CHI-SQUARE(CHI-SQUARE(XX22) ) DISTRIBUTIONDISTRIBUTION

Chi-Square TestChi-Square Test

CHI-SQUARE(CHI-SQUARE(XX22) DISTRIBUTION) DISTRIBUTION

• PROPERTIES:

1.It is one of the most widely used distribution in statistical applications

2.This distribution may be derived from normal distribution

3.This distribution assumes values from

( zero to + infinity)

CHI-SQUARE(CHI-SQUARE(XX22) DISTRIBUTION) DISTRIBUTION

4. X2 relates to frequencies of occurrence of individuals (or events) in the categories of one or more variables.

5. X2 test used to test the agreement between the observed frequencies with certain characteristics and the expected frequencies under certain hypothesis.

CHI-SQUARE(CHI-SQUARE(XX22) DISTRIBUTION) DISTRIBUTION

• CHI-SQUARE(X2) test of Goodness of fit

• CHI-SQUARE(X2) test of homogeneity

• CHI-SQUARE(X2) test of Independence

CHI-SQUARE(XCHI-SQUARE(X22) test of ) test of IndependenceIndependence

• It is used to test the null hypothesis that two criteria of classification when applied to the same set of entities are independent (NO ASSOCIATION)

CHI-SQUARE(XCHI-SQUARE(X22) test of ) test of IndependenceIndependence

• Generally , a single sample of size (n) can be drawn from a population, the frequency of occurrence of the entities are cross-classified on the basis of the two variables of interest( X &Y). The corresponding cells are formed by the intersections of the rows (r), and the columns (c).

The table is called the ‘contingency table’

CHI-SQUARE(XCHI-SQUARE(X22) test of ) test of IndependenceIndependence

• Calculation of expected frequency is based on the Probability Theory

• The hypotheses and conclusions are stated on in terms of the independence or lack of independence of the two variables.

CHI-SQUARE(XCHI-SQUARE(X22) test of ) test of IndependenceIndependence

• X2=∑(O-E)2/E

• df=(r-1)(c-1)

For 2x2 table, another formula to calculate X2

n(ad-bc)2

X2 =--------------------------------

(a+c)(b+d)(a+b)(c+d)

Steps in constructing XSteps in constructing X22 -test -test

1. Hypotheses

Ho: the 2 criteria are independent (no association)

HA: The 2 criteria are not independent (There is association)

2. Construct the contingency table

Steps in constructing XSteps in constructing X22 -test -test

3. Calculate the expected frequency for each cell

By multiplying the corresponding marginal totals of that cell, and divide it by the sample size

∑E = ∑O for each row or column

Steps in constructing XSteps in constructing X22 -test -test

4. Calculated the X2 value (calculated X2 c)

X2=∑(O-E)2/E X2=∑(O-E)2/E

For each cell we will calculate X2 value

X2 value for all the cells of the contingency table will be added together to find X2 c

Steps in constructing XSteps in constructing X22 -test -test

5. Define the critical value (tabulated X2)

This depends on alpha level of significance and degree of freedom The value will be determined from X2 table

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

r: no. of row

c: no. of column

Steps in constructing XSteps in constructing X22 -test -test

6. Conclusion

If the X2 c is less than X2 tab we accept Ho.

If the X2 c is more than X2 tab we reject Ho.

Observed frequencies in a fourfold Observed frequencies in a fourfold tabletable

Y1 Y2 Total

row total

X1 a b a+b

X2 c d c+d

Total

column total

a+c b+d N=a+b+c+d

For r X c table XFor r X c table X22 –test is not –test is not applicable if:applicable if:

1. The expected frequency of any cell is <1

2. The expected frequencies of 20% of the cells is < 5

For 2 X 2 table XFor 2 X 2 table X22 –test is not –test is not applicable if:applicable if:

The expected frequency of any cell is <5

EXERCISEEXERCISE

• A group of 350 adults who participated in a health survey were asked whether or not they were on a diet. The response by sex are as follows

EXERCISEEXERCISE

male female Total

On diet 14 25 39

Not on diet 159 152 311

Total 173 177 350

EXERCISEEXERCISE

• At alpha =0.05 do these data suggest an association between sex and being on diet?

