Degrees of freedom in HLM models
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- Brook Moore
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1 Degees o eedom n HLM models The vaous degees o eedom n a HLM2/HLM3 model can be calculated accodng to Table 1 and Table 2. Table 1: Degees o Feedom o HLM2 Models Paamete/Test Statstc Degees o Feedom Gammas - Least Squae tables Beta andom Beta xed T Vaance components K q()* Geneal Hypothess tests Statstcs o cuent covaance components Vaance-Covaance test ** Homogenety o level-1 vaance T (egadless o xng; not an appoxmaton) J q The numbe o contasts speced ( 1)/1 2 o all models; add 1 loats, add MLF. the d.. speced along wth the model speccaton the numbe o unts o OLS esdual vaances (OLSRSVAR n the esdual le) * I the message statng "Note: The ch-squaes epoted above ae based on K o K unts..." s pnted, K s used. ** The degees o eedom o "Statstcs o cuent covaance components"
2 Table 2: Degees o Feedom o HLM3 Models Paamete/Test Statstc Degees o Feedom Gammas - Least Squae T tables (egadless o xng; not an appoxmaton) Gammas - Beta andom K q () 2 Gammas - Beta xed/p andom Gammas - Beta xed/p xed J q() T Vaance components: Level-3 K q ()* Level-1 and level-2 J q () K q ()* Geneal Hypothess tests The numbe o contasts speced Statstcs o cuent covaance components: 1( 1 1) / 2 2( 2 1) / 2 ; add 1 Vaance-Covaance test **: 2 loats the d.. speced along wth the model speccaton * I the message statng "Note: The ch-squaes epoted above ae based on K/J o K/J unts..." s pnted, K/J s used. ** The degees o eedom o "Statstcs o cuent covaance components" whee T = numbe o level-1 ecods n MDM J = numbe o level-2 ecods n HLM3 MDM K = numbe o level-3 ecods n HLM3 (level-2 ecods n HLM2) MDM = numbe o level-2 andom eects n HLM2 1 = numbe o andom eects at -level n HLM3 2 = numbe o andom eects at -level HLM3 = total numbe o xed eects q () n a HLM2 model, a gven s assocated wth a cetan. q ( ) s the the numbe o s assocated wth ths ; n a HLM3 model, a gven and s assocated wth a cetan. q ( ) s the numbe o s assocated wth ths. q () a gven s assocated wth a cetan. 2 q () 2 s the numbe o s assocated wth ths.
3 q () the numbe o andom eects assocated wth a cetan. q () the numbe o xed eects assocated wth a cetan. The calculatons o these degees o eedom ae llustated n the ollowng HLM2 and HLM3 examples. Example 1: A HLM2 model Gven the ollowng andom-ntecept model om the HSB data, MATHACH ( SES) 1 ( SECTOR) ( MEANSES) u ( SECTOR) ( MEANSES) u We have: T = 7185, K = 16, = 1 ( u ), =6 (6 s), and q() q(1) = 3. The degees o eedom n the least-squaes tables ae: T = = The appoxmate degees o eedom o the Gammas ae calculated accodng to the andomness o the beta s they ae assocated wth and ae lsted n Table 3. Table 3: Degees o Feedom o the Gammas Fxed Eect Randomness o Beta q () Fomula o d.. Appox. d.. andom 3 K q() andom 3 K q() andom 3 K q() xed 3 T xed 3 T 7179
4 12 xed 3 T 7179 The degees o eedom o the vaance component u s 16 3 = 157. The degees o eedom o the geneal hypothess tests ae the numbe o contasts speced, whch s 2 n ou case. The degees o eedom o the Statstcs o cuent covaance components model s ( 1)/2 = 2. The degees o eedom o the Test o homogenety o level-1 vaance s 16 1 = 159. Example 2: A HLM3 model The HLM3 model wth x and andom slopes om the EG data s shown below: MATH ( YEAR) e 1 ( BLACK) ( HISPANIC) ( BLACK) ( HISPANIC) ( LOWINC) u ( LOWINC) u In ths example we have the values: T = 723, J = 1721, K = 6, 1 = 2 (, ), 1 2 = 2 (, 1 u u ), =8 (8 s), q() q(1) = 3 (o and 1), q2() q2(1) (o and 1), q () q (1) = 1, and q () q (1) = 2 The degees o eedom n the least-squaes tables ae: T = = 7222.
5 The appoxmate degees o eedom o the Gammas ae calculated accodng to the andomness o the s and s they ae assocated wth, whch ae lsted n Table 4. Table 4. Degees o Feedom o the Gammas Eect Randomness o Beta Randomness o P q () q () omula o d.. d.. 2 andom andom 3 2 K q () 2 58 andom andom 3 2 K q () xed andom 3 1 J q() xed andom 3 1 J q() andom andom 3 2 K q () andom andom 3 2 K q () xed andom 3 1 J q() xed andom 3 1 J q() The degees o eedom o the level-3 vaance components ae K q ( ), whee K can be K when the message Note: The ch-squae epoted above ae based on K o K unts s pnted n the output. In ou case, the degees o eedom o u s 6 2 = 58, and the same wth u 1. The degees o eedom o level-1 and level-2 vaance components ae J q ( ) Kq( ), whee J and K can be J and K as above. The degees o eedom o the vaance components s summazed n Table 5.
6 Table 5: Degees o Feedom o the Random Eects Vaance Component q () q () omula o d.. d J q() Kq() J q() Kq() 1659 u 1 2 K q () 58 u K q () 58
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