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simple regression of two indicator economic, Study Guides, Projects, Research of Economic Analysis

econometrie simple regression of two indicator economic

Typology: Study Guides, Projects, Research

2019/2020

Uploaded on 11/18/2021

asmae-sennad
asmae-sennad 🇲🇦

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Individual project
Presented by :
SENNAD ASMAE
Code apogée :
19510212
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Individual project

Presented by :

SENNAD ASMAE

Code apogée : 19510212

simple regression of two indicator economic :

1 – INDRODUCTION :

simple regression is the relation

between selected values of X and observed

values of Y ( from which the most probable

value of y can be predicted for any value of

x ) regression toward the mean.

So in this project im gonna analyse

the data of two indicator economic withe

simple regession on the following dependent

variable :

X : producter prices index

years X : Producer prices index PPI Y : costumer prices index CPI 2016 102 115. 2017 107 118 2018 108.5 120 2019 111 119 2020 109.2 120. TOTAL 537.7 592.

Table 1 :

Evolution of the main economic indicator in The

period 2016 TO 2020

SOURCE : https://mtataes.gov.ma/ar

 In statistics the analysis of variables that

are dependent on only one other variable.

Regression analysis uses regression equations ,

which shows the value of a dependent variable

as a function of an independent variabe. For

our project a simple regression equation could

take the form :

Y = β 0 ₊ β1 X ₊ ξ

So lets find the least square regression line :

Yˆ = βˆ₁ X ₊βˆ₀

and we now that : βˆ₁=

∑x²₋

β₀ = y ₋ βˆ₁ X

n ∑x.y ₋ ∑ x ∑ y ∑x ² n

and now we gonna calculat βˆ₀ :

First : Y ‗
and : X =
So βˆ₀ = 118.54 ₋0.5061 * 107.

AT last : : Y = 64.144 ₊ 0.5061 X

∑Y i. 1

n 5 ∑ X i. 1 n 5

This the graphe I draw it by exel and the least

square regression :

Y is determined by the value of X

100 102 104 106 108 110 112 112 113 114 115 116 117 118 119 120 121 f(x) = 0.51 x + 64. R² = 0. Series Linear (Series1)

Statistiques de la régression Coefficient de détermination multiple 0, Coefficient de détermination R^2 0, Coefficient de détermination R^2 0, Erreur-type 2, Observations 5 La cofficient par exel so now we calculat the solution of test slope coefficient H₀ : β₁ ‗ 0 H₁ : β₁ ≠ 0 α = 0. and we have S² = = 3.646245977 = 1. S = 1.215415326 = 1. SS = ∑xi² ₋ ‗ 57870.89 ₋ = 46. SSE n₋ 5₋ ( ∑xi )² n (537.7)² 5

SB 1 ‗ S

S = 1.

t =

Conclusion : There is evidence of a relationship

SSXX 57870.89 ₋ (537.7)² 5 βˆ₁ SB 1