(Soal 2) ANALISIS KORELASI dan REGRESI ~ MUNAWI INSIDE

(Soal 2) ANALISIS KORELASI dan REGRESI

Posted by Nursaptia Purwa Asmara on Saturday, November 08, 2014 with No comments

Buat analisis regresi linier sederhana ${Y}_{i}$ ke ${X}_{i}$ untuk data berikut ini:

	${X}_{i}$	${Y}_{i}$	${U}_{i}$	${V}_{i}$
	2	4	-	-
	3	7	-	-
	1	3	-	-
	5	9	-	-
	8	17	-	-
Jumlah	-	-	-	-
Rerata	-	-	-	-
Simp. Baku	-	-	-	-

Hitung koefisien regresi $Y$ ke $X$

Hitung koefisien korelasi antara $Y$ dan $X$

Isi dua kolom terakhir dimana ${U}_{i}=\frac{{Y}_{i}-\bar{Y}}{{S}_{Y}}$ dan ${V}_{i}=\frac{{X}_{i}-\bar{X}}{{S}_{X}}$ dengan $\bar{Y}$ dan $\bar{X}$ adalah rerata data $Y$ dan $X$ , sedangkan ${S}_{Y}$ dan ${S}_{X}$ adalah simpangan bakunya

Hitung koefisien regresi $U$ ke $V$

Hitung koefisien korelasi antara $U$ dan $V$

JAWABAN (Mohon Dikoreksi Kembali)

$\sum{{X}_{i}=\left(2+3+1+5+9\right)=20}$

$\bar{{X}_{i}}=\frac{\sum{{X}_{i}}}{n}=\frac{20}{5}=4$

$\sum{{X}_{i}^{2}=\left({{2}^{2}+{3}^{2}+{1}^{2}+{5}^{2}+{9}^{2}}\right)=120}$

${S}_{X}=\frac{\sqrt{\sum{{\left({X}_{i}-\bar{X}\right)}^{2}}}}{2}$

       $=\frac{\sqrt{\left[{\left(2-4\right)}^{2}+{\left(3-4\right)}^{2}+{\left(1-4\right)}^{2}+{\left(5-4\right)}^{2}+{\left(9-4\right)}^{2}\right]}}{2}=3,16$

$\sum{{Y}_{i}=\left(4+7+3+9+17\right)=40}$

$\bar{{Y}_{i}}=\frac{\sum{{Y}_{i}}}{n}=\frac{40}{5}=8$

$\sum{{Y}_{i}^{2}=\left({{4}^{2}+{7}^{2}+{3}^{2}+{9}^{2}+{17}^{2}}\right)=444}$

${S}_{Y}=\frac{\sqrt{\sum{{\left({Y}_{i}-\bar{Y}\right)}^{2}}}}{2}$

       $=\frac{\sqrt{\left[{\left(4-8\right)}^{2}+{\left(7-8\right)}^{2}+{\left(3-8\right)}^{2}+{\left(9-8\right)}^{2}+{\left(17-8\right)}^{2}\right]}}{2}=5,57$

$\sum{{X}_{i}{Y}_{i}=\left[\left(2\times4\right)+\left(3\times7\right)+\left(1\times3\right)+\left(5\times9\right)+\left(9\times17\right)\right]=230}}$

Koefisien regresi $Y$ ke $X$

${b}_{YX}=\frac{n\sum{{X}_{i}{Y}_{i}-\sum{{X}_{i}}}\sum{{Y}_{i}}}{n\sum{{X}_{i}^{2}-{\left(\sum{{X}_{i}}\right)}^{2}}}=\frac{\left(5\right)\left(230\right)-\left(20\right)\left(40\right)}{\left(5\right)\left(120\right)-{\left(20\right)}^{2}}=1,75$

Koefisien korelasi $Y$ dan $X$

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

         $=\frac{\left(5\right)\left(230\right)-\left(20\right)\left(40\right)}{\sqrt{\left(\left(5\right)\left(120\right)-{\left(20\right)}^{2}\right)\left(\left(5\right)\left(444\right)-{\left(40\right)}^{2}\right)}}=0,994$

Pengisian kolom ${U}_{i}$ dan ${V}_{i}$

${U}_{i}=\frac{{Y}_{i}-\bar{Y}}{{S}_{Y}}$

      ${U}_{1}=\frac{{Y}_{1}-\bar{Y}}{{S}_{Y}}=\frac{4-8}{5,57}=-0,7181$

      ${U}_{2}=\frac{{Y}_{2}-\bar{Y}}{{S}_{Y}}=\frac{7-8}{5,57}=-0,1796$

      ${U}_{3}=\frac{{Y}_{3}-\bar{Y}}{{S}_{Y}}=\frac{3-8}{5,57}=-0,8980$

      ${U}_{4}=\frac{{Y}_{4}-\bar{Y}}{{S}_{Y}}=\frac{9-8}{5,57}=0,1789$

      ${U}_{5}=\frac{{Y}_{5}-\bar{Y}}{{S}_{Y}}=\frac{17-8}{5,57}=0,6165$

$\sum{{U}_{i}=\left(-0,7181-0,1796-0,8980+0,1789+0,6165\right)=-1}\Rightarrow0$

$\bar{{U}_{i}}=\frac{\sum{{U}_{i}}}{n}=\frac{-1}{5}=-0,2\Rightarrow0$

$\sum{{U}_{i}^{2}=\left{\left(-0,718\right)}^{2}+{\left(-0,179\right)}^{2}+{\left(-0,898\right)}^{2}+{\left(0,178\right)}^{2}+{\left(0,616\right)}^{2}\right}=4$

