1. Introduction. 1.1 Aims. 2 characters. For each individual: 120 one pair (x, y) of values 100. Data series : 80
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1 TC Mathematics S2 1.1 Aims 1. Introduction Y : revenue (k ) Study 150? 2 characters For each individual: 120 one pair (x, y) of values 100 Data series : 80 2 variables X : expense (k ) Are Xand Yrelated? Modelthe relationship Estimatenew «correlation» «fitting», unknown values «regression» «prediction»
2 TC Mathematics S2 1. Introduction 1.2 Formatting «List» series Contingency tables Amountof fertilizer Harvested production plot n X(kg.ha -1 ) Y(q.ha -1 ) Time series Y : acuity X : age / / / X : year Y : expense
3 TC Mathematics S2 1. Introduction 1.3 Scatter plot «List» series X : year Y : expense Data seriesx i and y i Set of points M i (x i ; y i ) «point cloud»
4 TC Mathematics S2 1. Introduction 1.3 Scatter plot Contingency tables X : age / Y : acuity 6/ /
5 Introduction example 2. Independence χ² test Two variables are crossed. Test the hypothesis of their independence. Evaluate the difference between observation and theory by a global parameter : χ² calc e.g. : Let s test H 0 : «gender and IQ are independent (inside a population)» 1. Resultsfroma sample of 6 men and 8 women : 2. Results theoretically expected (on H 0 ) : 3. Differences: partial χ²s andtotal χ² <100 >100 <100 >100 <100 >100 F F F 1 1 H H H 4/3 4/ How to calculate a χ² : (obs th)² / th This value isthe χ² calc of our experiment.
6 χ² distribution 2. Independence χ² test Whatever the experiment, considering the nullhypothesis«h 0» (e.g. : «gender and IQ are independent»), the probability that a value of χ² be (or not) exceeded can be given. probability density reality 96.6 % 3.4 % probability density your possibilities? %? % our example possible values of χ² % 10% our example % 1% possible values of χ²
7 Methodology 2. Independence χ² test 1. Goal : test the nullhypothesisof independence«h 0» (e.g. : gender and IQ). Aim: canweaffordrejectingitor can twe? 2. Calculate the χ² from your sample : χ² calc. Aim: evaluatehow bigitis, accordingto the hypothesish 0 3. Locate your χ² calc (by probabilities) thanks to the form. Aim: compare itto the four values of χ²givenby the form Independence test usingrrows and k columns : ddl= (r 1)(k 1) probability density 4. Given a significance level α, a decision has to be made (rejecting H 0 or not). 0 χ² calc
8 Exercise 1 2. Independence χ² test observeddistribution theoreticaldistribution on H 0 Gm 1. Nullhypothesis«H 0» : gender and tobacco behaviour are independent. 2. χ² calc = Gw Bn Bs Bss number of dof : (3-1)(2-1) = 2 significancelevel: α= 10% χ² lim = At a 10% significancelevel, H 0 canberejected. Gm Gw Bn Bs Bss probability density 0 Bn Bs Bss? 4.61 values of the χ²s Gm χ² calc 10%? 5% 7.82 Gw % Form values 1%
9 3. Fitting, Mayer 3.1 Moving means Exercise 5 1 to 5 2 to turnover (M ) x: 1 to
10 3.2 Purpose of linear fitting 3. Fitting, Mayer Point clouds may show an infinite variety of shapes. Sometimes, a straight line correctly follows it. «linear fitting» For someothers, a curvedoes. «non-linearfitting»
11 3.2 Purpose of linear fitting 3. Fitting, Mayer Belonging to the point cloud: By definitionof the point M i, x i and y i are associated. y i y i e i Once drawna line: To x i isalsoassociated y i = ax i + b x i Definition: wenameresiduethe number e i = y i y i e i > 0 : M i isabovethe line e i < 0 : M i isbelowthe line (D) : regression line(or trendline) ; determining(d) : performing a linear fitting.
