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--- sequential_regression [2024/06/12 08:10] – [Enter] hkimscil
+++ sequential_regression [2024/06/12 08:30] (current) – [r] hkimscil
@@ Line 50: / Line 50: @@
 | .794  | -.692  | .565  | -.409  |
 |  zero-order square  ||  part (in spss) = semipartial (in general)  ||
-| x1 sq (x1sq)  | x2 sq (x1sq)  | x1 part sq (x1psq) | x2 part sq (x1psq)  |
+| x1 zsq (x1zsq)  | x2 zsq (x1zsq)  | x1 semi-partial (or part) sq (x1spsq) | x2 part sq (x1spsq)  |
 | .630436  | .478864  | .319225  | .167281  |
 | a+b / a+b+c+d  | b+c / a+b+c+d  | a / a+b+c+d  | c / a+b+c+d  |
-x1sq - x1psq  ~= x2sq - x2psq
+x1zsq - x1spsq  ~= x2zsq - x2spsq
 .311211 ~= 0.311583
-혹은 위의 아웃풋에서 part correlation값과 zero-order correlation값을 이용하여 공통부분을 구할 수도 있다.
+아래는 r 에서 계산한 것
 <code>
 > .794^2 - .565^2
@@ Line 279: / Line 279: @@
 <code>
-pcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
+pp.b.i <- pcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
-pcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
+p.b.i
+p.b.i$estimate
+p.b.f <- pcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
+p.b.f
+p.b.f$estimate
+sp.b.i <- spcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
+sp.b.i
+sp.b.i$estimate
+sp.b.f <- spcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
+sp.b.f
+sp.b.f$estimate
+zc.b.i <- cor(datavar$bankaccount, datavar$income)
+zc.b.i
+zc.b.f <- cor(datavar$bankaccount, datavar$famnum)
+zc.b.f
+zc.b.i^2 - (sp.b.i$estimate)^2
+zc.b.f^2 - (sp.b.f$estimate)^2
-spcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
-spcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
 </code>
 . . .
 <code>
-> pcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
+> pp.b.i <- pcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
-   estimate    p.value statistic  n gp  Method
+> p.b.i
-0.7825112 0.01267595  3.325102 10  1 pearson
+  estimate p.value statistic  n gp  Method
-> pcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
+   0.7825 0.01268     3.325 10  1 pearson
-   estimate    p.value statistic  n gp  Method
+> p.b.i$estimate
--0.672856 0.04702022 -2.406425 10  1 pearson
+[1] 0.7825
 >
-> spcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
+> p.b.f <- pcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
-   estimate  p.value statistic  n gp  Method
+> p.b.f
-0.5646726 0.113182  1.810198 10  1 pearson
+  estimate p.value statistic  n gp  Method
-> spcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
+  -0.6729 0.04702    -2.406 10  1 pearson
-    estimate   p.value statistic  n gp  Method
+> p.b.f$estimate
--0.4086619 0.2748117 -1.184655 10  1 pearson
+[1] -0.6729
+>
+> sp.b.i <- spcor.test(datavar$bankaccount, datavar$income, datavar$famnum)
+> sp.b.i
+  estimate p.value statistic  n gp  Method
+   0.5647  0.1132      1.81 10  1 pearson
+> sp.b.i$estimate
+[1] 0.5647
+> sp.b.f <- spcor.test(datavar$bankaccount, datavar$famnum, datavar$income)
+> sp.b.f
+  estimate p.value statistic  n gp  Method
+  -0.4087  0.2748    -1.185 10  1 pearson
+> sp.b.f$estimate
+[1] -0.4087
+>
+>
+> zc.b.i <- cor(datavar$bankaccount, datavar$income)
+> zc.b.i
+[1] 0.7944
+> zc.b.f <- cor(datavar$bankaccount, datavar$famnum)
+> zc.b.f
+[1] -0.6923
+>
+> zc.b.i^2 - (sp.b.i$estimate)^2
+[1] 0.3123
+> zc.b.f^2 - (sp.b.f$estimate)^2
+[1] 0.3123
+>
 >
 >
 </code>