Chi_ Squre Test

8/8/2019 Chi_ Squre Test

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PG students: Dr Amit Gujarathi

Dr Naresh Gill


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History.

y Chi Square (chi, the Greek letter pronounced "kye)

test is a Nonparametric statistical technique used todetermine if a distribution of observed frequencies

differs from the theoretical expected frequencies.

y It was developed by Prof. Karl Pearson in 1900


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Introductiony The Relationship between two or more continuous

variables can be studied by correlation and regression

y

But in medical research to test the association betweentwo Discrete variables we use Chi-Square test.

y Such as to see association between a continuous

variable grouped into categories( Hb level: Mild ,Mod,

Severe anemia) and a discontinuous variables( Socio

economic status) or between two continuous variablesgrouped into categories (Hb Level: Mild ,Mod, Severe

anemia and No of ANC visits i.e. None,1-3 or .3).


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Application of Chi Square testy Test of proportions.

y Test of association.

y Goodness of Fit test.


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Test Of Proportion

yAs an alternative test to find the significance in two

or more than two proportions .

y For comparing values of two binomial samples evenif they are small, less than 30.(provided correction

factor, Yates correction is applied and expected

value is not less than 5 in any cell.)

y For comparing the frequencies of multinomialsamples.


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Test Of Association

y Most important application of the test.

y Can be used between two discrete events in

binomial or multinomial samples.y It measures the probability of association between

two discrete variables.

y Two possibilities:

y Either influence each other ( Dependent)

y Or not influencing each other( Independent) i.e. No

association is there.


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Contd..

yAssumption of independence is made i.e. Null

hypothesis, that there is no association between two

event.

y Thus Chi square test measures the probability(p)or relative frequency of association due to chance

and also if the two events are dependant on each

other or associated with each other.

yAdded advantage : can be used to find associationbetween two discrete variables when they are

categorized into more than two classes as happens

in multinomial samples.


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Test Of Goodness Of Fity Idea behind is that Chi square goodness of fit test

is to see the if samples comes from the Population

with the Claimed distribution.y This goodness of fit test is used to determine

whether population has certain hypothesized

distribution, expressed as a proportions of

individuals in the population falling into variousoutcome categories.


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Hypothesizes for Goodness of Fit Testy Suppose that hypothesized distribution has K outcome

categories.

y H0 = the actual population proportions are equal to thehypothesized proportions.

y Ha = the actual proportions are different from thehypothesized proportions

y First calculate the chi-square value which has X2

distribution at ( k-1) degree of freedom.y For test ofH0 against alternative hypothesis at least

two of the actual population proportions differ from theirhypothesized proportions.


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Chi - square distributiony Chi square distribution are family of distribution that take

only positive values and are skewed to right. A specific chi square distribution is specified by one parameter is calleddegree of freedom.


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Requirementsy Requirements :

yA random sample

y

Qualitative datay Lowest expected frequency in any cell should not be

less than 5 (Chi square distribution)

y If the smallest expected frequency is less than 5 thenFishers exact probability test should be used.

y Chi square should be calculated using thefrequencies only and not with rates, proportions orpercentages.

y Frequencies should be independent.


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y Sum the X2 values of all the cells to get the total chi square value.

y X2df indicates the total X2 value at particular degrees

of freedom.

y Calculates the degree of freedom.

df = (c-1) (r-1)

y Refer to Fishers X2 table .Compare the calculatedvalue with highest obtainable by chance at thedesired degree of freedom given in the table under

different probabilities such as 0.05.0.01,0.001 etc.y If calculated value of X2df is higher than the value

given in the table , then its significant at thatparticular level of significance.


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exampleyA student wants to see whether the food preferences of

males and females differed. He tried to see whether

males or femalesh

ad a general difference in th

epreference for cooked and raw foods. A survey wasconducted with the following results:

y Twelve males preferred Cooked foods.

y

Eight males preferred Raw foods.y Five females preferred Cooked foods.

y Five females preferred Raw foods.


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Step 1: State the null hypothesis

and the alternative hypothesis.y Ho: There is no significant difference between the food

preferences of males and females.

