 Y = a + b X + e with b  0 Grup1: only X (when X =< 0) n1=46 Means -0.7087 S: 0.3187 GRup 2: X and Y when X > 0 n2: 54 Means: 0.6924 12.8277 S: 0.2183 0.9851 0.9851 8.8516 X<0 X>0 X Y (taking an exam if passes X exam) MAR: We want to estimate the mean of Y for the complete sample !! Sample: MCAR, MAR NON-MAR (Rubin, 1984)

3 Ignoring Type of Missingness /TITLE analysis ignoring missing data problem (complete cases) /SPECIFICATIONS CASES=54; VARIABLES=2; ANALYSIS=MOMENT; MATRIX=COVARIANCE; METHOD=ML; GROUPS=1; /EQUATIONS V1 = 3.9*V999 + F1; V2 = *V999 + F2; /VARIANCES F1 = *; F2 = *; /COVARIANCES F1,F2=.3*; /PRINT EFFECT=YES; /MATRIX 0.2183 0.9851 8.8516 /MEANS 0.6924 12.8277 /END

4 ... ignoring type of missingness V1 =V1 =.692*V999 + 1.000 F1.064 10.789 V2 =V2 = 12.828*V999 + 1.000 F2.409 31.389 V F --- --- I F1 - F1.218*I I.042 I I 5.148 I I I I F2 - F2 8.852*I I 1.719 I I 5.148 I V F --- --- I F2 - F2.985*I I F1 - F1.234 I I 4.209 I I I V F --- --- I F2 - F2.709*I I F1 - F1 I I I

5 ML estimation (MAR) /TITLE Example of multiple group and missing data /SPECIFICATIONS CASES=54; VARIABLES=2; ANALYSIS=MOMENT; MATRIX=COVARIANCE; METHOD=ML; GROUPS=2; /EQUATIONS V1 = 3.9*V999 + F1; V2 = *V999 + F2; /VARIANCES F1 = *; F2 = *; /COVARIANCES F1,F2=.3*; /MATRIX 0.2183 0.9851 8.8516 /MEANS 0.6924 12.8277 /END /TITLE group 2 with missing y /SPECIFICATIONS CASES=46; VARIABLES=1; ANALYSIS=MOMENT; MATRIX=COVARIANCE; METHOD=ML; /EQUATIONS V1 = 3.9*V999 + F1; /VARIANCES F1 = *; /COVARIANCES /MATRIX 0.3187 /MEANS -0.7087 /CONSTRAINTS (1,V1,V999)=(2,V1,V999); (1,F1,F1)=(2,F1,F1); /LMTEST /END

6 ML estimation (MAR) V1 =V1 =.049*V999 + 1.000 F1.088.560 V2 =V2 = 9.941*V999 + 1.000 F2.488 20.378 V F --- --- I F1 - F1.752*I I.107 I I 7.000 I I I I F2 - F2 19.578*I I 3.237 I I 6.049 I I I V F --- --- I F2 - F2 3.378*I I F1 - F1.543 I I 6.215 I I I correlation.880*I

7 If we had the complete data /TITLE analysis of complete data /SPECIFICATIONS CASES=100; VARIABLES=2; ANALYSIS=MOMENT; MATRIX=COVARIANCE; METHOD=ML; GROUPS=1; /EQUATIONS V1 = 3.9*V999 + F1; V2 = *V999 + F2; /VARIANCES F1 = *; F2 = *; /COVARIANCES F1,F2=.3*; /PRINT EFFECT=YES; /MATRIX 0.7543 3.1699 17.2721 /MEANS 0.0479 10.1720 /END

8 ... if we had the complete data V1 =V1 =.048*V999 + 1.000 F1.087.549 V2 =V2 = 10.172*V999 + 1.000 F2.418 24.353 V F --- --- I F1 - F1.754*I I.107 I I 7.036 I I I I F2 - F2 17.272*I I 2.455 I I 7.036 I V F --- --- I F2 - F2 3.170*I I F1 - F1.483 I I 6.566 I correlation.878*I


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