1 / 38

بسم الله الرحمن الرحيم

بسم الله الرحمن الرحيم. Hypothesis Testing. Two-samples tests, X 2. Dr. Mona Hassan Ahmed Prof. of Biostatistics HIPH, Alexandria University. Z-test (two independent proportions). P1= proportion in the first group P2= proportion in the second group n1= first sample size

lilia
Download Presentation

بسم الله الرحمن الرحيم

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. بسم الله الرحمن الرحيم

  2. Hypothesis Testing Two-samples tests, X2 Dr. Mona Hassan Ahmed Prof. of Biostatistics HIPH, Alexandria University

  3. Z-test(two independent proportions) P1= proportion in the first group P2= proportion in the second group n1= first sample size n2= second sample size

  4. Critical z = • 1.96 at 5% level of significance • 2.58 at 1% level of significance

  5. Example Researchers wished to know if urban and rural adult residents of a developing country differ with respect to prevalence of a certain eye disease. A survey revealed the following information Test at 5% level of significance, the difference in the prevalence of eye disease in the 2 groups

  6. P1 = 24/300 = 0.08 p2 = 15/500 = 0.03 Answer 2.87 > Z* The difference is statistically significant

  7. t-Test (two independent means) = mean of the first group = mean of the second group S2p = pooled variance

  8. Critical t from table is detected • at degree of freedom = n1+ n2 - 2 • level of significance 1% or 5%

  9. Example Sample of size 25 was selected from healthy population, their mean SBP =125 mm Hg with SD of 10 mm Hg . Another sample of size 17 was selected from the population of diabetics, their mean SBP was 132 mmHg, with SD of 12 mm Hg . Test whether there is a significant difference in mean SBP of diabetics and healthy individual at 1% level of significance

  10. Answer S1 = 12 S2 =11 State H0H0 : 1=2 State H1H1 : 12 Choose αα = 0.01

  11. Answer Critical t at df = 40 & 1% level of significance = 2.58 Decision: Since the computed t is smaller than critical t so there is no significant difference between mean SBP of healthy and diabetic samples at 1 %.

  12. Paired t- test(t- difference) Uses: To compare the means of two paired samples. Example, mean SBP before and after intake of drug.

  13. di = difference (after-before) Sd = standard deviation of difference n = sample size Critical t from table at df = n-1

  14. Example The following data represents the reading of SBP before and after administration of certain drug. Test whether the drug has an effect on SBP at 1% level of significance.

  15. Answer

  16. Answer

  17. Critical t at df = 6-1 = 5 and 1% level of significance = 4.032 Decision: Since t is < critical t so there is no significant difference between mean SBP before and after administration of drug at 1% Level. Answer

  18. Chi-Square test It tests the association between variables... The data is qualitative . It is performed mainly on frequencies. It determines whether the observed frequencies differ significantly from expected frequencies.

  19. Where E = expected frequency O = observed frequency

  20. Critical X2 at df = (R-1) ( C -1) Where R = raw C = column • I f 2 x 2 table X2*=3.84 at 5 % level of significance X2* = 6.63 at 1 % level of significance

  21. In a study to determine the effect of heredity in a certain disease, a sample of cases and controls was taken: Example Using 5% level of significance, test whether family history has an effect on disease

  22. Answer X2 = (80-88)2/88 + (120-112)2/112 + (140-132)2/132 + (160-168)2/168 = 2.165 < 3.84 Association between the disease and family history is not significant

  23. Odds Ratio (OR) • The odds ratio was developed to quantify exposure – disease relations using case-control data • Once you have selected cases and controls  ascertain exposure • Then, cross-tabulate data to form a 2-by-2 table of counts

  24. 2-by-2 Crosstab Notation • Disease status A+C = no. of cases B+D = no. of non-cases • Exposure status A+B = no. of exposed individuals C+D = no. of non-exposed individuals

  25. The Odds Ratio (OR) Cross-product ratio

  26. Example • Exposure variable = Smoking • Disease variable = Hypertension

  27. Interpretation of the Odds Ratio • Odds ratios are relative risk estimates • Relative risk are risk multipliers • The odds ratio of 9.3 implies 9.3× risk with exposure

  28. Interpretation Positive association Higher risk OR > 1 OR = 1 No association OR < 1 Negative association Lower risk (Protective)

  29. OR Confidence level • In the previous example OR = 9.3 • 95% CI is 1.20 – 72.14

  30. Multiple Levels of Exposure

  31. Multiple Levels of Exposure • k levels of exposure  break up data into (k – 1) 22 tables • Compare each exposure level to non-exposed • e.g., heavy smokers vs. non-smokers

  32. Multiple Levels of Exposure Notice the trend in OR (dose-response relationship)

  33. Small Sample Size Formula For the Odds Ratio It is recommend to add ½ to each cell before calculating the odds ratio when some cells are zeros

More Related