ANSWERANSWER

1. Ho: Being on diet and sex are independent ( no association)

HA: Being on diet and sex are not independent ( there is association)

2. Calculation of expected frequencies

173 x 39

Cell a =-------------=19.3

350

177 x 39

Cell b=--------------=19.7

350

2. Calculation of expected frequencies

173 x 311

Cell c =-------------=153.7

350

177 x 311

Cell d=--------------=157.3

350

Observed and (expected) Observed and (expected) frequenciesfrequencies

male female Total

On diet 14

(19.3)

25

(19.7)

39

Not on diet 159

(153.7)

152

(157.3)

311

Total 173 177 350

ANSWERANSWER

3. Calculate X2 :

X2=∑(O-E)2/E (14-19.3)2 (25-19.7)2 (159-153.7)2 (152-157.3)2

=-----------+-----------+--------------+-------------

19.3 19.7 153.7 157.3

=1.455+1.426+0.183+0.17

X2c =3.243

ANSWERANSWER

4. Find X2 tab

df= (r-1) (c-1)= (2-1)(2-1)=1

X20.95 df=1=3.841

ANSWERANSWER

5. Conclusion

Since X2 c < X2 tab we accept Ho ( No association between sex and being on diet)

Another solution Another solution

• Since this a 2x2 table we can use this formula:

n(ad-bc)2

X2 =--------------------------------

(a+c)(b+d)(a+b)(c+d)

350{(14 x 152)-(25 x 159)}2

=------------------------------------- =3.22

39 x 311 x 173 x 177

(Example)(Example)

Five hundred elementary school Five hundred elementary school children were cross classified by children were cross classified by

socioeconomic group and the socioeconomic group and the presence or absence of a certain presence or absence of a certain speech defect. The result were as speech defect. The result were as

followsfollows

Speech defect

Socioeconomic GroupUpper Upper

middleLower Middle

Lower Total

Present 8(9.1)

24(26.4)

32(30.9)

27(24.6)

91

Absent 42(40.9)

121(118.6)

138(139.1)

108(110.4)

409

Total 50 145 170 135 500

• Are these data compatible with the hypothesis that the speech defect is unrelated to socioeconomic status?

• 1) Ho :Speech defect and SE group are independent ( no Association)

• HA: Speech defect and SE group are not independent ( Association exist)

• 2)Calculate the expected frequencies• 3)Calculate the X2 value ( calculated value)

• X² = ∑ (0 –E)² / E

• X² = ∑ (8 – 9.1)² /9.1 + (24 – 26.4)²/26.4 + (32 – 30.9)² /30.9 + (27-24.6)² /34.6 + (121 – 118.6)²/118.6 + (138 -139.1)²/139.1 + (108 – 110.4)²/110.4

• X²=0.5

• Tab X²

• DF = (2-1) (4-1) =3 → X²0.95 = 7.815

((Example 2)Example 2)• Five hundred employees of a factory that

manufacture a product suspected of being associated with respiratory

disorders were cross classified by level of exposure to the product and weather or

not they exhibited symptoms of respiratory disorders. The results are

shown in following table:

SymptomLevel of exposure

High Limited No exposure

Total

Present 185(143.4)

33(49.8)

17(41.8)

235

Absent 120(161.6)

73(56.2)

72(47.2)

265

Total 305 106 89 500

• Do these data provide sufficient evidence, at the 0.01 level of significance to indicate a relationship between level of exposure and the presence of respiratory and the presence of respiratory disorder ?

• 1) Ho : The presence of respiratory symptoms and the level of exposure are independent.

• HA : The two criteria are not independent • 2)Calculate the expected frequencies • 3) Calculate the X2

• X² =∑ (185 – 143.4)²/143.4 + (33 – 49.8)²/49.8 + (17-41.8)²/41.8 + (120-161.6)²/161.6 + (73 -56.2)² /56.2 + (72-47.2)²/47.2 = 33.47

Tab X² 0.99 = 9.21 Reject Ho• Df = (3-1) (2-1) = 2

(Example 3)(Example 3)• In a clinical trial involving a potential

hypothesis drug, patients are assigned at random either to receive the active drug or placebo. The trial is double blind, that is

neither the patient nor the examining physician knows with of the 2 treatment the

patient is receiving. Patients response to treatment is categorized as favorable or unfavorable on the basis of degree and

duration of response in BP. There are 50 patients assigned to each group.

Out comeTreatment

Drug Placebo Total

Favorable 34 9 43

Unfavorable 16 41 57

Total 50 50 100

• X² = n (ad – bc)²/(a+b)(c+d)(a+c)(b+d)

• = 100[(34x41) – (9x16)]²/(50) (50)(43)(57)

• =25.5

(Example 4)(Example 4)

• A study found that mongolism in babies is associated with hepatitis A

injection of the mother during pregnancy. Suppose a study of 2000

randomly selected mothers to be yielded the following table after the

births of their babies.

Hepatitis A.Baby

Mongoloid Non- Mongoloid

Total

+ 26 34 60

- 4 1936 1940

Total 30 1970 2000