${S}_{U}=\frac{\sqrt{\sum{{\left({U}_{i}-\bar{U}\right)}^{2}}}}{2}$

       $=\frac{\sqrt{{\left(\left(-0,718\right)-0\right)}^{2}+{\left(\left(-0,179\right)-0\right)}^{2}+{\left(\left(-0,898\right)-0\right)}^{2}+{\left(\left(0,178\right)-0\right)}^{2}+{\left(\left(0,616\right)-0\right)}^{2}}}{2}$

      $=1$

${V}_{i}=\frac{{X}_{i}-\bar{X}}{{S}_{X}}$

      ${V}_{1}=\frac{{X}_{1}-\bar{X}}{{S}_{X}}=\frac{2-4}{3,16}=-0,6329$

      ${V}_{2}=\frac{{X}_{2}-\bar{X}}{{S}_{X}}=\frac{3-4}{3,16}=-0,3165$

      ${V}_{3}=\frac{{X}_{3}-\bar{X}}{{S}_{X}}=\frac{1-4}{3,16}=-0,5494$

      ${V}_{4}=\frac{{X}_{4}-\bar{X}}{{S}_{X}}=\frac{5-4}{3,16}=0,3165$

      ${V}_{5}=\frac{{X}_{5}-\bar{X}}{{S}_{X}}=\frac{9-4}{3,16}=1,5823$

$\sum{{V}_{i}=\left(-0,6329-0,3165-0,5494+0,3165+1,5823\right)=-1}\Rightarrow0$

$\bar{{V}_{i}}=\frac{\sum{{V}_{i}}}{n}=\frac{-1}{5}=-0,2\Rightarrow0$

$\sum{{V}_{i}^{2}=\left{\left(-0,633\right)}^{2}+{\left(-0,317\right)}^{2}+{\left(-0,594\right)}^{2}+{\left(0,317\right)}^{2}+{\left(1,582\right)}^{2}\right}=4$

${S}_{V}=\frac{\sqrt{\sum{{\left({V}_{i}-\bar{V}\right)}^{2}}}}{2}$

       $=\frac{\sqrt{{\left(\left(-0,633\right)-0\right)}^{2}+{\left(\left(-0,317\right)-0\right)}^{2}+{\left(\left(-0,594\right)-0\right)}^{2}+{\left(\left(0,317\right)-0\right)}^{2}+{\left(\left(1,582\right)-0\right)}^{2}}}{2}$

      $=1$

	${X}_{i}$	${Y}_{i}$	${U}_{i}$	${V}_{i}$
	2	4	-0,7181	-0,6329
	3	7	-0,1796	-0,3165
	1	3	-0,8980	-0,5494
	5	9	0,1789	0,3165
	8	17	0,6165	1,5823
Jumlah	20	40	0	0
Rerata	4	8	0	0
Simp. Baku	3,16	5,57	1	1

Koefisien regresi $U$ ke $V$

${b}_{UV}=\frac{n\sum{{V}_{i}{U}_{i}-\sum{{V}_{i}}}\sum{{U}_{i}}}{n\sum{{V}_{i}^{2}-{\left(\sum{{V}_{i}}\right)}^{2}}}=\frac{\left(5\right)\left(3,975\right)-\left(0\right)\left(0\right)}{\left(5\right)\left(4\right)-{\left(0\right)}^{2}}=0,994$

Koefisien korelasi $U$ dan $V$

${r}_{UV}=\frac{n\sum{{V}_{i}{U}_{i}-\sum{{V}_{i}}}\sum{{U}_{i}}}{\sqrt{\left(n\sum{{U}_{i}^{2}-{\left(\sum{{U}_{i}}\right)}^{2}}\right)\left(n\sum{{V}_{i}^{2}-{\left(\sum{{V}_{i}}\right)}^{2}}\right)}}$

$=\frac{\left(5\right)\left(3,975\right)-\left(0\right)\left(0\right)}{\sqrt{\left(\left(5\right)\left(4\right)-{\left(0\right)}^{2}\right)\left(\left(5\right)\left(4\right)-{\left(0\right)}^{2}\right)}}=0,994$

Categories: SOAL, soal ANAREG

MUNAWI INSIDE

MATERI BIOLOGI

Saturday, November 8, 2014

(Soal 2) ANALISIS KORELASI dan REGRESI

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