卡方分布英文课件

Reject the Null !
Example
A game involves flipping 3 times a coin (having a tail and a head). Suppose a gambler plays the game 80 times, with the following observed counts:
No. of head 0 1 2 3
No. of times
4
32
38
6
Do the observed number of results fit the expected ones at 5% level? What can we conclude?
Solution
• Theoretical distribution
Solution
• Theoretical distribution
No of sixes No of Rolls 0 58 1 35 2 7 3 1
(2df 3)
(48 58)2 (35 35)2 (15 7)2 (3 1)2 14.9 58 35 7 1
Chi-Square test for independence
•How to calculate the “expected frequencies”:
fe
Tot column Tot row Total
Others Hollande Sarkozy
95 95 x 25 210
Male Female
25
210
Chi-Square test for independence
•Now we can calculate the chi square value :
FO
10
15
50
60
35
40
FE
11.31 13.69
49.76 60.24
33.93 41.07
( f f ) 0 e 2 fe
The casino becomes suspicious of the gambler and wishes to determine whether the dice are fair. Do the observed number of rolls fit the expected ones? What do they conclude?
Testing for Independence
Source: Based on a Fox News Poll conducted January, 1999
And now your answer!
• Please answer this question: “Which one is the most important in your life?” 1)Money 2)Health 3)Love • Is there any differences between females and males?
•You can test your sample against any possible expected values
25% 10%
25% 10%
25% 10%
25% 70%
H0
H0
Example
A new casino game involves rolling 3 dice. The winnings are directly proportional to the total number of sixes rolled. Suppose a gambler plays the game 101 times, with the following observed counts:
Chi-Square test for goodness of fit
2 26.6
df C 1 4 1 3
2(3,n=120) = 26.66,
p< 0.001 Critical value = 7.81
26.6
Chi-Square test for Goodness of fit
Pearson Chi-Square:
•Frequencies No mean and SD 2 statistics No assumption of normality
Chi-Square test for goodness of fit
Suppose we have 50 blue balls, 30 orange balls, 30 yellow balls and 10 green balls. -Is the frequency of balls with different colors equal in our bag?
Observed Frequencies
ห้องสมุดไป่ตู้
50
30
30
10
Expected Frequencies
25%
25%
25%
25%
Chi-Square test for goodness of fit
Observed Frequencies Total
50
30
30
10
120
Expected Frequencies
2
df (C 1)(R 1) (31)(2 1) 2
2 (15 13.69)2 ( 10 11 . 31 ) 2 11.31 ... 0.34 13.69
Chi-Square test for independence
2(2) = 0.34, p= 0.83
Your answers...
Money Males Females Health Love
5 2
2 11
3 12
Solution and answer
• This is the theoretical distribution under the independence assumption:
Chi-Square test for independence
•When we have two or more sets of categorical data the Chi-square test works
Others
Male
Hollande 50
60 110
Sarkozy 35
40 75 95
No. of heads No. of times 0 10 1 30 2 30 3 10
(2df 3)
(4 10)2 (32 30)2 (38 30)2 (6 10)2 7.466 10 30 30 10
ACCEPT the Null ! We can conclude that the coin is regular and the game is fair
Theoretical
2 5
3.71 9.29
4.29 10.71
2 ( 2 3.71) 2 (3 4.29) 2 ( 5 2 ) (2df 2 ) 2 3.71 4.29 (2 5) 2 (11 9.29) 2 (12 10.71) 2 7.95 5 9.29 10.71
2
( fObs f e ) fe
(50 30)2 (30 30)2 (30 30)2 (10 30)2 2 26.6 30 30 30 30
Chi-Square test for goodness of fit
2 26.6
df C 1 4 1 3
10
15 25
Female
115
210
Chi-Square test for independence
•Also called contingency table analysis •H0: There is no relation between gender and voting preference (sort of lack of correlation) OR •H0: There is no difference between the voting preference of males and females AGAINST H1 : There is difference…or there is a relation… •The logic is the same as the goodness of fit test: Comparing observed frequencies and Expected frequencies if the two variables were independent
25%
25%
25%
25%

120
=
Expected Frequencies
30
30
30
30
H0
Chi-Square test for goodness of fit
Observed Frequencies
50 30
30 30
2
30 30
10 30
Difference Normalize
Expected Frequencies
There is no significant effect of gender on vote preference Or We cannot reject the null hypothesis that gender and vote preference are independent
EXAMPLE