T test

Paired and Unpaired t test

1. t-test 1) Unpaired/ Independent t- test 2) Paired t-test Jagdish D. Powar Statistician cum Tutor Community Medicine SMBT, IMSRC, Nashik JDP-CM-SMBT 1

2. JDP-CM-SMBT 2 Competency SLOs(Core) CM6.3, Describe, discuss and demonstrate the application of elementary statistical methods including test of significance in various study designs   The student should be able to Write the null and alternative hypothesis for independent t-test and paired t-test Test if two sample means are significantly different or not for small sample sizes To compare difference between paired observations To test whether intervention or treatment is effective or not for paired data. Competency & Learning objectives

3. Unpaired/Independent t-test:- Let x̅ 1 and x ̅2 be the two sample means of two independent random samples of sizes n 1 and n 2 drawn from two normal populations having mean µ 1 and µ 2 . To test the whether two population mean are equal (n 1 <30, n 2 <30) Null Hypothesis Ho:- There is no significant difference between mean of two populations i.e. µ 1 =µ 2 H1 :- There is significant difference between mean of two populations i.e. µ 1 ≠µ 2 Test statistics t = with df= (n1-1)+(n2-1) SE of (x̅ 1 -x̅ 2 ) = Find t-table value for (n 1 -1)+(n 2 -1 ) df Decision rule, if I t I< tab t then accept Ho otherwise reject it.   JDP-CM-SMBT 3

4. JDP-CM-SMBT 4

5. A group of 7 patients treated with medicine ‘ A’ had mean weight found to be 56 kg with SD of 8.22 kg. Another group of 9 patients from the same ward of a hospital had mean weight 45 kg with SD of 8.56. Do you claim that the medicine increases the weight significantly? Ans- Given values n1=7 = 56kg SD1= 8.22kg n2 =9 =45kg SD2= 8.56kg Ho:- There is no significant difference between mean weight of two groups H1 :- There is significant difference between mean weight of two groups i.e. the medicine increases the weight significantly Test statistics t= SE of (x̅1-x̅2) = = =4.25   JDP-CM-SMBT 5

6. t = = =2.59 t cal = 2.59 df= (7-1)+(9-1)= 14, 5% t tab =2.145 t cal < t tab hence Reject H Conclusion There is significant difference between mean weight of two groups i.e. the medicine increases the weight significantly   JDP-CM-SMBT 6

7. Paired t-test Paired observation for same individuals( n1=n2=n) Used to compare the effect of two drugs, given to same individuals. If there are n number of paired observation in the data set (x= before, y=after) To test whether there is no difference between the mean of before and after the observations. N ull hypothesis, Ho: D̅=0 i.e. there is no significant difference between the means before and after the observations. (Treatment is not effective in case of drug). Alternative Hypothesis, H1: D̅ ‡ 0( treatment is effective) JDP-CM-SMBT 7

8. Test Procedure:- Let D =( x -y) be the difference between each set of paired observation before and after the experiment. Calculate mean of the difference D̅ = Calculate SD of difference SD= Test statistics, t= , ) = df =n-1. Decision rule, if ι t ι < tab t then accept Ho otherwise reject it.   JDP-CM-SMBT 8

9. 2) Systolic blood pressure of 6 hypertensive patients were 183, 179,165, 190,175 and 180 mm of Hg. After administration of a particular drug for 1 week the BP’s were 185, 175, 150, 180, 172 and 170 mm of Hg respectively. Test whether drug is effective or not. Ans:- Null hypothesis, Ho : D̅=0 i.e. Drug is not effective. H1: Drug is effective Let us calculate D and SD D ̅ = =40/5 =6.67 SD = = =6.12 Test statistics, t = ,   X=Before Y=After   D= X-Y   183 185 -2 75.17 179 175 4 7.13 165 150 15 69.39 190 180 10 11.09 175 172 3 13.47 180 170 10 11.09 ∑ 40 187.33 X=Before Y=After   D= X-Y 183 185 -2 75.17 179 175 4 7.13 165 150 15 69.39 190 180 10 11.09 175 172 3 13.47 180 170 10 11.09 ∑ 40 187.33 JDP-CM-SMBT 9

10. ) = = = 2.74 t= = = 2.43 t df=5,0.05 =2.571 Here t cal < t tab accept H Drug is effective.   JDP-CM-SMBT 10

11. Thank You JDP-CM-SMBT 11