chapter-8 chi-square test. Ⅰ the mathematical properties of chi-square distribution types of...

Chapter-8 Chi-square test

Ⅰ The mathematical properties

of chi-square distribution

Types of chi-square tests

Chi-square test

Chi-square distribution

1. Tests of goodness-of-fit

Observed frequencies of one variable are significantly

different from the expected frequencies of the same

variable.

E.g. occurrences of heads and tails while flipping a coin.

2. Chi-SquareChi-Square tests of independence( or relationship)

Two variables are associated or independent of the other.

E.g. association between smoking and lung cancer.

Types of chi-square tests

The chi-square test of independence is probably the

most frequently used hypothesis test in the medicine.

In this chapter, we will use chi-square test to evaluate

differences among population when the test variable is

nominal, dichotomous, ordinal, or grouped interval.

Chi-square test

Independence Defined

Two variables are independent if, for all cases, the

classification of a case into a particular category of one

variable (the group variable) has no effect on the

probability that the case will fall into any particular

category of the second variable (the test variable).

When two variables are independent, there is no

relationship between them. We would expect that the

frequency breakdowns of the test variable to be similar

for all groups.

Independence Demonstrated

Suppose we are interested in the relationship between gender and attending college.

If there is no relationship between gender and attending college and 40% of our total sample attend college, we would expect 40% of the males in our sample to attend college and 40% of the females to attend college.

If there is a relationship between gender and attending college, we would expect a higher proportion of one group to attend college than the other group, e.g. 60% to 20%.

Displaying Independent and Dependent Relationships

Independent Relationship between Gender and College

40% 40% 40%

0%

20%

40%

60%

80%

100%

Males Females TotalPo

po

rtio

n A

tte

nd

ing

Co

lleg

e

Dependent Relationship between Gender and College

60%

20%

40%

0%

20%

40%

60%

80%

100%

Males Females TotalPo

po

rtio

n A

tte

nd

ing

Co

lleg

e

When the variables are independent, the proportion in both groups is close to the same size as the proportion for the total sample.

When group membership makes a difference, the dependent relationship is indicated by one group having a higher proportion than the proportion for the total sample.

Independent and Dependent Variables

The two variables in a chi-square test of independence each play a specific role. The group variable is also known as the independent

variable because it has an influence on the test variable.

The test variable is also known as the dependent variable because its value is believed to be dependent on the value of the group variable.

The chi-square test of independence is a test of the influence or impact that a subject’s value on one variable has on the same subject’s value for a second variable.

Chi square distribution

E

EO 22 )(

Expected frequency

observed frequency

Expected frequency are

computed as if there is no

difference between the groups,

i.e. both groups have the same

proportion.

This formula compute how the pattern of observed frequency differs from the pattern of expected frequency.

2. Chi-square distributions are determined by degree of freedom

Chi square distribution

1. Chi-square distribution is a nonsymmetrical distribution

Chi square test statistic Cannot be negative because all discrepancies are

squared.

Will be zero only in the unusual event that each

observed frequency exactly equals the corresponding

expected frequency.

Larger the discrepancy between the expected

frequencies and their corresponding observed

frequencies, the larger the observed value of chi-square.

Table 2.1 Partial Table of Critical Values of Chi-Square

Probability for chi square test statistic can be obtained

from the chi-square probability distribution.

0.05

reject region

The decision rule

The quantity will be small if the observed and

expected frequency are close together and will be large if

the differences are large.

The computed value of χ2 is compared with the tabulated

value of with K-1 degrees of freedom. The decision rule,

then is: reject H0 if χ2 is greater than or equal to the

tabulated χ2 for the chosen value of α.

E

EO 2)(

ⅡⅡ Chi-Square test Chi-Square test

(tests of goodness-of-fittests of goodness-of-fit)

E

EO 22 )(

Model assumptions: No

cell has an expected

frequency less than 5.

E

EO 22 )5.0(

At least one cell has an

expected frequency less

than 5.

Degrees of Freedom: k - 1

Number of outcomesNumber of outcomes

tests of goodness-of-fittests of goodness-of-fit

Example 1

As personnel director, you want to test the perception of

fairness of three methods of performance evaluation. Of 180

employees, 63 rated Method 1 as fair. 45 rated Method 2 as

fair. 72 rated Method 3 as fair. At the 0.05 level, is there a

difference in perceptions?


H0: p1 = p2 = p3 = 1/3

H1: At least 1 is different

a = 0.05


3.6

60

6072

60

6045

60

6063

O

6031180

222

cells all i

2i2

321

E

E

EEE

i

3.6

60

6072

60

6045

60

6063

O

6031180

222

cells all i

2i2

321

E

E

EEE

i

Reject Reject H0 at a = 0.05 at a = 0.05. There is evidence of a difference in There is evidence of a difference in

proportions proportions

Exercise 1

Ask 100 People (n) Which of 3 Candidates (k) They Will

Vote For. At the 0.05 level, is there a difference in

candidates?

