Hypothesis Testing and Comparison of Two Populations

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 % Height Z 60 0 57 63 57 - 60 Z= -1.5 = .4332 10 / 25 .8664 = 86.8% x -  Z = s / n 63 - 60 Z= 1.5 = .4332 10 / 25

4. 7.2b % 58 -1.0 60 0 Height Z x -  58 - 60 Z = Z = -1.0 = .5000 - .3413 s / n 10 / 25 .1587 = 16%

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

6. Hypothesis Testing Hypothesis: A statement of belief… Null Hypothesis, H0: …there is no difference between the population mean  and the hypothesized value 0. Alternative Hypothesis, Ha: …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 Truth H0 True H0 False a b Type II Error Correct results Accept H0  Test result 1 -  c d Type I Error Correct results Reject H0  1 -  Differences H0 True = statistically insignificant H0 False = statistically significant Accept H0 = statistically insignificant Reject H0 = statistically significant http://en.wikipedia.org/wiki/False_positive

8.  = 0.05 0.025 0.025 Probability Distribution for a two-tailed test 1.96 SE -3 -2 -1 1 2 3 0 SE SE Magnitude of (XE – XC) XE < XC XE > XC

9. Probability Distribution for a one-tailed test  = 0.05 1.645 SE -3 -2 -1 0 1 2 3 SE SE Magnitude of (XE – XC) XE < XC XE > XC

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

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

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

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 xA - xB - 0 t = [ ( 1/NA ) + ( 1/NB) ] Sp d f = N A + N B - 2

16. Sample A (A – Mean)2 26 34.34 24 14.90 18 4.58 17 9.86 18 4.58 20 .02 18 4.58 Mean = 20.14 A2 = 2913 N = 7 (A – Mean)2 = 72.86 Var = 12.14 s = 3.48 Sample B (B – Mean)2 38 113.85 26 1.77 24 11.09 24 11.09 30 7.13 22 28.41 Mean = 27.33 B2 = 4656 N = 6 (B – Mean)2 = 173.34 Var = 34.67 s = 5.89 Independent Sample Means

17. Standard error of the difference between the means (SED) Theoretical  A 2  B 2 SED of  E - C = + Population NB NA Estimate of the s A 2 s B 2 Sample SED of xE - xC = + NB NA

18. Pooled estimate of the SED (SEDp) Estimate of the 1 1 SEDp of xA - xB = Sp + NB NA s2(nA-1) + s2 (nB – 1) Sp = nA + n B - 2 +34.67(5) 12.14 (6) Sp = = 22.38 = 4.73 7 + 6 - 2

19. t-Test(Two Tailed) xA - xB - 0 t = [ ( 1/NA ) + ( 1/NB) ] Sp 20.14 - 27.33 - 0 = = -2.73 ( 1/7 ) + ( 1/6) 4.73 d f = N E + N C - 2 = 11 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. d =  D/N =  d 2 / N - 1 S d2  d 2 =  D 2 – ( D) 2 / N Pairedt-test d - 0 d - 0 = ------------- t paired = t p = S d2 Standard error of d N df = N - 1

22. Pre/post attitude assessment

23. = 3.7 / 41.5667 / 10 = 3.7 / 4. 15667 df = N – 1 = 9 0.05 > 1.833 Pre/post attitude assessment N = 10 Student Before After Difference D squared Total 171 208 37 511 d - 0 d - 0 = ------------- t paired = t p = S d2 Standard error of d N d =  D/N = 37/10 = 3.7  d 2 =  D 2 – ( D) 2 / N = 511 - 1369/10 = 374.1 = 3.7 / 2.0387 =  d 2 / N - 1 S d2 = 1.815 = 374.1 / 10 – 1 = 41.5667

24. Probabilities of Type I and Type II errors Truth H0 True H0 False Type II Error Correct results Accept H0  Test result 1 -  Type I Error Correct results Reject H0  1 -  Differences H0 True = statistically insignificant H0 False = statistically significant Accept H0 = statistically insignificant Reject H0 = statistically significant

25. Disease status Present Absent Total Present a b a + b Risk Factor Status Absent c d c + d a + c b + d a+b+c+d Total 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

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

27. Diabetic Screening Program Disease status Diabetic Nondiabetic Total >125mg/100ml 5 13 18 Risk Factor Status 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.) <125mg/100ml 1 81 82 6 94 100 Total