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## University of Jaffna Wk 3 Impact of Maternal Health on Child Survival Case Study

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Florida National University HSA-6752 Statistic in Health Care Management: Assignment Week 3 Case Study: Chapters 5 and 6. Objective: The students will complete a Case study assignments that give the occasion to create and apply the thoughts learned in this and previous project to examine a real-world scenario. This set-up will illustrate through example the practical importance and implications of various roles and functions of a Health Care Administrator in probability and interval Estimates. The investigative trainings will advance students’ understanding and ability to think critically about basic concepts of probability and introduction to estimation. ASSIGNMENT GUIDELINES (10%): Students will critically measure the readings from Chapters 5 and 6 in your textbook. This assignment is planned to help you examination, evaluation, and apply the readings and strategies to your Health Care organization. You need to read the chapters assigned for week 4 and develop a 3-4 page paper reproducing your understanding and capability to apply the readings to your Health Care organization. Each paper must be typewritten with 12-point font and double-spaced with standard margins. Follow APA style 7th edition format when referring to the selected articles and include a reference page. EACH PAPER SHOULD INCLUDE THE FOLLOWING: 1. Introduction (25%) Provide a brief synopsis of the meaning (not a description) of each Chapter you read, in your own words that will apply to the case study presented. 2. Your Critique (50%) Case Studies The Effect of Maternal Healthcare on the Probability of Child Survival in Azerbaijan Abstract This study assesses the effects of maternal healthcare on child survival by using nonrandomized data from a cross-sectional survey in Azerbaijan. Using 2SLS and simultaneous equation bivariate probit models, we estimate the effects of delivering in healthcare facility on probability of child survival taking into account self-selection into the treatment. For women who delivered at healthcare facilities, the probability of child survival increases by approximately 18%. Furthermore, if every woman had the opportunity to deliver in healthcare facility, then the probability of child survival in Azerbaijan as a whole would have increased by approximately 16%. 1.

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Introduction Poor child outcomes are usually associated with underutilization of maternal healthcare. Given unusually high mortality rates in countries of Central Asia and Caucasus, poor child outcomes and maternal healthcare should become important topics for research. Nevertheless, there are a very few studies on these topics in the region. The available studies can be divided into two broader groups. The first group explored determinants of child mortality. The second group explored determinants of maternal healthcare utilization. Although these studies have important contributions, their main limitation is that the most important question on whether healthcare has an effect on the reduction of child mortality is overlooked. However, designing and implementing effective health policy require concrete information on the effectiveness of the existing maternal healthcare. The contribution of the presented study is that it attempts to fill the gap in the existing literature by quantifying the direct effect of delivery in healthcare facility on probability of child survival. The robust evaluation of program effect on population usually involves randomized control trials (RCT). In many cases, including evaluation of maternal healthcare, conducting a RCT is not possible from an ethical perspective, withholding vital service, and from technical perspective, lack of money and time required to conduct a countrywide RCT. To overcome these difficulties, we assess the effect of healthcare and homecare on child survival by using quasiexperimental evaluation of nonrandomized data from a cross-sectional survey. In this way, this study contributes to the recent discussion on appropriate methods for the evaluation of effect of healthcare programs when RCT is not feasible. Azerbaijan, a low-income transitional country on Caucasus, is an interesting setting for examining the above-mentioned issues for several reasons. First, Azerbaijan has the highest infant mortality rate and one of the highest proportions of child deliveries outside of healthcare facilities even compared with other transitional countries in the region. Second, by studying Azerbaijan, we benefit from recently available 2006 Azerbaijan Demographics and Health Survey that contains high-quality nationally representative data on the issues of our interest. Third, there is a current theoretical debate on the actual effectiveness of maternal healthcare in transitional countries. On the one hand, maternal healthcare is universal, officially free of charge, fully funded, and operated by the government. It has an extensive network of facilities which is adequately staffed with qualified personnel. Hence, a fairly strong positive impact on child survival could be expected and some authors underscore the importance of maternal healthcare utilization in transitional countries to improve child outcomes. On the other hand, the system is characterized by chronic underfunding, lack of drugs and supplies, dilapidated facilities, lack of systematic and effective treatments, and high levels of unofficial out-of-pocket expenditures for personnel. Hence, no or only weak impact on the child survival could be expected. Therefore, by focusing on Azerbaijan, a transitional country, this study provides necessary empirical evidence which will contribute to the current theoretical debate on the effectiveness of maternal healthcare in transitional countries. 