Candidate

Tom Bill Mary Total

35 20 45 100

Candidate

Tom Bill Mary Total

35 20 45 100


Ⅲ Ⅲ Chi-Square test Chi-Square test

(tests of independence or relationshiptests of independence or relationship)

1. hypothesis test for 2hypothesis test for 2××2table2table

n≥40 and E≥5

n≥40 and 1≤ E < 5

n<40 or E<1

E

EO 22 )(

n≥40 and E≥5

))()()((

)( 22

dbcadcba

nbcad


Pearson chi- square

E

EO 22 )5.0|(|

n ≥ 40 and 1≤E<5

))()()((

)2/|(| 22

dbcadcba

nnbcad


Continuity correction of chi- square

!!!!!

)!()!()!()!(

ndcba

dbcadcbaP

n<40 or E<1


Fisher’ exact test

Example 2

A sample of 200 college students participated in

a study designed to evaluate the level of college

students’ knowledge of a certain group of

common diseases. The following table shows

the students classified by major field of study

and level of knowledge of the group of diseases:


major good poor total

premedical 16 24 40

other 20 140 160

total 36 164 200

Do these data suggest that there is a relationship between

knowledge of the group of diseases and major field of study

of the college students from which the present sample was

drawn? Let α=0.05.


major good poor total

premedical a b R1

other c d R2

total C1 C2 n

Four cells four-fold table

16 24

20 140

131.228.8

32.87.2

Observed cells

Expected cells

200

164160;

200

36160200

16440;

200

3640

2221

1211

EE

EE

H0: there is no relationship (independent) between knowledge and

major field

H1: there is a relationship between knowledge and major field

a = 0.05

396.162.131

2.131140

8.28

8.2820

8.32

8.3224

2.7

2.716

)(

2222

22

）（）（）（）（E

EO

131.228.8

32.87.2

14020

2416

396.16

)1643616040/(200202414016

))()()((

)(

2

22

）（

dbcadcba

nbcad

1. Chi-Square test for 2Chi-Square test for 2××2table2table

df=(R-1)(C-1)=1

84.31,05.02

Reject HReject H00 at a=0 .05 at a=0 .05

There is relationship between knowledge of the group of

diseases and major field of study of the college students.

The students major in premedical has higher knowledge

rates of diseases.

1. Chi-Square test for 2Chi-Square test for 2××2table2table

Exercise 2

A study was conducted to determine whether the

antibody status in wives is related with antibody

status in their husband. 48 couples were examined, the

data regarding the incidence of anti- sperm antibodies

is as follows:

Ab of wife Ab of husband

- + Total

- 8 10 18

+ 4 23 27

total 12 33 45

Question: Is the antibody status in wives

related with antibody status in their husband?

H0: the antibody status in wives is related with antibody

status in their husband

H1: the antibody status in wives is not related with antibody

status in their husband

a = 0.05

452.3))()()((

)2/( 22

dbcadcba

nnbcad

Not reject H0, we can not think the antibody status in

wives is related with antibody status in their husband.

2. hypothesis test for R×C tablehypothesis test for R×C table

E

EO 22 )( )1(

22

CRnn

On

Model assumptions :

The expected frequency should be greater than 5 in

more than 4/5 cells;

The expected frequency in any cell should be greater

than 1.

Pearson chi- square for R×C R×C table table

Example3 To study menstrual dysfunction in distance runners. Somebody did an observational study of three groups of women. The first two groups were volunteers who regularly engaged in some form of running, and the third, a control group, consisted of women who did not run but were otherwise similar to the other two groups. The runners were divided into joggers who jog "slow and easy" 5 to 30 miles per week, and runners who run more than 30 miles per week and combine long, slow distance with speed work. The investigators used a survey to show that the three groups were similar in the amount of physical activity (aside from running), distribution of ages, heights, occupations, and type of birth control methods being used.

Are these data consistent with the hypothesis that running

does not increase the likelihood that a woman will consult

her physician for a menstrual problem?

Table 5-6 shows these expected frequencies, together with the expected frequencies of women who did not consult their physicians.

58.22165

546911

E

627.9...42.31

)42.3140(

58.22

)58.2214( 222

2)12)(13()1)(1( cr

H0: π1 = π2 = π3

H1: At least 1 is different from the other

a = 0.05

99.52,05.02

Reject H0 at 0.05 level, so we can think that running increases

the likelihood a woman will consult her physician for a

menstrual problem.

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