Ejemplo de análisis factorial

Análisis Factorial
Objetivo: Identificar las características laborales de un conjunto de empleados para establecer aquellas que son más compatibles.
Pregunta de investigación: ¿Cuáles son las características laborales, más compatibles, para un conjunto de empleados?
Hipótesis nula: No existen un conjunto de variables más compatibles que compongan las características laborales de un conjunto deempleados.
Hipótesis alternativa: Existen un conjunto de variables más compatibles que compongan las características laborales de un conjunto de empleados.

Matriz de correlacionesa |
| | Educational Level (years) | Employment Category | Current Salary | Beginning Salary | Months since Hire | Previous Experience (months) | Date of Birth |
Correlación | Educational Level (years) | 1.000| .515 | .661 | .633 | .050 | -.252 | .281 |
| Employment Category | .515 | 1.000 | .780 | .755 | .004 | .062 | -.010 |
| Current Salary | .661 | .780 | 1.000 | .880 | .084 | -.097 | .144 |
| Beginning Salary | .633 | .755 | .880 | 1.000 | -.018 | .045 | .009 |
| Months since Hire | .050 | .004 | .084 | -.018 | 1.000 | .002 | -.053 |
| Previous Experience (months) | -.252 | .062 |-.097 | .045 | .002 | 1.000 | -.802 |
| Date of Birth | .281 | -.010 | .144 | .009 | -.053 | -.802 | 1.000 |
Sig. (Unilateral) | Educational Level (years) | | .000 | .000 | .000 | .137 | .000 | .000 |
| Employment Category | .000 | | .000 | .000 | .468 | .088 | .414 |
| Current Salary | .000 | .000 | | .000 | .033 | .017 | .001 |
| Beginning Salary | .000 | .000 | .000 | | .344 |.162 | .423 |
| Months since Hire | .137 | .468 | .033 | .344 | | .485 | .127 |
| Previous Experience (months) | .000 | .088 | .017 | .162 | .485 | | .000 |
| Date of Birth | .000 | .414 | .001 | .423 | .127 | .000 | |
a. Determinante = .012 |

Observamos las variables que no sean significativas.

KMO y prueba de Bartlett |
Medida de adecuación muestral de Kaiser-Meyer-Olkin.| .724 |
Prueba de esfericidad de Bartlett | Chi-cuadrado aproximado | 2075.310 |
| gl | 21 |
| Sig. | .000 |

EL KMO DICE QUE TAN FUERTE ES LA VARIABLE DEBE SER MAYOR A 0.5
EN LA PRUEBA DE ESFERICIDAD DE BARTLET, LA SIGNIFICANCIA MENOR A .05 PARA QUE SEA ESTADISTICAMENTE SIGNIFICATIVO.

Matrices anti-imagen |
| | Educational Level (years) | Employment Category | Current Salary |Beginning Salary | Months since Hire | Previous Experience (months) | Date of Birth |
Covarianza anti-imagen | Educational Level (years) | .493 | -.010 | -.042 | -.070 | -.039 | .049 | -.044 |
| Employment Category | -.010 | .357 | -.095 | -.038 | .043 | -.030 | .012 |
| Current Salary | -.042 | -.095 | .158 | -.115 | -.088 | .023 | -.023 |
| Beginning Salary | -.070 | -.038 | -.115 |.182 | .085 | -.033 | .022 |
| Months since Hire | -.039 | .043 | -.088 | .085 | .931 | .017 | .065 |
| Previous Experience (months) | .049 | -.030 | .023 | -.033 | .017 | .342 | .259 |
| Date of Birth | -.044 | .012 | -.023 | .022 | .065 | .259 | .343 |
Correlación anti-imagen | Educational Level (years) | .921a | -.024 | -.149 | -.235 | -.058 | .119 | -.106 |
| Employment Category| -.024 | .881a | -.398 | -.151 | .074 | -.085 | .035 |
| Current Salary | -.149 | -.398 | .723a | -.678 | -.230 | .098 | -.101 |
| Beginning Salary | -.235 | -.151 | -.678 | .743a | .206 | -.131 | .086 |
| Months since Hire | -.058 | .074 | -.230 | .206 | .098a | .029 | .116 |
| Previous Experience (months) | .119 | -.085 | .098 | -.131 | .029 | .538a | .757 |
| Date of Birth |-.106 | .035 | -.101 | .086 | .116 | .757 | .548a |
a. Medida de adecuación muestral |

Se elimina la variable que menor valor de Medida de adecuación muestral tenga. AQUELLAS MARCADAS CON a. AQUELLA QUE ES MENOR A 0.5

Una vez eliminada la variable menos compatible se analiza nuevamente los factores, con la rotación de variables y calculo de varimax.

Matriz de correlacionesa |
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