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Data Analysis Using R Introduction to the R language - Vietsciences文档
Descriptive analysisCategorical data * Comparison of two proportions - theory Group 1 2 ____________________________________________ Sample size n1 n2 Number of events e1 e2 Proportion of events p1 p2 Difference: D = p1 – p2 SE difference: SE = [p1(1–p1)/n1 + p2(1–p2)/n2]1/2 Z = D / SE 95% CI: D + 1.96(SE) With (n1 + n2) 20, and if Z 2, it is possible to reject the null hypothesis. * Comparison of two proportions - example Group Heroine Cocaine __________________________________________ Sample size 100 100 Number of deaths 90 36 Mortality rate 0.90 0.36 Thirty-day mortality rate (%) of 100 rats who had been exposed to heroine or cocain. Analysis Difference: D = 0.90 – 0.36 = 0.54 SE (D) = [0.9(0.1)/100 + 0.36(0.64)/100]1/2 = 0.057 Z = 0.54 / 0.057 = 9.54 95% CI: 0.54 + 1.96(0.057) 0.43 to 0.65 Conclusion: reject the null hypothesis. * Comparison of two proportions - R events - c(90, 36) total - c(100, 100) prop.test(events, total) 2-sample test for equality of proportions with continuity correction data: deaths out of total X-squared = 60.2531, df = 1, p-value = 8.341e-15 alternative hypothesis: two.sided 95 percent confidence interval: 0.4190584 0.6609416 sample estimates: prop 1 prop 2 0.90 0.36 * Comparison of 2 proportions – Chi square analysis table(sex, ethnicity) ethnicity sex African Asian Caucasian Others Female 4 43 22 0 Male 4 17 8 2 females - c(4, 43, 22, 0) total - c(8, 60, 30, 2) prop.test(females, total) * Comparison of 2 proportions – Chi square analysis 4-sample test for equality of proportions without continuity correction data: females out of total X-squared = 6.2646, df = 3, p-value = 0.09942 alternative hypothesis: two.sided sample estimates: prop 1 prop 2 prop 3 prop 4 0.5000000 0.7166667 0.7333333 0.0000000 Warning message: Chi-squared approximation may be incorrect in: prop
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