2. Materials and Methods 2.1. Conceptual Framework We are guided by Mosley and Chen’s framework for studying the determinants of child survival. According to the framework, socioeconomic determinants at individual (e.g., women’s education), household (e.g., household income), and community (e.g., healthcare input) levels affect a total of 14 proximate determinants of mortality which are grouped into several categories, namely, maternal factors, environmental contamination, nutrient deficiency, and personal illness control. However, the model has a few limitations for applied research. Some proximate determinants, for instance, environmental contamination, are notoriously difficult to define and measure adequately, especially in population-based surveys. Furthermore, if a model includes all socioeconomic and all proximate determinants, then the coefficients on the socioeconomic variables should not be statistically significant given that the proximate determinants will pick up all significance by definition. Consequently, we reduced the number of independent variables to women’s age at birth and education, birth order, low child birthweight, household wealth, and healthcare input. As a result, we used the reduced set of independent variables which is similar to previous studies on child survival in the region and international comparative studies. 2.2. Method We are interested in estimating effect of treatment, having child delivery at a healthcare facility, on the outcome, probability of child survival. Thus, we face a problem of self-selection—the sampled individuals who receive the treatment are different from those who do not receive it in unobservable ways which are also simultaneously correlated with outcome . To address the selfselection we use simultaneous equation regression that tackles the endogeneity by specifying and estimating a joint model of the treatment and outcome. Since both treatment and outcome variable in our case are binomial, we use a simultaneous equation bivariate probit, so-called biprobit. The model consists of first and main equations. In the first equation, a dummy treatment variable is regressed on all control variables and one or more instruments. In the main equation, a dummy outcome variable is regressed on all control variables and the value of the treatment variable estimated in the first stage. Importantly, the instruments are excluded from the main equation. This statistical specification is estimated using biprobit command in Stata software package. After biprobit was estimated, we compute the average treatment effect (ATE) and the average treatment effect on the treated (ATT). The value of the ATE indicates the expected mean effect of the treatment for a woman drawn at random from the population. By contrast, the value of ATT indicates the expected mean effect of the treatment for a woman who actually participates in the program and receives treatment. ATT permits us to evaluate the effect on women who received treatment and who can be considered as a more relevant subpopulation for the purposes of evaluating effect of a specific program. The full details of biprobit, ATE, and ATT computations can be found in Greene and Wooldridge . 2.3. Data This study uses data from the 2006 Azerbaijan Demographic and Health Survey (the AZDHS). The AZDHS is conducted by the national statistical authority, the State Statistical Committee of Azerbaijan, with technical assistance of Macro International, USA, and with financial support from USAID and UNICEF. The AZDHS is a cross-sectional survey of 8,444 women aged 15 to 49 from 7,180 households. Field work was conducted from July to November 2006. The household gross response rate exceeds 90 percent. The AZDHS gathered information on demographics, educational level, household wealth, healthcare utilization, and child mortality. The AZDHS collected information about the outcome of each respondent’s pregnancy for the period, whether the pregnancy ended in a live birth, a stillbirth, a miscarriage, or an induced abortion. The survey used the international definition of child mortality, under which any birth in which a child showed any sign of life such as breathing, beating of the heart, or movement of voluntary muscles is defined as a live birth. The AZDHS collected information on child mortality for births in 2001 or later, covering a period of 5 years before the date of the survey only. Among recorded 13,565 observations, about 92% of children survived between birth and their fifth birthday and about 8% died. However, our sample is further reduced since the questions about place of delivery asked only about the most recent birth delivered during the the last 5 years before the date of the survey. It means that if a women had multiple births during the last 5 years, the questions about place of delivery was asked only about the latest birth. Consequently, our final sample consists of 2,285 observations for analysis. 