Test of-significance : Z test , Chi square test

Test of significance and qualitative tests

1. Tests of significance Qualitative data Z & Chi square Test Dr. Shaikh B.M. JRII Dept. of Community Medicine

2. outline Basic terms (recap)………. Sampling Variation , Null hypothesis, Level of significance and confidence, P value, power of the test, degree of freedom Tests of significance and type Selection of the test of significance Steps in hypothesis testing Z test ( SEDP) Chi Square test Limitations of the tests of significance 2

3. Sampling Variation Research done on samples and not on populations. Variability of observations occur among different samples. This complicates whether the observed difference is due to biological or sampling variation from true variation. To conclude actual difference, we use tests of significance 3

4. Test of significance The test which is done for testing the research hypothesis against the null hypothesis. Why it is done? To assist administrations and clinicians in making decision. The difference is real ? or Has it happened by chance ? 4

5. Null Hypothesis (H O ) 1st step in testing any hypothesis. Set up such that it conveys a meaning that there exists no difference between the different samples. Eg : Null Hypothesis – The mean pulse rate among the two groups are same (or) there is no significant difference between their pulse rates. 5

6. By using various tests of significance we either: – Reject the Null Hypothesis ( or ) – Accept the Null Hypothesis Rejecting null hypothesis → difference is significant. Accepting null hypothesis → difference is not significant. 6

7. Level of significance and confidence Significance means the percentage risk to reject a null hypothesis when it is true and it is denoted by 𝛼. Generally taken as 1%, 5%, 10% ( 1 − 𝛼) is the confidence level in which the null hypothesis will exist when it is true. 7

8. Level of Significance – “P” Value   p -value  is a function of the observed sample results (a  statistic) that is used for  testing a statistical hypothesis. It is the probability of null hypothesis being true . It can accept or reject the null hypothesis based on P value. Practically, P < 0.05 (5%) is considered significant. 8

9. P = 0.05 implies, – We may go wrong 5 out of 100 times by rejecting null hypothesis . – Or , We can attribute significance with 95% confidence. 9

10. 5 % Significance level & 95% confidence level Acceptance and Rejection regions 𝑅𝑒𝑗𝑒𝑐𝑡𝑖 𝑜 𝑛 𝑟𝑒 𝑔 𝑖 𝑜 𝑛 / 𝑠𝑖𝑔𝑛 𝑖 𝑓 𝑖 𝑐𝑎𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙 (𝛼 = 0.025 𝑜 𝑟 2.5 % ) 𝑅𝑒𝑗𝑒𝑐𝑡𝑖 𝑜 𝑛 𝑟𝑒 𝑔 𝑖 𝑜 𝑛 / 𝑠𝑖𝑔𝑛𝑖𝑓 𝑖 𝑐𝑎 𝑛 𝑐𝑒 𝑙𝑒𝑣𝑒𝑙 (𝛼 = 0.025 𝑜𝑟 2.5%) 𝑇𝑜 𝑡 𝑎 𝑙 𝐴𝑐𝑐𝑒 𝑝 𝑡𝑎𝑛𝑐𝑒 𝑟 𝑒 𝑔 𝑖 𝑜𝑛 𝑜𝑟 𝑐𝑜𝑛 𝑓 𝑖𝑑𝑒𝑛 𝑐 𝑒 𝑙𝑒 𝑣𝑒𝑙 ( 1 − 𝛼 ) = 95% 10 10

11. Power of the test & Degree of freedom Power of the test : Type II error is   β and 1 − β is called power of the test. Probability of rejecting False H0, i. e. taking correct decision Degree of freedom: Number of independent observations used in statistics (d. f.) 11

12. Various tests of significance 1. Parametric – Data is or assumed to be normal distributed. 2. Non parametric – Data is not normal distributed. Parametric tests : For Qualitative data:- Z test 2. Chi-square test or X 2 For Quantitative data:- Unpaired ‘t’ test Paired ‘t’ test ANOVA 12

13. Selection of the test Base Type of data Size of sample Number of samples 13

14. Selection of the test 14

15. Steps in Testing a Hypothesis General procedure in testing a hypothesis Set up a null hypothesis ( H O ). Define alternative hypothesis (H A ) . Calculate the test statistic ( Z, X 2 , t etc.). Find out the corresponding probability level (P Value) for the calculated test statistic from relevant tables. Accept or reject the Null hypothesis depending on P value. P> 0.05 H O accepted P < 0.05 H O rejected 15

16. Z test [Standard error of difference between proportions (SE p1-p2 )] For comparing qualitative data between 2 groups. Used for large samples only. (> 30 in each group ). Standard error of difference between proportions (SE p1-p2 ) is calculated by using the formula 16

