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Lecture 2

The analysis of cross-tabulations. Lecture 2. Cross-tabulations. Tables of countable entities or frequencies Made to analyze the association, relationship, or connection between two variables This association is difficult to describe statistically

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Lecture 2

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  1. The analysis of cross-tabulations Lecture 2

  2. Cross-tabulations • Tables of countable entities or frequencies • Made to analyze the association, relationship, or connection between two variables • This association is difficult to describe statistically • Null- Hypothesis: “There is no association between the two variables” can be tested • Analysis of cross-tabulations with larges samples

  3. Delivery and housing tenure

  4. Delivery and housing tenure • Expected number without any association between delivery and housing tenure

  5. Delivery and housing tenureIf the null-hypothesis is true • 899/1443 = 62.3% are house owners. • 62.3% of the Pre-terms should be house owners: 99*899/1443 = 61.7

  6. Delivery and housing tenureIf the null-hypothesis is true • 899/1443 = 62.3% are house owners. • 62.3% of the ‘Term’s should be house owners: 1344*899/1443 = 837.3

  7. Delivery and housing tenureIf the null-hypothesis is true • 258/1443 = 17.9% are council tenant. • 17.9% of the ‘preterm’s should be council tenant: 99*258/1443 = 17.7

  8. Delivery and housing tenureIf the null-hypothesis is true • In general

  9. Delivery and housing tenureIf the null-hypothesis is true • In general

  10. Delivery and housing tenureIf the null-hypothesis is true

  11. Delivery and housing tenuretest for association • If the numbers are large this will be chi-square distributed. • The degree of freedom is (r-1)(c-1) = 4 • From Table 13.3 there is a 1 - 5% probability that delivery and housing tenure is not associated

  12. Chi Squared Table

  13. Delivery and housing tenureIf the null-hypothesis is true • It is difficult to say anything about the nature of the association.

  14. 2 by 2 tables

  15. 2 by 2 tables

  16. Chi Squared Table

  17. Chi-squared test for small samples • Expected valued • > 80% >5 • All >1

  18. Chi-squared test for small samples • Expected valued • > 80% >5 • All >1

  19. Fisher’s exact test • An example

  20. Fisher’s exact test • Survivers: • a, b, c, d, e • Deaths: • f, g, h • Table 1 can be made in 5 ways • Table 2: 30 • Table 3: 30 • Table 4: 5 • 70 ways in total

  21. The properties of finding table 2 or a more extreme is: Fisher’s exact test • Survivers: • a, b, c, d, e • Deaths: • f, g, h • Table 1 can be made in 5 ways • Table 2: 30 • Table 3: 30 • Table 4: 5 • 70 ways in total

  22. Fisher’s exact test

  23. Yates’ correction for 2x2 • Yates correction:

  24. Chi Squared Table

  25. Yates’ correction for 2x2 • Table 13.7 • Fisher: p = 0.001455384362148 • ‘Two-sided’ p = 0.0029 • χ2: p = 0.001121814118023 • Yates’ p = 0.0037

  26. Odds and odds ratios • Odds, p is the probability of an event • Log odds / logit

  27. Odds • The probability of coughs in kids with history of bronchitis. p = 26/273 = 0.095 o = 26/247 = 0.105 The probability of coughs in kids with history without bronchitis. p = 44/1046 = 0.042 o = 44/1002 = 0.044

  28. Odds ratio • The odds ratio; the ratio of odds for experiencing coughs in kids with and kids without a history of bronchitis.

  29. Is the odds ratio different form 1? • We could take ln to the odds ratio. Is ln(or) different from zero? • 95% confidence (assumuing normailty)

  30. Confidence interval of the Odds ratio • ln (or) ± 1.96*SE(ln(or)) = 0.37 to 1.38 • Returning to the odds ratio itself: • e0.370 to e1.379 = 1.45 to 3.97 • The interval does not contain 1, indicating a statistically significant difference

  31. Chi-square for goodness of fit • df = 4-1-1 = 2

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