chi-square( x 2 ) distribution chi-square test. chi-square( x 2 ) distribution properties: 1.it is...
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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
• 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
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.
(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.