17.  And then the test statistic ‘Z Score’ value is calculated by using the formula: If Z > 1.96 P < 0.05 SIGNIFICANT - H0 rejected If Z < 1.96 P > 0.05 NOT SIGNIFICANT- H0 accepted Finding p value from Z table Go to the row that represents the ones digit and the first digit after the decimal point (the tenths digit) of your  z -value. Go to the column that represents the second digit after the decimal point (the hundredths digit) of your  z -value. Intersect the row and column from Steps 1 and 2. 1.0000 - 0.9772 =0.0228 < 0.05 17

18. Example Consider a hypothetical study where cure rate of Typhoid fever after treatment with Ciprofloxacin and Ceftriaxone were recorded to be 90% and 80% among 100 patients treated with each of the drug . How can we determine whether cure rate of Ciprofloxacin is better than Ceftriaxone? 18

19. Solution Step – 1: Set up a null hypothesis – H0: “ There is no significant difference between cure rates of Ciprofloxacin and Ceftriaxone .” Step – 2: Define alternative hypothesis – Ha : “Ciprofloxacin is 1.125 times better in curing typhoid fever than Ceftriaxone.” Step – 3: Calculate the test statistic – ‘Z Score’ 19

20. Step – 3: Calculating Z score Here , P1 = 90, P2 = 80 – SE P1-P2 will be given by the formula: So, Z = 10/5 = 2 20

21. Step – 4: Find out the corresponding P Value – Since Z = 2 i.e ., > 1.96, hence , P < 0.05 Step – 5: Accept or reject the Null hypothesis – Since P < 0.05, So we reject the null hypothesis(H0 ) There is no significant difference between cure rates of Ciprofloxacin and Ceftriaxone – And we accept the alternate hypothesis (Ha) that , Ciprofloxacin is 1.125 times better in curing typhoid fever than Ceftriaxone 21

22. Chi square ( X 2 ) test This test is also for testing qualitative data . Its advantage over Z test is: – Can be applied for smaller samples as well as for large samples . Prerequisites for Chi square ( X 2 ) test to be applied: – The sample must be a random sample – None of the observed values must be zero. – Adequate cell size 22

23. Steps in Calculating ( X 2 ) value Make a contingency table mentioning the frequencies in all cells. Determine the expected value (E) in each cell. Calculate the difference between observed and expected values in each cell (O-E). 23

24. 4. Calculate X 2 value for each cell 5. Sum up X 2 value of each cell to get X 2 value of the table . 6. Find out p value from x2 table 7. If p > 0.05 , difference is not significant , null hypothesis accepted; If p < 0.05 , difference is significant , null hypothesis rejected. 24

25. Example C onsider a study done in a hospital where cases of breast cancer were compared against controls from normal population against with a family history of Ca Breast. 100 in each group were studied for presence of family history. 25 of cases and 15 among controls had a positive family history. Comment on the significance of family history in breast cancer. 25

26. Solution From the numbers, it suggests that family history is 1.66 (25/15) times more common in Ca breast. So is it a risk factor in population? We need to test for the significance of this difference. We shall apply X 2 test . 26

27. Solution Step – 1: Set up a null hypothesis – H0: “There is no significant difference between incidence of family history among cases and controls .” Step – 2: Define alternative hypothesis – Ha: “Family history is 1.66 times more common in Ca breast” 27

28. Step – 3: Calculating X 2 Make a contingency table mentioning the frequencies in all cells 28

29. Step – 3: Cont.. 2. Determine the expected value (E) for each cell. 29

30. O – observed values , E – expected value 30

31. Step – 4: Determine degree of freedom. DoF is determined by the formula: DoF = (r-1) x (c-1) where r and c are the number of rows and columns respectively Here , r = c = 2. Hence , DoF = (2-1) x (2-1) = 1 31

32. Step – 5: Find out the corresponding P Value – P values can be derived by using the X 2 distribution tables 32

33. Step – 6: Accept or reject the Null hypothesis In given scenario , X 2 = 3.125 This is less than 3.84 (for P = 0.05 at dof = 1) Hence Null hypothesis is Accepted, i.e., “ There is no significant difference between incidence of family history among cases and controls” 33

34. Limitations of tests of significance Testing of hypothesis is not decision making itself; but it help s for decision making Test does not explain the reasons as why the difference exist s , tests do not tell about the reason causing the difference . Tests are based on the probabilities and as such can not be expressed with full certainty . Statistical inferences based on the significance tests can not be said to be entirely correct evidences concerning the truth of the hypothesis. 34

35. THANK YOU 35