*Chi-Square Test(one) Chapter 8

*Content test of fourfold data test of RC table Multiple comparison of sample rates test of paired fourfold data Fisher probabilities in fourfold data test of goodness of fit

objection to deduce if there is any discrimination of the ratio or structure ratio between two populations or among more than two populations multiple comparison of the ratio of multi-samples to deduce if there is any correlation between two class variables test of goodness of fit fit for :qualitative data

*Section one test of fourfold data

*objectiveto judge if there is any discrimination of the rate or structure ratio between two populations equal to the u-testdemandthe number of individuals from the two samples classified into two categories should be transformed into a fourfold data

* 1 distribution is a continuous distribution 2 one of the basic characters is that it can be plus to others 1 The basic idea of test distribution

3critical value of

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*2. The basic idea of test eg 8-1 one hospital want to compare the curative effect of drug AOmeprazole and drug B (Ranitidinein treating the peptic ulcer. They devided 169 patients with peptic ulcer into two groups at randomthe results are as follows (table 8-1). According to the data, whether the effective ratio is different?

*Table 8-1 the comparison of the healing ratio between two groups in healing peptic ulcer H0 1 = 2 H1 1 2

group

healing

Not healing

total

the ratio of healing%

Omeprazole

64(57.84) a

21(27.16) b

85(a+ b)

75.29

Ranitidine

51(57.16) c

33(26.84) d

84 (c+d)

60.71

total

115(a+c)

54(b+d)

169 (n)

68.05

*chart 8-2 the basic form of fourfold table data

treatment group

occur

not occur

total

first

a

b

a+b

second

c

d

c+d

total

a+c

b+d

n

*The respected frequencies can be calculated by the following formulaTRC refers to the expected frequencies in Row R and Line C nR refers to the frequency sum of a certain row nC refers to the frequency sum of a certain column

*Table 8-1 the comparison of the healing ratio between two groups in healing peptic ulcer

group

healing

Not healing

total

the ratio of healing%

Omeprazole

64(57.84) a

21(27.16) b

85(a+ b)

75.29

Ranitidine

51(57.16) c

33(26.84) d

84 (c+d)

60.71

total

115(a+c)

54(b+d)

169 (n)

68.05

*Basic idea can be showed with the basic formula of testA means actual frequencywhile T means theoretical frequency

*Basic idea of 2 testGiven a set of observed frequency distribution A1, A2, A3 to test whether the data follow certain theory.If the theory is true, then we will have a set of theoretical frequency distribution: T1, T2, T3 Comparing A1, A2, A3 and T1, T2, T3 If they are quite different, then the theory might not be true; Otherwise, the theory is acceptable.

* the respected frequency is set by the hypothesis and by the ratio after merging

* the test statistic :the value of reflects the fitness of actual frequency and expected frequency

* from formula 8-2,we can see that the value of also depends on the size of (exactly the size of ). is decided by the number of the grids, but not the sample size .

*1 establish hypothesis, and set the criteria of the testH0:1=2 H1:12 =0.05

3. The process of hypothesis test

*2to calculate the test statistic

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is smaller than 0.05according to the test criteria 0.05 ,we should reject

and accept

that is to say that the healing ratios of the two groups are different in curing the peptic ulcer the former is better than the latter .

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*4.The specific formula fourfold table data

* distribution is a continuous one, while the fourfold table data is dispersible, the value of calculated by the latter is also dispersible, in order to improve the continuousness of the statistic distribution ,the continuousness correcting is needed.

*

*5.The corrected formula

*the conditions in choosing test formula for the fourfold table data special formula

corrected formula

Fishier exact probabilities method

* eg 8-2 one doctor want to compare the effect of treatment A (simple chemotherapy) and treatment B (complex chemotherapy) in curing Lymphoid tumorhe divided 40 patients with such illness into two groups at random ,the results are in table 8-3, whether the remission rate of the two treatments are the same ?

* in this case, n=40, 1

* If not corrected then

the conclusion is on the contrary

*Section 2 -test of paired fourfold table

* Table 8-7 the result of the two testing methods Example 8-5

B method

total

80a

10b

90

31c

11d

42

total

111

21

132

*a, d are the agreement of the two methodsb, c are not agreement of the two methods Statistic:

If b+c

40,

,

If b+c

*Steps of the test

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look up the

critical value table

.

According to the level

,reject

accept

.there is difference between the two methods, the positive ratio of B method is higher than that of A method.

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*Section 3Chi square test for RC table

*RC tablemultiple rate comparingR 2 tabletwo constituent ratio comparing, 2C tablemultiple constituent ratio comparing, R C table

*genereal formula

* eg8-3 table 8-5 the cure rates of three treatments for chronic bronchitis

group

effective

ineffective

total

Effective rate(%)

Drug A

35

5

40

87.50

Drug B

20

10

30

66.67

Drug C

7

25

32

21.88

Total

62

40

102

60.78

*

the effective rates of three treaments are not all same

look up the

critical value table

.According to the level

,reject

accept

.there is difference between the three drugs.

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* eg8-4Table 8-6 the blood type distribution between children acute leukemia patients and adults acute leukemia

group

Type A

Type B

Type O

Type AB

Total

Children

30

38

32

12

112

Adults

19

30

19

9

77

total

49

68

51

21

189

*Testing steps

the constituent ratios of population between two groups are same

the constituent ratios of population between two groups are different

look up the

critical value table

.According to the level

,we can not reject

.we can not reach the conclusion that there is difference between the two group.

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*Caution: (1) Either 22 table or RC table are all called contingency table. 22 table is a special case of RC table (2) When R>2, H0 is rejectedonly means there is difference among some groups. Does not necessary mean that all the groups are different. (3) The 2 test requires large sample, By experience, The theoretical frequencies should be greater than 5 in more than 4/5 cells;The theoretical frequency in any cell should be greater than 1. Otherwise, we can not use chi-square test directly.

*

If the above requirements are violated, what should we do?(1) Increase the sample size.(2) Re-organize the categories, Pool some categories, or Cancel some categories

*Multiple comparison between multiple rates

*Bonferroni method

When the comparison among three groups results in significant difference, multiple comparison is needed to know which pairs are different.

We can use 22 four fold chi square test for two of them, but we have to adjust the testing level according to the comparing times in order to increase the probobility of I type error.

Formula for

K is the number of sample rates.

If we only compare the multiple experimental groupswith the control group, you can adjust the testing level by the formula below:

Formula for compute

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