2.4. Outcome and Treatment Variables The outcome variable of this study is child survival defined as probability to survive during 60 months or 5 years. This variable is binomial; it takes the value of 1 if the child survives 60 months and takes the value of 0 if otherwise. There are two endogenous instrumented variables of interests which denote treatment and serve to gauge healthcare input. The instrumented treatment variable is “delivery in a healthcare facility” that takes the value of 1 if the child was delivered in a healthcare facility and takes the value of 0 if otherwise. The healthcare facility is defined as a government or private hospital, maternity home, polyclinic, woman’s consultation, and primary healthcare posts. Overall, from the sample of 2,285 women who answered the questions about place of delivery in the AZDHS, approximately 79% delivered babies in a healthcare facility. 2.5. Instrumental Variables The instrumental variables used to estimate the endogenous treatment variables are taken from a previous study that used instrumental variables to estimate the effect of prenatal healthcare utilization on child birthweight in Azerbaijan. There are two instrumental variables—“women from wealthier households” and “birth order.” The AZDHS contains a variable representing 5 quintiles of household wealth—poorest, poor, middle, richer, and richest. We create a “wealthier households” dummy variable which denotes women from richest and richer households, and this variable is used in our model 1 and model 2. Finally, “birth order” is a straightforward continuous variable denoting number of births. 2.6. Exogenous Variables The exogenous variables used to explain child survival are taken from the previous studies on the determinants of child mortality conducted in the countries of the region of Caucasus and Central Asia. We have two dummy variables representing women’s age: variable “age 20” indicates women aged 20 or younger at the time of delivery, while variable “age 36” indicates women aged 36 and older at the time of delivery. Dummy variable “low birthweight” indicates if a child’s birthweight was 2500 grams or lower. Dummy variable “higher education” indicates women with bachelor education or higher. Previous studies reported that having delivery at age 35 is associated with higher probability of child mortality. Likewise, previous studies reported that having low birthweight is associated with higher probability of child mortality, while having higher educational achievements is associated with lower probability of child mortality. 2.7. Estimation We commence with 2SLS model because the tests for overidentifying restrictions and the adequacy of the instruments are readily available for the 2SLS but not for biprobit [11, 12, 32]. Since the number of instrumental variables exceeds the number of endogenous variables in our case, the Hansen statistic is employed to evaluate overidentifying restrictions. If Hansen statistics cannot reject the null hypothesis, then the selected instrumental variables are exogenous. In addition, Kleibergen-Paap LM statistic is used to test the adequacy of the instruments. If the test rejects the null hypothesis, the instruments are adequate to identify the equation. Lastly, we conduct Durbin-Wu-Hausman test for potential endogeneity. The significance of the test confirms the presence of endogeneity and suggests that estimation of equations without taking into account endogeneity will lead to biased results. All the abovedescribed tests have been passed in all estimated models. Next we estimate biprobit which is more relevant model due to the binary nature of outcome and treatment variables. A straightforward Wald test of endogeneity is available in biprobit. If result of the test is significantly different from zero, then biprobit should be estimated due to the presence of endogeneity. In all estimated models, the Wald tests have confirmed endogeneity. After biprobit model estimation, ATE and ATT are computed and reported. 3. Results The results are reported in Table 1. In the first equation four variables are significant with predicted directions in 2SLS estimation. Having birth at the age of 20 or earlier and having a higher value of birth order are associated with lower probability of delivery in a healthcare facility, while having higher educational achievements and being from a wealthier household are associated with higher probability of delivery in a healthcare facility. Looking at the main equation in 2SLS, we can see that having a delivery in the facility improves the chances of child survival. Results of biprobit estimation are consistent with the results of the 2SLS estimation. The same four variables are significant in the first equation and with the same direction. 