Finansiell Statistik, GN, 15 hp, VT2008 Lecture 12::(1) Analysis of Variance (2) Chi-Square Tests for Independence and for Goodness-of- t

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1 Finansiell Statistik, GN, 15 hp, VT2008 Lecture 12::(1) Analysis of Variance (2) Chi-Square Tests for Independence and for Goodness-of- t Gebrenegus Ghilagaber, PhD, Associate Professor 1

2 1 Introduction In the previous lectures we outlined methods for testing hypotheses on one parameter, such as the mean (H 0 : = 0 ) or comparison of two parameters (H 0 : 1 = 2 ): Suppose, we have studies involving more than two populations and we want to test if all population means are equal: H 0 : 1 = 2 = 3 = ::: = J H 0 : i 6= j for at least one pair (i; j); i 6= j 2

3 2 One way Analysis of Variance Aim: To study the e ect of di erent levels of the same (single) factor (treatment) on a response variable. This is equivalent to comparing the means of the di erent levels of the treatment. The model is where Y ij = + j + " ij; i = 1; 2; :::; n j ; j = 1; 2:::J = + j + Y ij j Y ij = value of response variable for the ith individ. who received treatment j = Common (overall) e ect (overall mean) j = j = Treatment e ect (e ect of j th treatment) " ij = Y ij j = random error 3

4 Design layout Treatment (j) observation (i) 1 2 : : : J 1 y 11 y y 1J 2 y 21 y y 2J :. :. :. n y n1 y n2 : : : y nj Total T: 1 T: 2 : : : T: J T:: (Grand total) Mean y 1 y 2 : : : y J y:: (Grand mean) 4

5 where y: j = T: j n j = y:: = T:: N = n X j y ij i=1 n j j=1 n X j i=1 T: j n j = n X j y ij n X j n j i=1 5

6 Analysis of variance: Let s consider the total deviation of each observation from the overall mean: y ij y:: = y ij y: j + y: j y:: = y ij y: j + y:j y:: Thus, yij y:: 2 = h yij y: j + y:j y:: i 2 so that = y ij y: j 2 + y:j y:: yij y: j y:j y:: 6

7 n j X yij y:: 2 = n j X n X j +2 yij n 2 X j y: j + yij y: j y:j y:: y:j y:: 2 It can be shown that n j X yij y: j y:j y:: = 0: 7

8 Thus, n j X yij y:: 2 {z } SS T otal = n j X yij y: j 2 j=1 i=1 {z } SS W ithin + j=1 n j y:j y:: 2 {z } SS Between In other words, Total Sum of Squares = Sum of Sq. within treatm. + Sum of Sq. between treatm. SS Total = SS Within + SS Between = SS Error + SS Treatment = SS Unexplained + SS Explained 8

9 where SS Treatment = SS Between = = n j X T: 2 j T 2 :: n j N SS Error = SS Within = SS Total = = y:j y:: 2 = J X j=1 j=1 n X j n 2 X j yij y: j = yij 2 j=1 i=1 n X j yij y:: n 2 X j = y 2 T 2 :: ij N 0 n X y 2 T: j T: 2 A ij j T 2 1 :: A j=1 n j j=1 n j N = SS Error + SS Treatment = (SS Within + SS Between ) n j y:j y:: 2 j=1 T: 2 j n j 9

10 ANOVA Table Source of Degrees Sum of Mean F-ratio variation of freedom Squares Squares Treatment ( j ) J 1 SS B MS B = SS B J 1 F = MS B MS W Error (" ij ) N J SS W MS W = SS W N J Total N 1 SS T - 10

11 Example: 4 advertising methods applied on 5 companies (y ij = sales) Method (j) Company (i) Total Method (j) Company (i) Total (T: j ) Here, J = 4; n j = 5 for all j; and N = j=1 n j = 20: Thus, 11

12 SS Total = n X j yij 2 = 1340 (6) 2 T 2 :: N = (6)2 + ::: + ( 9) 2 (6) = 1338:2 SS Treatment = j=1 = 1135 T: 2 j n j T 2 :: N = (42)2 5 + (36)2 5 + ( 32)2 5 + ( 40)2 5 (6) 2 20 SS Error = SS Total SS Treatment = 1338: = 203:2 12

13 The resulting ANOVA Table is then given by Source of Degrees Sum of Mean F-ratio variation of freedom Squares Squares Treatment (Method) Error (" ij ) = 378:33 F = 203:2 16 = 12:7 378: 33 12: 7 = 29:79 Total The tabulated value of F at 5% signi cance level is F (3;16;0:05) = 3:24 which is by far less than the calculated value of F. Thus, we reject the null hypothesis H 0 : 1 = 2 = 3 = 4 and conclude that the average e ects of at least one pair of adv-methods are signi cantly di erent from each other. 13

14 3 Simple and Simultaneous Con dence Intervals Having concluded that there is signi cant treatment e ect (some of the means di er from each other), a natural question would be which treatment levels? A simple con dence interval for i j is given by yi y j t(n J; 2 ) v u t Sp 2 1 n i + 1 n j! where Sp 2 = MS W is the error variance. Thus, a 95% con dence interval for 2 is given by 1 14

15 (y 1 y 2 ) t (16;0:025) = 2: = 1:2 4:7782 = ( 3:5782; 5:9782) s v u t Sp 2 12:7 1 1 n n ! Similarly, a general (simultaneous) con dence interval for i Sche e s multiple comparison j is given by yi y j q (J 1)) F(J 1;N J;) v uut S 2 p 1! n n 2 15

