Transcript The Chi-Square Distribution
Goodness Of Fit
Goodness Of Fit
The purpose of a
chi-square goodness-of-fit test
is to compare an observed distribution to an expected distribution. For example, suppose there are four entrances to a building. You want to know if the four entrances are equally used. You observe 400 people entering the building on a random basis:
Entrance
Main Back Side 1 Side 2
Total Observed Frequency
140 120 90 50
400 Expected Frequency
100 100 100 100
400
H H
0 1 : :
p M = p B = p S1 = p S2 The proportions are not all equal.
If the entrances are equally utilized, we would expect each entrance to be used approximately 25% of the time. Is the difference shown above statistically significant?
Chi Square Test
If the observed frequencies are obtained from a random sample and each expected frequency is at least 5, the sampling distribution for the goodness of-fit test is a chi-square distribution with
k
-1 degrees of freedom. (where
k
= the number of categories) 2
Test Statistic
f o
f e f e
2 O = observed frequency in each category E = expected frequency in each category
Goodness-of-Fit Test: Equal Expected Frequencies
Let
f
0 and
f
e be the observed and expected frequencies, respectively.
H
0 : There is no difference between the observed and expected frequencies.
H
0 :
p 1 = p 2 = p 3 = p 4 H
1 : There is a difference between the observed and the expected frequencies.
H
1 :
The proportions are not all equal.
df = 3 df = 5
k
-1 degrees of freedom. (where
k
= the number of categories) See Table P.495
df = 10
2
EXAMPLE
The following information shows the number of employees absent by day of the week at a large manufacturing plant. At the .05 level of significance, is there a difference in the absence rate by day of the week? Day Monday Tuesday Wednesday Thursday Friday
Total
Frequency 120 45 60 90 130
445
EXAMPLE
continued
The expected frequency is: (120+45+60+90+130)/5=89.
The degrees of freedom is (5-1)=4.
The critical value is 9.488.
(Appendix B, P.495)
Example
continued
Day Monday Tuesday Wednesday Thursday Friday
Total
Freq.
120 45 60 90 130
445
Because the computed value of chi-square is greater than the critical value,
H
0 rejected.
is We conclude that there is a difference in the number of workers absent by day of the week.
Expec.
89 89 89 89 89
445
(
f
o
–
f
e
)
2
/
f
e
10.80
21.75
9.45
0.01
18.89
60.90
2
f o
f e f e
2
Example
Goodness of Fit
A seller of baseball cards wants to know if the demand for the following 6 cards is the same.
Tom Seaver Nolan Ryan Ty Cobb George Brett Hank Aaron Johnny Bench
Cards Sold
13 33 14 7 36 17
120
MegaStat
Tom Seaver Nolan Ryan Ty Cobb George Brett Hank Aaron Johnny Bench
Goodness of Fit Test Observed
13 33 14 7 36 17 120
Expected
34.40 chi-square 5 df 1.98E-06 p-value 20 20 20 20 20 20 120 O - E -7.000
13.000
-6.000
-13.000
16.000
-3.000
0.000
(O - E)² / E 2.450
8.450
1.800
8.450
12.800
0.450
34.400
% of chisq 7.12
24.56
5.23
24.56
37.21
1.31
100.00
Goodness Of Fit
(unequal frequencies)
Example - Goodness Of Fit
(unequal frequencies) The Bank of America (BoA) credit card department knows from national US government records that 5% of all US
VISA
card holders have no high school diploma, 15% have a high school diploma, 25% have some college, and 55% have a college degree. Given the information below, at the 1% level of significance can we conclude that (BoA) card holders are significantly different from the rest of the nation?
Education
Some HS HS Diploma Some College College Degree
Total Observed Frequency
50 100 190 160
500 Expected Frequency
25 75 125 275
500
= (500)(.05) = (500)(.15) = (500)(.25) = (500)(.55)
2
f o
f e f e
2 115 .
22
C
2 11 .
345
df = (4 - 1) = 3
Reject
H
0
Limitations of Chi-Square
Limitations of Chi-Square
1.) If there are only 2 cells, the expected frequency in each cell should be at least 5. 2.) For more than 2 cells, chi-square should not be used if more than 20% of
f e
cells have expected frequencies less than 5.
Roll-Of-The-Die Experiment
Outcome
1 2 3 4 5 6
TOTAL Observed Frequency
3 6 2 3 9 7
30 Expected Frequency
5 5 5 5 5 5
30
Two-thirds of the computed chi-square value is accounted for by just two categories (outcomes). Although the expected frequency is not less than 5, too much weight may be given to these categories. More experimental trials should be conducted to increase the number of observations.
MegaStat
Goodness of Fit Test observed 3 6 2 3 9 7 30 expected 5.000
5.000
5.000
5.000
5.000
5.000
30.000
7.60 chi-square 5 df .1797 p-value O - E -2.000
1.000
-3.000
-2.000
4.000
2.000
0.000
(O - E)² / E 0.800
0.200
1.800
0.800
3.200
0.800
7.600
% of chisq 10.53
2.63
23.68
10.53
42.11
10.53
100.00
Independence & Contingency Tables
Contingency Table Analysis
A contingency table is used to investigate whether two traits or characteristics are related. Each observation is classified according to two criteria.
The
degrees of freedom
is equal to:
df
= (# rows - 1)(# columns - 1).
The
expected frequency
is computed as: Expected Frequency = (row total)(column total)/Grand Total
EXAMPLE
Is there a relationship between the location of an accident and the gender of the person involved in the accident? A sample of 150 accidents reported to the police were classified by type and gender. At the .05 level of significance, can we conclude that gender and the location of the accident are related?
Gender
Work Home Other
Total
Male 60 Female 20
Total 80
20 30
50
10 10
20 90 60 150
EXAMPLE
continued
Gender
Work Home Other
Total
Male 60 Female 20
Total 80
20 30
50
10 10
20 90 60 150
The expected frequency for the work-male intersection is computed as (90)(80)/150=48. Similarly, you can compute the expected frequencies for the other cells.
H
0 : Gender and location are not related.
H 1
: Gender and location are related.
EXAMPLE
continued
H
0 is rejected if the computed value of χ 2 is greater than 5.991. There are (3- 1)(2-1) = 2 degrees of freedom.
2
Find the value of
χ
2
.
60 48 2 48 10 8 2 8 16 .
667
H 0
is rejected. We conclude that gender and location are related.
MegaStat Example
Contingency Tables
A crime agency wants to know if a male released from prison and returned to his hometown has an easier (or more difficult) time adjusting to civilian life .
Residence After Release From Prison Adjustment to Civilian Life
Hometown Not Hometown
Total Outstanding
27 13
40 Good
35 15
50 Fair
33 27
60 Unsatisfactory
25 25
50 Total 120 80 200
MegaStat
Chi-square Contingency Table Test for Independence
Hometown Observed Expected Not Hometown Observed Expected Total Observed Expected Outstanding
27
24.00
13
16.00 40 40.00 Good
35
30.00
15
20.00 50 50.00 Fair Unsatisfactory
33
36.00
27
24.00 60 60.00
25
30.00
25
20.00 50 50.00 Total 120 120.00 80 80.00 200 200.00
5.73 chi-square
3 df
.126 p-value