Presentation on theme: "Biostatistics course Part 11 Comparison of two proportions Dr. Sc. Nicolas Padilla Raygoza Department of Nursing and Obstetrics Division of Health Sciences."— Presentation transcript:
Biostatistics course Part 11 Comparison of two proportions Dr. Sc. Nicolas Padilla Raygoza Department of Nursing and Obstetrics Division of Health Sciences and Engineering Campus Celaya-Salvatierra Universidad de Guanajuato Mexico
Biosketch Medical Doctor by University Autonomous of Guadalajara. Pediatrician by the Mexican Council of Certification on Pediatrics. Postgraduate Diploma on Epidemiology, London School of Hygiene and Tropical Medicine, University of London. Master Sciences with aim in Epidemiology, Atlantic International University. Doctorate Sciences with aim in Epidemiology, Atlantic International University. Associated Professor B, School of Nursing and Obstetrics of Celaya, university of Guanajuato. firstname.lastname@example.org
Competencies The reader will apply a Z test to obtain inferences from two independent proportions. He (she) will calculate confidence interval from two independent proportions.
Introduction Often, we make comparisons of two proportions from independent samples. In class we learned earlier to calculate confidence intervals and hypothesis test for a proportion; we can use the same methods to make inferences on proportions, if the sample size is large. For a large sample we can use a Normal approximation to the binomial distribution.
Examples In a study of urinary tract infection not complicated, patients were assigned to be treated with trimethoprim / sulfamethoxazole and fosfomycin / trometamol. 92 of 100 treated with fosfomycin / trometamol showed bacteriological cure while 61 of 100 treated with trimethoprim / sulfamethoxazole were cured infection.
Introduction When comparing proportions of independent samples, we must first calculate the difference in proportions. Analysis to compare two independent proportions is similar to that used for two independent means. We calculate a confidence interval and hypothesis test for difference in proportions.
Notation The notation we use for analysis of two proportions is the same as that for a proportion. The numbers below are for distinguishing the two groups. ParametersPopulation 1 2 Sample 1 2 Proportionπ1 π2p1 p2 Standard deviation √π1(1-π2) √π2(1-π2)√p1(1-p1) √p2(1-p2)
Inferences from two independent proportions The square of the standard error of a proportion is known as the variance of proportion. The variance of the difference between two independent proportions is equal to the sum of the variances of the proportions of each sample. The variances are summed because each sample contributes to sampling error in the distribution of differences.
Inferences from two independent proportions SE = √p(1-p)/n Variance = p(1-p)/n p1(1- p1) p2(1- p2) Variance (p1-p2) = variance of p1 + variance of p2 = --------- + ---------- n1 n2 The standard error of the difference between two proportions is given by the square root of the variances. SE (p1-p2) = √[p1(1-p1)/n1 + p2(1-p2)/n2]
Confidence intervals for two independent proportions To calculate the confidence interval we need to know the standard error of the difference between two proportions. The standard error of the difference between two proportions is the combination of the standard error of two independent distributions, ES (p1) and (p2). We estimated the magnitude of the difference of two proportions from the samples; now, calculate the confidence interval for this estimate.
The general formulae for confidence interval 95% is: Estimate ±1.96 x SE The formulae for IC 95% of two proportions should be: (p1-p2) ± 1.96 SE (p1-p2) Confidence intervals for two independent proportions
In the study of urinary tract infection, the proportion in the group of fosfomycin / trometamol was 0.92 and trimethoprim / sulfamethoxazole was 0.61 Difference in proportions = 0.92-0.61 = 0.31 ES = √ [(0.92 (1-0.92) / 100 + 0.61 (1-0.61) / 100] = 0056 The confidence interval at 95% would be: 0.31 ± 1.96 (0,056) = 0.31 ± 0.11 = 0.2 to 0.42 Confidence intervals for two independent proportions
The confidence interval at 95% would be: 0.31 ± 1.96 (0,056) = 0.31 ± 0.11 = 0.2 to 0.42 I have 95% confidence that the difference in the proportions in the population would be between 0.2 and 0.42. As the difference does not include 0, we are confident that the proportion of the population treated with fosfomycin / trometamol is different than with trimethoprim sulfamethoxazole. Confidence intervals for two independent proportions
Hypothesis test for two independent proportions A hypothesis test uses the difference and standard error of difference. However, we use a slightly different standard error to calculate the hypothesis test. This is because we are assessing the probability that the observed data assume that the null hypothesis is true. The null hypothesis is that there is no difference in the proportions of both samples and both groups have a common π.
The best estimate we can get from π is the common proportion, p of the two proportions of the sample. P = r1 + n2 + r2/n1+n2 Where: r1 and r2 are numbers of positive responses in each sample n1 and n2 are the sample sizes in each sample. Common proportion will be between two individual proportions. Hypothesis test for two independent proportions
The standard error can be calculated by replacing p by p1 and p2. SE (p1-p2) =√p(1-p)(1/n1 +1/n2) This is known as a pooled standard error. Hypothesis test for two independent proportions
Example In the study of urinary tract infection, the proportion in the group of fosfomycin / trometamol was 0.92 and trimethoprim / sulfamethoxazole was 0.61 100 integrants were in each group. Common p = 92 + 61/100 + 100 = 153/200 = 0.765 SE (p1-p2) = √ 0.77 (1-0.77) (1 / 100 +1 / 100) = √ 0.1771 x 0.002 = 0.019
Example Assuming a normal approximation to the binomial distribution, we calculate the Z test, as before. To calculate the hypothesis test, we must: 1.- Identify the null hypothesis Ho 2.- Identify the alternative hypothesis H1 3.- Calculate the hypothesis test Z.
Example Null hypothesis: when comparing two independent proportions of populations is usually the two proportions are equal. Ho: π1 = π2 It is as if the difference in the proportions of the two populations is 0. Ho: π1 - π2 = 0 Alternative hypothesis: is usually that the two proportions are not equal. H1: π1 ≠ π2 This is the same as the difference in proportions is not equal to zero. H1: π1 - π2 ≠ 0
Z statistic test The general formula for the Z test is the same as for the difference in two means. (p1-p2) – 0 z= -------------- SE (p1-p2) When the null hypothesis is that the difference in two proportions is zero estimate: (p1-p2) – 0 p1-p2 z= -------------- = -------- SE (p1-p2) SE (p1-p2)
Example 0.92 success for fosfomycin / trometamol and 0.61 for trimethoprim / sulfamethoxazole SE = 0.019 (p1-p2) – 0 0.31 - 0 z= -------------- = -----------= 16.32 SE (p1-p2) 0.019 P<0.05
Bibliografía 1.- Last JM. A dictionary of epidemiology. New York, 4ª ed. Oxford University Press, 2001:173. 2.- Kirkwood BR. Essentials of medical ststistics. Oxford, Blackwell Science, 1988: 1- 4. 3.- Altman DG. Practical statistics for medical research. Boca Ratón, Chapman & Hall/ CRC; 1991: 1-9.