16 where J is the number of means to be compared. In our present case, a 95% simultaneous con dence interval for 1 2 is given by v u t Sp 2 (y 1 y 2 ) q 1 (4 1)F (3;16;0:05) + 1 n 1 n 2 s q 1 = (3)(3:24) 12: = 1:2 7: = ( 5: 826 9; 8: 226 9)! 16

17 4 Two-way ANOVA with one observation per cell: Randomized Blocks Suppose the ve companies in the above example were from di erent locations (geographical) and we want to account for geographical variations (e ects of location). Then, the model is Y ij = + i + j + " ij; i = 1; 2; :::; n j ; j = 1; 2:::J = + ( i ) + j + Y ij j 17

18 where Y ij = value of response variable for the i th company (location) who did adv. j = Common (overall) e ect (overall mean) i = i = Block e ect (e ect of i th location/block) j = j = Treatment e ect (e ect of j th treatment/adv.) " ij = Y ij j = random error n j X yij y:: 2 {z } SS T otal = IX (y i : y::) 2 {z } + IX + IX SS Block yij y i : y: j : + y:: 2 y:j : y:: 2 {z } SS T reatment {z } SS Error 18

19 and the corresponding ANOVA table will be given by Source of Degrees Sum of Mean F-ratio variation of freedom Squares Squares Block ( i ) I 1 SS B MS B = SS B I 1 F = MS B MS E Treatment ( j ) J 1 SS T MS T = SS T J 1 F = MS T MS E Error (" ij ) N I J + 1 SS E MS E = SS E N I J+1 Total N 1 SS T otal - 19

20 5 Two-way ANOVA with multiple observations per cell 20

21 6 Chi-Square as a Test of Independence Example 1: Smoking and Cancer Smoking Cancer Yes No Total Yes O 11 O 12 O 1+ No O 21 O 22 O 2+ Total O +1 O +2 n Smoking Cancer Yes No Total Yes E 11 E 12 E 1+ No E 21 E 22 E 2+ Total E +1 E +2 n where E ij = O i+o +j n = (T otal of Row i) (T otal of Column j) G rand T otal 21

22 Then, 2 = RX CX Oij E ij 2 E ij s 2 (R 1)(C 1) 22

23 Example 2: trängselskatten A survey on trängselskatten in Stockhom gave the following results: Kön Åsikt Man Kvinna Total Mycket bra Ganska bra Ganska dåligt Mycket dåligt Total Do the results indicate that Opinion depends on Gender? 23

24 Solution H 0 : No relationship between sex and opinion H 1 : Sex and Opinion are related (dependent) Under the null hypothesis (assuming H 0 is true), the expected frequencies in each cell may be obtained as usual: Expected frequency in cell (i; j) = (Total in Row i)*(total in Column j) Grand Total Thus, E 11 = = 114:02; E 12 = = 115:98 24

25 E 21 = = 171:53; E 22 = = 174: E 31 = E 41 = = 144:76; E 32 = = 147: = 268:69; E 42 = = 273: Thus, 2 2 4X 2X Oij E ij = E ij ( :02)2 ( :98)2 = + 114:02 115:98 = 21:938 (p value = ) + ::: + ( :31)2 273:31 25

26 Since 2 cal = 21:938 > 2 tab = 2 (3;0:05) = 7:815; we reject the null hypothesis (H 0 ) and conclude that the data provides evidence on strong relationship between Gender and Opinion on trängselskatt. 26

27 7 Chi-Square as a Test of Goodness of Fit Example 1: Mendelian Theory (proportions in hybridizations) 9:3:3:1 Example 2: An analyst got the following results while recording antalet aktieköp under 30 enminutesperioder and would like to test if the number of aktieköp under 30 enminutesperioder per minute follows a Poisson distribution with expected value 1.5: Antal handlade Observerade antal

28 Solution: H 0 : The number of sales (antal handlade) follows a Poisson distribution with parameter (expected value) 1.5 : P (X = r) = e 1:5 (1:5) r ; r = 0; 1; 2; 3 r! H 1 : H 0 is not true Under the Null-hypothesis (when H 0 is true), the probabilities of 0, 1, 2, and 3 sales per minute may be obtained as follows: 28

29 3X r=0 P (X = 0) = e 1:5 (1:5) 0 = 0: ! P (X = 1) = e 1:5 (1:5) 1 = 0: ! P (X = 2) = e 1:5 (1:5) 2 = 0: ! P (X = 3) = e 1:5 (1:5) 3 = 0: ! P (X = r) = 0: : : :12551 = 0:93436 so that P (X 4) = 1 0:93436 = 0:

30 Thus, the table of observed frequencies may be expanded as follows: Antal (i) Observed (O i ) Expected (E i ) :22313 = 6: :33470 = 10: :25102 = 7: :12551 = 3: :06564 = 1:969 Total

31 The appropriate test statistics is, then, computed as 2 = 4X i=0 (O i E i ) 2 E i (10 6:6939)2 (9 10:041)2 = + 6: :041 (0 1:969)2 + 1:969 = 5:887 < 2 (4; 0:05) = 9:49 + (5 7:5306)2 7: (6 3:7653)2 3:7653 Do Not Reject H 0 31

32 8 Applications 32

Finansiell Statistik, GN, 15 hp, VT2008 Lecture 10-11: Statistical Inference: Hypothesis Testing

Finansiell Statistik, GN, 15 hp, VT008 Lecture 10-11: Statistical Inference: Hypothesis Testing Gebrenegus Ghilagaber, PhD, Associate Professor April 1, 008 1 1 Statistical Inferences: Introduction Recall: