1. Hypothesis Testing and Comparison of Two Populations Dr. Burton

2. If the heights of male teenagers are normally distributed with a mean of 60 inches and standard deviation of 10, And the sample size was 25, what percentage of boy’s heights in inches would be: Between 57 and 63 Lass than 58 61 or larger

3. 7.2a 60 0 Height Z 5763 % Z = x -  s / n 57 - 60 10 / 25 63 - 60 10 / 25 Z= -1.5 =.4332 Z= 1.5 =.4332.8664 = 86.8%

4. 7.2b 60 0 Height Z 58 % Z = x -  s / n 58 - 60 10 / 25 Z = -1.0 =.5000 -.3413.1587 = 16%

5. 7.2c 60 0 0.5 Height Z 61 % Z = x -  s / n 61 - 60 10 / 25 Z = 0.50 -.1915 =.3085 = 30.9%= 0.50

6. Hypothesis Testing Hypothesis: A statement of belief… Null Hypothesis, H 0 : …there is no difference between the population mean  and the hypothesized value  0. Alternative Hypothesis, H a : …reject the null hypothesis and accept that there is a difference between the population mean  and the hypothesized value  0.

7. Probabilities of Type I and Type II errors H 0 TrueH 0 False Accept H 0 Reject H 0 Type I Error Type II Error Correct results Correct results Truth Test result  1 -   1 -  H 0 True = statistically insignificant H 0 False = statistically significant Accept H 0 = statistically insignificant Reject H 0 = statistically significant Differences ab cd http://en.wikipedia.org/wiki/False_positive

8. -3 -2 0 123 SE Probability Distribution for a two-tailed test SE Magnitude of (X E – X C ) 1.96 SE X E < X C X E > X C  = 0.05 0.025

9. -3 -2 0 123 SE Probability Distribution for a one-tailed test SE Magnitude of (X E – X C ) 1.645 SE X E < X C X E > X C  = 0.05

10. Box 10 - 5 t = A Distance between the means Variation around the means

11. Box 10 - 5 t = A B Distance between the means Variation around the means

12. Box 10 - 5 t = A B C Distance between the means Variation around the means

13. t-Tests Students t-test is used if: –two samples come from two different groups. –e.g. A group of students and a group of professors Paired t-test is used if: –two samples from the sample group. –e.g. a pre and post test on the same group of subjects.

14. One-Tailed vs. Two Tailed Tests The Key Question: “Am I interested in the deviation from the mean of the sample from the mean of the population in one or both directions.” If you want to determine whether one mean is significantly from the other, perform a two-tailed test. If you want to determine whether one mean is significantly larger, or significantly smaller, perform a one-tailed test.

15. t-Test (Two Tailed) Independent Sample means x A - x B - 0 t = Sp [ ( 1/N A ) + ( 1/N B ) ] d f = N A + N B - 2

16. Independent Sample Means Sample A(A – Mean) 2 2634.34 2414.90 184.58 179.86 184.58 20.02 184.58 Mean = 20.14  A 2 = 2913 N = 7 (A – Mean) 2 = 72.86 Var = 12.14 s = 3.48 Sample B(B – Mean) 2 38113.85 261.77 2411.09 307.13 2228.41 Mean = 27.33  B 2 = 4656 N = 6 (B – Mean) 2 = 173.34 Var = 34.67 s = 5.89

17. Standard error of the difference between the means (SED) SED of  E -  C = s A 2 Estimate of the s B 2 N AN A N BN B + SED of x E - x C =  A 2  B 2 N AN A N BN B + Theoretical Population Sample

18. Pooled estimate of the SED (SEDp) 1 Estimate of the 1 N AN A N BN B + SEDp of x A - x B = Sp s 2 (n A -1) + s 2 (n B – 1) Sp= n A + n B - 2 12.14 ( 6 ) + 34.67 ( 5 ) Sp= 7 + 6 - 2 = 22.38 = 4.73

19. t-Test (Two Tailed) d f = N E + N C - 2 = 11 x A - x B - 0 t = Sp [ ( 1/N A ) + ( 1/N B ) ] 20.14 - 27.33 - 0 = 4.73 ( 1/7 ) + ( 1/6) = -2.73 Critical Value 95% = 2.201

20. One-tailed and two-tailed t-tests A two-tailed test is generally recommended because differences in either direction need to be known.

21. Paired t-test t paired = t p = d - 0 Standard error of d = ------------- d - 0 S d 2 N df = N - 1 d =  D/N  d 2 =  D 2 – (  D) 2 / N S d 2 =  d 2 / N - 1

22. Pre/post attitude assessment StudentBeforeAfterDifferenceD squared 1252839 22319-416 33034 416 4710 39 536 39 62226 416 71213 11 83047 17289 9516 11121 10149-525 Total171208  D = 37  D 2 = 511

23. Pre/post attitude assessment StudentBeforeAfterDifference D squared Total171208 37 511 t paired = t p = d - 0 Standard error of d = ------------- d - 0 S d 2 N d =  D/N N = 10  d 2 =  D 2 – (  D) 2 / N S d 2 =  d 2 / N - 1 = 37/10 = 3.7 = 511 - 1369/10 = 374.1 = 374.1 / 10 – 1 = 41.5667 = 3.7 / 2.0387 = 1.815 = 3.7 / 41.5667 / 10 = 3.7 / 4. 15667 df = N – 1 = 9 0.05 > 1.833

24. Probabilities of Type I and Type II errors H 0 TrueH 0 False Accept H 0 Reject H 0 Type I Error Type II Error Correct results Correct results Truth Test result  1 -   1 -  H 0 True = statistically insignificant H 0 False = statistically significant Accept H 0 = statistically insignificant Reject H 0 = statistically significant Differences

25. Standard 2 X 2 table a = subjects with both the risk factor and the disease b = subjects with the risk factor but not the disease c = subjects with the disease but not the risk factor d = subjects with neither the risk factor nor the disease a + b = all subjects with the risk factor c + d = all subjects without the risk factor a + c = all subjects with the disease b + d = all subjects without the disease a + b + c + d = all study subjects PresentAbsent Present Absent Disease status Risk Factor Status ab cd a + b c + d a + c b + d a+b+c+d Total

26. Standard 2 X 2 table Sensitivity = a/a+c Specificity = d/b+d PresentAbsent Present Absent Disease status Risk Factor Status ab cd a + b c + d a + c b + d a+b+c+d Total

27. Diabetic Screening Program Sensitivity = a/a+c = 100 X 5/6 = 83.3% (16.7% false neg.) Specificity = d/b+d = 100 X 81/94 = 86.2%(13.8% false pos.) DiabeticNondiabetic >125mg/100ml <125mg/100ml Disease status Risk Factor Status 513 181 18 82 694100 Total