4. Discussion and Policy Implications In this study, we identified and then attempted to fill the important gap in the literature regarding the effectiveness of maternal healthcare in reducing under-five child mortality in the region of the Central Asia and the Caucasus. We assessed the effects of delivering in a healthcare facility on child survival by using a quasiexperimental evaluation based on nonrandomized data from a cross-sectional survey in Azerbaijan, a low-income country in transition. The empirical evidence presented in this paper allows for drawing several conclusions. First, delivering children in healthcare facilities increases the probability of survival. Since reducing child mortality is raison d’être for maternal healthcare programs, such a funding could be expected. However, we were able to confirm that the effect of delivering at a healthcare facility on child survival is statistically significant on the national level. We also quantified the positive effect of such treatment. For women who delivered at healthcare facilities the probability of child survival increases by approximately 18%. Furthermore, if every woman in Azerbaijan had the opportunity to deliver in a healthcare facility, then the probability of child survival in the country would have increased by approximately 16%. These findings suggest that utilization of maternal services in transitional countries should be encouraged and promoted in spite of the limitations and deficiencies in the current maternal healthcare system. Second, our study demonstrates that the wealth gradient is an important barrier for utilization and hence influences the child outcomes. Since maternal healthcare is officially free, the prior studies explained the effect of wealth gradient by high level of unofficial out-of-pocket expenditures for healthcare personnel, supplies, and medication .As a result, the wealthier use healthcare facilities which the poorer cannot afford. This is in line with our finding that the wealthier deliver in healthcare facilities, while the poorer have to deliver outside of healthcare facilities. While the poorer have to deliver at home. In this context, one of the promising ways to reduce effect of wealth gradient to utilization is to introduce the benefits for pregnant women which could be linked to receipt of targeted social assistance programs. Third, our study demonstrates that the risk of not delivering at a healthcare facility increased for less educated women. Women with higher education are strongly associated with delivering in medical settings and hence with higher chances of child survival. Habibov reported that there is no significant gender gap in the level of literacy and education in general in Azerbaijan and concluded that increase in nonacademic educational activities promoting antenatal care should be a priority. Habibov and Fan confirmed these conclusions showing the example of Tajikistan, another transitional country that having limited knowledge on matters related to sex is associated with a lower probability of maternal healthcare utilization. The authors underlined that significant effect of knowledge about sex is independent of formal educational level and it persisted even if formal educational level is controlled for. Effectiveness of communication campaigns designed to explain the benefits of maternal healthcare and encourage healthcare utilization is well documented in developing countries. In addition, intensive communication campaigns aimed at encouraging healthcare utilization slowly but steadily became appreciated in some transitional countries. This positive experience should be shared across the region. Finally, the population based nationally representative surveys such as the Demographic and Health Surveys by Macro International and Living Standards Measurement Surveys by the World Bank became an important tool for measuring policy effect on health outcomes in many transitional and developing countries. Most of these surveys include modules on healthcare utilization and childbirth outcome. Having high-quality microdata to conduct evaluation of healthcare programs is an effective way to save time, effort, and costs while providing nationally representative results. From this standpoint, our results are illustrative to empirical strategies for evaluation of nonrandomized data from cross-sectional surveys using a standard statistical software package. CASE STUDY CHALLENGE 1. Students should be asked to read the case and discuss all procedures done and suggest a solution program. 3. Conclusion (15%) Briefly summarize your thoughts & conclusion to your critique of the case study and provide a possible outcome for. The Effect of Maternal Healthcare on the Probability of Child Survival in Azerbaijan. How did these articles and Chapters influence your Maternal Healthcare on the Probability? Evaluation will be based on how clearly you respond to the above, in particular: a) The clarity with which you critique the case study; b) The depth, scope, and organization of your paper; and, c) Your conclusions, including a description of the impact of these Case study on any Health Care Setting ASSIGNMENT RUBRICS Assignments Guidelines 1 Points 10% Introduction 2.5 Points 25% Your Case Study Critique 6 Points 50% Conclusion 1.5 Points 15% Total 11 points 100% ASSIGNMENT GRADING SYSTEM A 90% – 100% B+ 85% – 89% B 80% – 84% C+ 75% – 79% C 70% – 74% D 60% – 69% F 