Lecture 11 Paired t test.pptx

OBJECTIVES:
Run the test of hypothesis for mean difference using paired samples. Construct a confidence interval for the difference in population means using paired samples.
Observation of interest will be the difference in the readings
before and after intervention called paired difference observation.
Paired t test:
A paired t-test is used to compare two means where you have two samples in which observations in one sample can be paired with observations in the other sample.
Examples of where this might occur are:
Before-and-after observations on the same subjects (e.g. students’ test
results before and after a particular module or course).
A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a sphygmomanometer and a dynamap).
When there is a relationship between the groups, such as identical twins.
This test is concerned with the pair-wise differences
between sets of data.
This means that each data point in one group has a related data point in the other group (groups always have equal numbers).
ASSUMPTIONS:
The sample or samples are randomly selected
The sample data are dependent
The distribution of differences is approximately normally
distributed.
Note: The under root is onto the entire numerator and denominator, so you should take the root after solving it entirely
where “t” has (n-1) degrees of freedom and “n” is
the total number of pairs.

1. PAIRED T-TEST Shakir Rahman BScN , MScN , MSc Applied Psychology, PhD Nursing (Candidate) University of Minnesota USA. Principal & Assistant Professor Ayub International College of Nursing & AHS Peshawar Visiting Faculty Swabi College of Nursing & Health Sciences Swabi Nowshera College of Nursing & Health Sciences Nowshera

2. OBJ E C T I V ES By the end of this session the students should be able to: Run the test of hypothesis for mean difference using paired samples Construct a confidence interval for the difference in population means using paired samples.

3. T-test for one sample s n X   t  

4. T-test for two dependent sample (Paired sample)

5. POPULATION Sample Before I n ter v e n t i on Intervention Sample After Intervention Observation of interest will be the difference in the readings before and after intervention called paired difference observation

6. WHAT IS PAIRED T- TEST? A paired t-test is used to compare two means where you have two samples in which observations in one sample can be paired with observations in the other sample. Examples of where this might occur are: Before-and-after observations on the same subjects (e.g. students’ test results before and after a particular module or course). A comparison of two different methods of measurement or two different treatments where the measurements/treatments are applied to the same subjects (e.g. blood pressure measurements using a sphygmomanometer and a dynamap). When there is a relationship between the groups, such as identical twins.

7. 8 P A I R E D D I F F E R E N C E S A M P L E S “ T T E S T ” A N E X P L A N A T I O N This test is concerned with the pair-wise differences between sets of data. This means that each data point in one group has a related data point in the other group (groups always have equal numbers).

8. A SS U M P T I O NS The sample or samples are randomly selected The sample data are dependent The distribution of differences is approximately normally distributed.

9. 10 Reading 1 (Before intervention/Case) Reading 2 (After i n t e r ven t i o n / Ma t c h e d control) Means of Differences (X D) Difference d= X1- X2 D 2 OR (X1-X2) 2 X1 X2 X1+X2/2 D 1 =X1-X2 2 (X1- X2) X1 X2 X1+X2/2 D 2 =X1-X2 2 (X1-X2) X1 X2 X1+X2/2 D 3 =X 1 -X2 . . . X1+X2/2 . . . . . . X1 X2 D i =X1-X2 . . . . . . . . . X1 X2 D n =X1-X2 . ∑X1 ∑X2 Xd ∑d 2 ∑d What is a Paired Difference Observation?

10. 11 X D = Mean of Differences S D = Standard deviation of Differences n= number of pairs D = Difference between the population means Where D is mostly zero   S D    n  t  x D TEST STATISTIC AND ASSUMPTIONS FOR PAIRED DIFFERENCE METHOD - In this method, the test statistic becomes: D

11. FORMULA FOR STANDARD DEVIATION Sd =√ n∑d 2 –(∑d) 2 n (n-1) Note: The under root is onto the entire numerator and denominator, so you should take the root after solving it entirely

12. 13 (1-  )% CONFIDENCE INTERVAL FOR PAIRED MEAN DIFFERENCE (  D ) where “t” has (n-1) degrees of freedom and “n” is the total number of pairs. F o r m u l a : n S D D , df 2 x  t 

13. 14 HYPOTHESIS TEST FOR PAIRED MEAN DIFFERENCE (  D ) One-tailed Test: – H : µ D ≤ H a : µ D > 0 (Right tailed) – H o : μ D ≥ H a : μ D < 0 (Left tailed) Two-tailed Test: – H : µ D = H a : μ D  – Step Two: α = 0.05

14. Step 3 Test statistic: Sd = √ n∑d 2 –(∑d) 2 n (n-1) where “t” has (n-1) degrees of freedom (df) and “n” is the total number of pairs. Step 4: Critical Region: Reject Ho if: t cal > t tab OR t cal < - t tab Step 5: Conclusion   n  t  D  s D   x  D

15. EX A MPLE BLOOD SAMPLES FROM 10 PATIENTS WERE SENT TO EACH OF TWO LABS FOR CHOLESTEROL DETERMINATION. MEASUREMENTS WERE AS FOLLOWS: Participants Lab 1 Lab 2 1 296 318 2 268 287 3 244 260 4 272 279 5 240 245 6 244 249 7 282 294 8 254 271 9 244 262 10 262 285

16. Q U E S T I O N - Is there a statistically significant difference at α = .05 in the cholesterol levels reported by lab 1 and lab 2.

17. SO L U T ION Step1: H0: µ D = Ha: µ D ≠ Step 2: α= .05 d.f. = n-1, 10-1 =9 Critical value = + 2.26 Step 3: t = -14.4-0 = -6.73 6.77/√10 Sd = √ 1 x 2486-(- 144) 2 10x 9 = 6.77 -6.73 -2.26 2 . 2 6

18. SO L U T ION Step 4: Reject Ho if: t cal > t tab OR if t cal < - t tab Since in this case, t cal < -t tab = -6.73 < -2.26, so we reject Ho Step 5 (Conclusion): Reject Ho at 5 % level of significance, and we have sufficient evidence to conclude that results from the both labs are different than each other.

19. C O N F I D E N C E I N T E R V A L 14.4 + 3.25 X6.77/√10 I N T E R P R E T A T I O N : W E A R E 9 9 % C O N F I D E N T T H A T T H E M E A N D I F F E R E N C E W I L L F A L L B E T WE E N L O W E R A N D U P P E R L I M I TS 99% C.I n S D D , df 2 x  t 

20. Acknowledgements Dr Tazeen Saeed Ali RM, RM, BScN, MSc ( Epidemiology & Biostatistics), Ph.D. (Medical Sciences), Post Doctorate (Health Policy & Planning) Associate Dean School of Nursing & Midwifery The Aga Khan University Karachi. Kiran Ramzan Ali Lalani BScN, MSc Epidemiology & Biostatistics (Candidate) Registered Nurse (NICU) Aga Khan University Hospital