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Dr. Amjad El-Shanti MD, PMH,Dr PH University of Palestine 2016

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Presentation on theme: "Dr. Amjad El-Shanti MD, PMH,Dr PH University of Palestine 2016"— Presentation transcript:

1 Dr. Amjad El-Shanti MD, PMH,Dr PH University of Palestine 2016
Chi-Square test Dr. Amjad El-Shanti MD, PMH,Dr PH University of Palestine 2016

2 Chi square test Used to measure the association between two qualitative variables. Example: *the association between miscarriage and exposure for pesticide. *the association between vaccination and Influenza. *the association between hypertension and smoking

3 Questions Is there relation or association between miscarriage and pesticide exposure? Is there relation or association between vaccination and Influenza? Is there relation or association between hypertension and smoking?

4 Hypothesis Null Hypothesis: There is no statistically significant correlation between miscarriage and pesticide exposure. Alternative Hypothesis: There is statistically significant correlation between miscarriage and pesticide exposure.

5 Steps for Hypothesis testing
Construct observed table. Construct Expected table. E= C t * R t /Grand total 3.Calculate chi square (X2). X2 = Σ ( {O-E}2/ E). 4. d.f. = (C-1)*(R-1). P value (d.f., X2(. 6. Compare p value with significant level (α).

6 Example: If the sample is composed of 200 women 100 of them from village and exposed to pesticide. 30 women of village women had miscarriage while 10 of the city had miscarriage. Is there association between the pesticide exposure and miscarriage?

7 1.,2. Construct observed and expected table.
Total Not-exposed Exposed Observed 40 10 30 Miscarriage 160 90 70 No Miscarriage 200 100 E= C t * R t /Grand total Total Not-exposed Exposed Expected 40 40*100 /200 (20) 40*100 / 200 Miscarriage 160 160*100/200 (80) No Miscarriage 200 100

8 3. Calculate chi square (X2).
X2 = Σ ( {O-E}2/ E). X2 = ({30-20}2 /20) + ({10-20}2 /20)+ ({70-80}2 /80)+ ({90-80}2 /80) = 100/ / / /80 = = 12.5

9 Degree of freedom and compare p value with significant level
4. d.f. = (C-1)*(R-1). = (2-1) * (2-1). = 1 * 1= 1 p > p >α P value significant reject null hypothesis There is statistically significant association between pesticide exposure and miscarriage.


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