有这样一个案例：调查一个公司雇佣员工是否存在性别歧视，收集了28个求职者的数据，其中包含雇佣（y=1)或未雇佣(y=0)；受高等教育的时间，工作经验时间，和性别(male=1,female=0）。数据在附件。我用SAS9.0中的分析家(analyst)，逻辑回归，得出的结果如下： 其中，性别的系数为-2.8526，意味着当性别为1，受雇佣机会更小，明显与数据中反映出的情况不符（男性明显受雇的几率大），我反复运算都是如此，请教各位高手这是为什么！！感激不尽！！
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The LOGISTIC Procedure

Model Information

Data Set _PROJ_.DISCRIM
Response Variable hired hired
Number of Response Levels 2
Number of Observations 28
Model binary logit
Optimization Technique Fisher's scoring


Response Profile

Ordered Total
Value hired Frequency

1 1 9
2 0 19

Probability modeled is hired=1.


Class Level Information

Design
Class Value Variables

sex 0 1
1 -1


Model Convergence Status

Convergence criterion (GCONV=1E-8) satisfied.


Model Fit Statistics

Intercept
Intercept and
Criterion Only Covariates

AIC 37.165 21.915
SC 38.497 27.244
-2 Log L 35.165 13.915


Testing Global Null Hypothesis: BETA=0

Test Chi-Square DF Pr > ChiSq

Likelihood Ratio 21.2493 3 <.0001
Score 15.6824 3 0.0013
Wald 5.5996 3 0.1328
1 17:47 Saturday, February 15, 2003 2

The LOGISTIC Procedure

Type 3 Analysis of Effects

Wald
Effect DF Chi-Square Pr > ChiSq

education 1 3.8720 0.0491
experience 1 4.5207 0.0335
sex 1 4.9405 0.0262


Analysis of Maximum Likelihood Estimates

Standard Wald
Parameter DF Estimate Error Chi-Square Pr > ChiSq

Intercept 1 -11.1897 4.8002 5.4339 0.0197
education 1 1.1540 0.5865 3.8720 0.0491
experience 1 0.8777 0.4128 4.5207 0.0335
sex 0 1 -2.8526 1.2834 4.9405 0.0262


Odds Ratio Estimates

Point 95% Wald
Effect Estimate Confidence Limits

education 3.171 1.005 10.008
experience 2.405 1.071 5.402
sex 0 vs 1 0.003 <0.001 0.509


Association of Predicted Probabilities and Observed Responses

Percent Concordant 94.2 Somers' D 0.883
Percent Discordant 5.8 Gamma 0.883
Percent Tied 0.0 Tau-a 0.399
Pairs 171 c 0.942