 CHI-SQUARE(X2) DISTRIBUTION

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CHI-SQUARE(X2) DISTRIBUTION
Chi-Square Test

CHI-SQUARE(X2) 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(X2) 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(X2) DISTRIBUTION
CHI-SQUARE(X2) test of Goodness of fit CHI-SQUARE(X2) test of homogeneity CHI-SQUARE(X2) test of Independence

CHI-SQUARE(X2) test of Independence
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(X2) test of Independence
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(X2) test of Independence
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(X2) test of Independence
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 X2 -test
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 X2 -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 X2 -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 X2 -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 X2 -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 table
Y1 Y2 Total row total X1 a b a+b X2 c d c+d column total a+c b+d N=a+b+c+d

For r X c table X2 –test is not applicable if:
The expected frequency of any cell is <1 The expected frequencies of 20% of the cells is < 5

For 2 X 2 table X2 –test is not applicable if:
The expected frequency of any cell is <5

EXERCISE 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

EXERCISE male female Total On diet 14 25 39 Not on diet 159 152 311
173 177 350

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

ANSWER 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
Cell a = =19.3 350 177 x 39 Cell b= =19.7

2. Calculation of expected frequencies
Cell c = =153.7 350 177 x 311 Cell d= =157.3

Observed and (expected) frequencies
male female Total On diet 14 (19.3) 25 (19.7) 39 Not on diet 159 (153.7) 152 (157.3) 311 173 177 350

ANSWER 3. Calculate X2 : X2=∑(O-E)2/E
( )2 ( )2 ( )2 ( )2 = = X2c =3.243

ANSWER 4. Find X2 tab df= (r-1) (c-1)= (2-1)(2-1)=1 X20.95 df=1=3.841

ANSWER 5. Conclusion Since X2 c < X2 tab we accept Ho ( No association between sex and being on diet)

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) Five hundred elementary school children were cross classified by socioeconomic group and the presence or absence of a certain speech defect. The result were as follows

Speech defect Socioeconomic Group Upper Upper middle Lower Middle
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 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/ (32 – 30.9)² / ( )² / (121 – 118.6)²/ ( )²/ (108 – 110.4)²/110.4 X²=0.5 Tab X² DF = (2-1) (4-1) =3 → X²0.95 = 7.815

(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:

Level of exposure Symptom Present Absent 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) 265 305 106 89 500

Do these data provide sufficient evidence, at the 0
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
Tab X² 0.99 = Reject Ho Df = (3-1) (2-1) = 2

(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.

Treatment Out come Drug Placebo Total Favorable 34 9 43 Unfavorable 16
41 57 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) 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.

Baby Hepatitis A. what is the conclusion. Mongoloid Non- Mongoloid
Total + 26 34 60 - 4 1936 1940 30 1970 2000 what is the conclusion.