1. Statistical Fridays J C Horrow, MD, MS STAT Clinical Professor, Anesthesiology Drexel University College of Medicine

2. Previous Session Review Tests involve a NULL hypothesis (H 0 ) an ALTERNATIVE hypothesis (H A ) Try to disprove H 0 There are 4 steps in hypothesis testing –Identify the test statistic –State the null and alternative hypotheses –Identify the rejection region –State your conclusion

3. Session Outline Student’s t test. Frequency Data. Chi-square contingency tables.

4. Student’s t Test Normal distribution applies if  2 known When  2 unknown, can estimate by s 2 Bad news: (xbar-  )/(s/  n) not ~ N(0,1) Good news: (xbar-  )/(s/  n) ~ t(n-1) –(n-1) is called the “degrees of freedom

5. Do the patients in the C-section cohort have initial systolic BPs that are too low, i.e., less than 85 mmHg? STEP #1: Identify the T.S. T.S. = x-bar SBP-init Performing Student’s t Test

6. Student’s t Distribution  -3s -2s -s 0 s 2s 3s t=2.39 R.R.

7. Do the patients in the C-section cohort have initial systolic BPs that are too low, i.e., less than 85 mmHg? STEP #2: State the hypotheses: H 0 :   85H A :  < 85 Note: this is a “one-sided” test Worked Example

8. Do the patients in the C-section cohort have initial systolic BPs that are too low, i.e., less than 85 mmHg? STEP #3: Identify the rejection region R.R. = (x-bar SBP-init – 85)/(s/  n) < t.05;n-1 Worked Example R.R. = (80.25 – 85)/(5.877/  25) < -2.06

9. Do the patients in the C-section cohort have initial systolic BPs that are too low, i.e., less than 85 mmHg? STEP #4: State your conclusion R.R. -4.04 < -2.06  outside R.R. We reject H 0. Data are consistent with initial systolic BPs that are too low. Worked Example

10. Frequency Data Not continuous, but “counting” Example (could be continuous) –Instead of age in yearss –# patients with age > 75 years Example (can’t be continuous) –Number of pregnant patients

11. Frequency Data Can be expressed as a proportion –Example: 12 of 25 patients female Are distributed “binomially”: b(n,p) Approximated by Normal distribution Can derive (Obs-Exp) 2 /Exp ~  2 1

12. Chi-square Contingency Tables MaleFemaleTOTAL StrokeABA+B No stroke CDC+D TOTALA+CB+DA+B+C +D

13. Chi-square Contingency Tables MaleFemaleTOTAL Stroke6612 No stroke 543488 TOTAL6040100

14. H 0 : No effect of sex on stroke Expected MaleFemaleTOTAL Stroke12*60/ 100=7.2 12*40/ 100=4.8 12 No stroke88*60/ 100=52.8 88*40/ 100=35.2 88 TOTAL6040100

15. H 0 : No effect of sex on stroke (O-E) 2 /E MaleFemaleTOTAL Stroke0.2000.3000.500 No stroke0.0270.0410.068 TOTAL0.2270.3410.568

16. Chi-square Contingency Table Calculated T.S. = 0.568 Rejection Region =  2.05;1 > 3.84 Cannot reject H 0 (p=0.451)

17. Session Outline Student’s t test. Frequency Data. Chi-square contingency tables.

18. Session Homework Drug X induces cancer remission in 35 of 50 Asians and in 115 of 250 Caucasians. Does race affect the action of drug X? Use p<.05 for significance. Set-up using 4 steps of hypothesis testing.