(Soal 1) ANALISIS KORELASI dan REGRESI ~ MUNAWI INSIDE

(Soal 1) ANALISIS KORELASI dan REGRESI

Posted by Nursaptia Purwa Asmara on Friday, November 07, 2014 with No comments

Kalau ${b}_{YX}$ adalah penduga koefisien regresi $Y$ ke $X$ , dan ${b}_{XY}$ adalah penduga koefisien regresi $X$ ke $Y$ , tunjukkan bahwa koefisien korelasi antara $Y$ dan $X$ adalah ${b}_{YX}$ kali ${b}_{XY}$

[Petunjuk: Ingat rumus koefisien korelasi dan rumus koefisien regresi]

JAWABAN:

${b}_{XY}=\frac{n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}\sum{{Y}_{i}}}}{n\sum{{X}_{i}^{2}}-{\left(\sum{{X}_{i}}\right)}^{2}}$

${b}_{YX}=\frac{n\sum{{Y}_{i}{X}_{i}-\sum{{Y}_{i}}\sum{{X}_{i}}}}{n\sum{{Y}_{i}^{2}}-{\left(\sum{{Y}_{i}}\right)}^{2}}$

      $r=\left[{b}_{YX}\right]\left[{b}_{XY}\right]$

   $=\left[\frac{n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}\sum{{Y}_{i}}}}{n\sum{{X}_{i}^{2}}-{\left(\sum{{X}_{i}}\right)}^{2}}\right]\left[\frac{n\sum{{Y}_{i}X_{i}-\sum{{Y}_{i}}\sum{X_{i}}}}{n\sum{{Y}_{i}^{2}}-{\left(\sum{Y_{i}}\right)}^{2}}\right]$

         $=\frac{\left(n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}\sum{{Y}_{i}}}\right)\left(n\sum{{Y}_{i}X_{i}-\sum{{Y}_{i}}\sum{X_{i}}}\right)}{\left(n\sum{{X}_{i}^{2}}-{\left(\sum{{X}_{i}}\right)}^{2}\right)\left(n\sum{{Y}_{i}^{2}}-{\left(\sum{Y_{i}}\right)}^{2}\right)}$

   $=\frac{{\left(n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}\sum{{Y}_{i}}}\right)}^{2}}{\left(n\sum{{X}_{i}^{2}}-{\left(\sum{{X}_{i}}\right)}^{2}\right)\left(n\sum{{Y}_{i}^{2}}-{\left(\sum{Y_{i}}\right)}^{2}\right)}$

   $=\sqrt{\frac{{\left(n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}\sum{{Y}_{i}}}\right)}^{2}}{\left(n\sum{{X}_{i}^{2}}-{\left(\sum{{X}_{i}}\right)}^{2}\right)\left(n\sum{{Y}_{i}^{2}}-{\left(\sum{Y_{i}}\right)}^{2}\right)}}$

   $=\frac{\left(n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}\sum{{Y}_{i}}}\right)}{\sqrt{\left(n\sum{{X}_{i}^{2}}-{\left(\sum{{X}_{i}}\right)}^{2}\right)\left(n\sum{{Y}_{i}^{2}}-{\left(\sum{Y_{i}}\right)}^{2}\right)}}$

Categories: SOAL, soal ANAREG

MUNAWI INSIDE

MATERI BIOLOGI

Friday, November 7, 2014

(Soal 1) ANALISIS KORELASI dan REGRESI

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