12 3. Fitting, Mayer 3.3 Mayer s method Mayer smethod(to finda relevant line) : Mayer s principle: n i= 1 e i = 0 1. Fairly divide the cloud(horizontally) 2. DetermineG 1 and G 2, meanpoints of each semi-cloud 3. (D) = (G 1 G 2 ) (D) G Equivalent to : G (D) notice : In anycase, G (G 1 G 2 ). Thus, (D) isa Mayer s line. G 1 G G 2
13 3.3 Mayer s method turnover (M ) 3. Fitting, Mayer Exercise5 + x: 1 à G 2 G 1 Mayer s line xg = = yg = = xg = = yg = = tri1 tri2 tri3 tri4 tri1 tri2 tri3 tri4 tri1 tri2 tri3 tri4 tri1 tri2 tri3 tri4 (M )
14 3.3 Mayer s method 3. Fitting, Mayer Exercise 6 x y G G = = = = 42 3 x y G G = = = = y: harvest (q/ha) G G D M x: fertilizer (kg/ha) fertilizer harvest (kg.ha -1 ) (q.ha -1 ) a = ; b = yg ax G
15 TC Mathematics S2 4. Linear fitting: least square 4.1 Parameters of a bivariant series Means Variances n r x y = = i= 1 n n i= 1 n Calculator: x y i i V V ( X ) ( Y ) x 2 i i= 1 2 = x n k y 2 j j= 1 2 = n enter CALC > SET CALC > 2Var X : List 1 Y : List 2 (n ij : List 3) y Standard deviations ( X ) V ( X ) σ = ( Y ) V ( Y ) σ = CALC > Stat2Var L 1,L 2,(L 3 ) Covariance (, ) Cov X Y results xy n x y i i i= 1 = n 2 x, x, x, σ, 2 y, y, y, σ, n X Y x y
16 TC Mathematics S2 4. Linear fitting: least square 4.1 Parameters of a bivariant series fertilizer harvest (kg.ha -1 ) (q.ha -1 ) Σxy = Cov(X,Y) = E(X) = 134 Σx = 670 Σx² = σ X = V(X) = 2384 E(Y) = 44.6 Σy = 223 Σy² = σ Y = V(Y) = Exercise 8 Y X [15 ; 25[ [25 ; 50[ 50 + none à à Σxy = Cov(X,Y) = E(X) = Σx = Σx² = σ X = V(X) = E(Y) = Σy = Σy² = σ Y = V(Y) =
17 TC Mathematics S2 4.2 Least square method 4. Linear fitting: least square Principle : n i= 1 e 2 i G Is minimum (D) equivalent to: formulas for a and b Results: A. The regression line is unique B. In any case, G (D) C. (, ) ( ) Cov X Y a = b = y ax V X
18 TC Mathematics S2 4. Linear fitting: least square 4.2 Least square method Exercise 9 expense (k ) (D) year(1 : 2002) V(X) = V(Y) = (, ) ( ) CALC > REG > X Cov(X,Y) = Cov X Y a = b V X (D) : y = x CALC > LinReg L 1,L 2
19 TC Mathematics S2 4. Linear fitting: least square 4.3 Linear correlation coefficient 0.95 < r 1 : Strong linear correlation Linear model: one of the most relevant (, ) ( X) σ ( Y ) Cov X Y r = R = ρ = σ 0.75 < r < 1 0 < r < < r < < r < 0-1 < r < < r < < r < < r < 0.5
20 TC Mathematics S2 4. Linear fitting: least square 4.3 Linear correlation coefficient
21 5. Non-linear fitting Methodology Study X, Y 1. replace Xor Y by a new variable (e.g.: T) 2. Linear fitting on the new pair Let s take: X Y Wedefinethe variable change T= X². T Y Wehave to determinean expression y= at+ b. of the fittingline, for the pair (T, Y). e.g. using the least square method: (D Y/T ) : y= t Final curved model of the relationship betweenxand Y 3. y= x²
22 5. Non-linear fitting Methodology Study X, Y 1. replace Xor Y by a new variable (e.g.: T) 2. Linear fitting on the new pair 3. Final curved model of the relationship betweenxand Y Exercise 13 X T Y (D Y/T ): y= t y= (x 60)²
23 6. Prediction 6.1 Point estimate Methodology Study X. Y 1. Find the expression of the fitting line (or curve): y = ax+ b(or y = f(x)) 2. Note the estimating value x 0 or y 0 3.Calculate. thanksto the expression. the estimated value y 0 or x 0 Exercise 18 Exercise 9 1. y = x Exercise 6 2. x 0 = y = x y 0 = k 2. y 0 = 60 q/ha 3. x 0 = kg/ha Exercise y = (x 60)² x 0 = 100 km/h 3. y 0 = L/100km
24 6. Prediction 6.2 Confidence interval Exercice 19 Methodology Exercice 9 1. a. b. 1. c. z= σ Z = a. Calculatethe values y i = ax i + b b.calculatethe values z i = y i /y i of a variable Z c.getz smeanand standard deviation 2. Calculate the point estimate y 0 3.Givethe coefficient u following the confidence level 4. Interval: given formula 2. y 0 = k 3. Confidence level: 95% u= y 0 z uα z y0 z u ( σ ) ; ( + σ ) [136.3( ) ; 136.3( )] = [102.8 ; 169.8] α z
25 6. Prediction 6.2 Confidence interval Methodology 1. a. Calculatethe values y i = ax i + b b.calculatethe values z i = y i /y i of a variable Z c.getz smeanand standard deviation 2. Calculate the point estimate y 0 3.Givethe coefficient u following the confidence level 4. Interval: given formula Exercise 20 Exercise 6 1. c. z= σ Z = a. b. 2. y 0 = x = Confidence level: 99% u= y z u ; y z + u ( σ ) ( σ ) 0 α z 0 [57.29( ) ; 57.29( )] = [50.25 ; 64.22] α z
26 6. Prediction 6.2 Confidence interval Exercise 21 x i y j n ij y i z i y = x z= y/y List1 List2 List3 List4 List z= σ Z = x 0 = 80 y 0 = x = Confidence level: 99% u = y 0 ( z uασ ) ; z y0 ( z + uασ z ) [0.368( ); ( )] = [ ; ]
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