Corrigé Epreuve 2002: Corrélation et droite de régression (5 points)

 

Corrélation et droite de regression (5 points-2002)

1.


\begin{array}
[c]{ccccc}
\begin{array}
[c]{c}
xi\end{array}&\begin{array}[c]{c}Math\\zi\end{array}
& xi^2& zi{{}^2}& ni\text{ }zi\\6 & 4.28 & 36 & 18.23 & 25.68\\8 & 6 & 64 & 36 & 48\\10 & 9.86 & 100 & 97.22 & 98.6\\12 & 12.46 & 144 & 155.25 & 149.52\\14 & 14.8 & 196 & 219.04 & 207.2\end{array}

 

2. a )


b)

\bar{x}=\frac{6+8+10+12+14}{5}=10

\bar{z}=\frac{4.28+6+9.86+12.46+14.8}{5}=9.48

 

le coefficient de correlation r:

 

r=\frac{Cov(X,Z)}{\sqrt{V(X)\text{ }V(Z)}}

Cov(X,Z)=\frac{1}{5}%
%TCIMACRO{\dsum \limits_{i-1}^{5}}%
%BeginExpansion
{\displaystyle\sum\limits_{i-1}^{5}}
%EndExpansion
xi zi-\bar{x} \bar{z}

=\frac{1}{5}(25.68+.48+98.6+149.52+207.2)-94.8

 

=105.8-94.8

 

 

V(x)=\frac{1}{5}%
%TCIMACRO{\dsum \limits_{i-1}^{5}}%
%BeginExpansion
{\displaystyle\sum\limits_{i-1}^{5}}
%EndExpansion
xi%
%TCIMACRO{\U{b2}}%
%BeginExpansion
{{}^2}%
%EndExpansion
-(\bar{x})%
%TCIMACRO{\U{b2}}%
%BeginExpansion
{{}^2}%
%EndExpansion

=\frac{1}{5}(36+64+100+144+196)-100

V(Z)=1/5(18,32+36+97,22+155,25+219,04)-89,8704

r=\frac{11}{\sqrt{8\ast15.2956}}

r\simeq0.99

c)

D_{z/x}:z=ax+b

a=\frac{Cov(X,Z)}{V(X)}=\frac{11}{8}

\bar{z}=a\bar{x}+b

b=\bar{z}-a\bar{x}=9.48-\frac{11}{8}\ast10

=\frac{-34.16}{8}


 

d) Voir figure

 

OIF
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