Ory Food preference is independent of gender.

y Ha: There is a significant difference between the food

preferences of males and females.Or

y Food preference is affected by gender.


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Step 2: State the level of

significance.y = 0.05

y 0.05 is the level of significance for most scientific

experiments


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Step 3: Set up a contingency tablePreference Male Female Total (row)

Cooked 12 5 17

Raw 8 5 13

Total(Column) 20 10 30


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Step 4: Compute for the expected frequencies

The chi-square test for independence usually uses the third method of gettingexpected frequencies.

Expected Frequency = R1X C

1

N

This expected frequency is computed for EACH cell.

Preference Male female Total (row)Cooked food (20) (17) /30

= 11.33

(10) (17) / 30

=5.67

17

Raw food (13) (20) / 30

=8.67

(13) (10) / 30

= 4.33

13

Total (

column)

20 10 30


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y Where O is the observed frequencies

E is the expected frequencies

And x2 is the chi-square value


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Step 5: Rearrange the table to show the observed and expected

frequen

cies on

the column

s, an

d the subcategories on

the rows.

Preference Observed expected Chi - square

Cooked food,Male

12 11.33 0.0396

Cooked food ,

female

5 5.67 0.0792

Raw food, male 8 8.67 0.0518

Raw food,

Female

5 4.33 0.1037

Total 0.2793


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Step 7: Check the tabular Chi-squared

value with your df and level of significance.

y Checking the table, we see that the tabular chi-

squared value for df = 1, and = 0.05 is 3.841.

y Since our calculated chi-square is less than this,

means there is the difference is not significant , its

due chance , null hypothesisisaccepted.

y Hence, food preference is independent of gender.

y If it were greater, we would reject the null hypothesis.


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y Example 2:

Attack rates among the vaccinated and unvaccinatedagainst measles . Protective value of Vaccination??

Group Results Total

Attacked Not-attacked

Vaccinated 10 90 100

Unvaccinated 26 74 100

Total 36 164 200


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Null Hypothesis(Ho) states that there is no difference in

attack rate between two groups.

While the alternate hypothesis(Ha) states that there issignificant difference in attack rates in two groups.

Group Total

Attacked Not-attacked

Vaccinated Observed 10 90 100

Expected 18 82

Unvaccinated Observed 26 74 100

Expected 18 82

Total 36 164 200


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After putting these values, Chi square value : 8.670

df = 1,

Table value of chi square at the df 1, and 5%significance level=3.84

Calculated value of Chi Square is greater than table

value, Hence its significant.

Null hypothesis is rejected and alternate hypothesis isaccepted.

-

!

E

EO2

2 )(G


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Test Of Goodness Of Fit: example

y Ques:In a sample of 100 persons, blood group

proportions as observed and expected are given

below. Find if the observed distribution fits the

hypothetical(expected) distribution.

Blood

group A

B AB O

Observed 23 35 5 37

Expected 42 9 3 46


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!'exp

exp)( 22 obs

There are 4 classes (k), Hence df=k-1, df=3

cesignificanoflevel5%at82.723 !'

Chi Square value calculated = 86.8

At the calculated value 86.8, P is far less than 0.001.

Hence , the value is highly significant.

The observed distribution does not fit to the

hypothetical distribution.


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Contd..

4. Interpret X2 test with caution if sample total or totalof values in all cells is less than 50.

5. X2 test tells the presence or absence of an

association between two events but does not

measures the strength of association.6. The statistical finding or relationship, does not

indicate the cause and effect .


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Alternate formulae for calculation

Chi - Square value

GroupResults(Measles)

TotalAttacked Not-attacked

Vaccinated a b a+b

Unvaccinated c d c+d

Total a+c b+d a+b+c+d

-

!

))()()((

)()( 22

dbcadcba

dcbabcadG


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Formulae with Yates correction:

-

!

))()()((NX)(

dbcdcbabcadG

-

! ))()()((

NXN/2)( 22

dbcadcba

bcad

G


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Documents

Chi_ Squre Test