50% – 59% Or less. Dr. Gisela Llamas Chapter 6 Confidence Interval Estimates Learning Objectives (1 of 2) • Define point estimate, standard error, confidence level, and margin of error • Compare and contrast standard error and margin of error • Compute and interpret confidence intervals for means and proportions • Differentiate independent and matched or paired samples Learning Objectives (1 of 2) • Compute confidence intervals for the difference in means and proportions in independent samples and for the mean difference in paired samples • Identify the appropriate confidence interval formula based on type of outcome variable and number of samples Statistical Inference (1 of 2) • There are two broad areas of statistical inference: estimation and hypothesis testing. • Estimation—the population parameter is unknown, and sample statistics are used to generate estimates of the unknown parameter. Statistical Inference (2 of 2) • Hypothesis testing—an explicit statement or hypothesis is generated about the population parameter; sample statistics are analyzed and determined to either support or reject the hypothesis about the parameter. • In both estimation and hypothesis testing, it is assumed that the sample drawn from the population is a random sample. Estimation (1 of 2) • Process of determining likely values for unknown population parameter • Point estimate is best single-valued estimate for parameter. • Confidence interval is range of values for parameter. point estimate ± margin of error Estimation (2 of 2) • A point estimate for a population parameter is the “best” single number estimate of that parameter. • A confidence interval estimate is a range of values for the population parameter with a level of confidence attached (e.g., 95% confidence that the range or interval contains the parameter). Confidence Interval Estimates point estimate ± margin of error point estimate ± Z SE (point estimate) where Z = value from standard normal distribution for desired confidence level and SE (point estimate) = standard error of the point estimate Confidence Intervals for m • Continuous outcome • One sample n ≥ 30 s XZ n s n < 30 X t [next slide], n (Find Z in Table 1B) (Find t in Table 2 df = n – 1) Table 2. Critical Values of the t Distribution • Table entries represent values from t distribution with upper tail area equal to a. Confidence level Two-sided test a One-sided test a 80% .20 .10 90% 95% 98% 99% .10 .05 .02 .01 .05 .025 .01 .005 df 1 3.078 6.314 12.71 31.82 63.66 2 1.886 2.920 4.303 6.965 9.925 3 1.638 2.353 3.182 4.541 5.841 4 1.533 2.132 2.776 3.747 4.604 5 1.476 2.015 2.571 3.365 4.032 6 1.440 1.943 2.447 3.143 3.707 7 1.415 1.895 2.365 2.998 3.499 8 1.397 1.860 2.306 2.896 3.355 9 1.383 1.833 2.262 2.821 3.250 10 1.372 1.812 2.228 2.764 3.169 Example 6.1. Confidence Interval for m (1 of 2) • In the Framingham Offspring Study (n = 3534), the mean systolic blood pressure (SBP) was 127.3 with a standard deviation of 19.0. • Generate a 95% confidence interval for the true mean SBP. XZ s n 19.0 127.3 1.96 3534 127.3 ± 0.63 (126.7, 127.9) Example 6.2. Confidence Interval for m (2 of 2) • In a subset of n = 10 participants attending the Framingham Offspring Study, the mean SBP was 121.2 with a standard deviation of 11.1. • Generate a 95% confidence interval for the true mean SBP. s Xt n 11.1 121.2 2.262 10 df = n – 1 = 9, t = 2.262 121.2 ± 7.94 (113.3, 129.1) New Scenario • Outcome is dichotomous (p = population proportion) – Result of surgery (success, failure) – Cancer remission (yes/no) • One study sample • Data – On each participant, measure outcome (yes/no) – n, x = number of positive responses x p̂ = n Confidence Intervals for p • Dichotomous outcome • One sample min[n p̂, n(1 − p̂)] 5 p̂(1 – p̂) p̂ ± Z n (Find Z in Table 1B) Example 6.3. Confidence Interval for p • In the Framingham Offspring Study (n = 3532), 1219 patients were on antihypertensive medications. • Generate a 95% confidence interval for the true proportion on antihypertensive medication. 1219 p̂(1 – p̂) p̂ = = 0.345 p̂ ± Z 3532 n 0.345 ± 0.016 0.345(1 – 0.345) 0.345 ±1.96 3532 (0.329, 0.361) New Scenario • Outcome is continuous – SBP, weight, cholesterol • Two independent study samples • Data – On each participant, identify group and measure outcome – n , X , s 2 (or s ), n , X , s 2 (or s ) 1 1 1 1 2 2 2 2 Two Independent Samples (1 of 2) RCT: Set of Subjects Who Meet Study Eligibility Criteria Randomize Treatment 1 Mean Trt 1 Treatment 2 Mean Trt 2 Two Independent Samples (2 of 2) Cohort Study: Set of Subjects Who Meet Study Inclusion Criteria Group 1 Group 2 Mean Group 1 Group 2 Mean Confidence Intervals for (m1 − m2) • Continuous outcome • Two independent samples n1 ≥ 30 and n2 ≥ 30 1 1 (X1 – X 2 ) ± ZSp + n1 n 2 n1 < 30 or n2 < 30 1 1 (X1 – X 2 ) ± tSp + n1 n 2 (Find Z in Table 1B) (Find t in Table 2, df = n1 + n2 – 2) Pooled Estimate of Common Standard Deviation, Sp • Previous formulas assume equal variances (s12 = s22) • If 0.5 ≤ s12/s22 ≤ 2, assumption is reasonable Sp = (n 1 − 1)s + (n 2 − 1)s n1 + n 2 − 2 2 1 2 2 Example 6.5. Confidence Interval for (m1 − m2) • Using data collected in the Framingham Offspring Study, generate a 95% confidence interval for the difference in mean SBP between men and women. n Men 17.5 Women Mean 1623 1911 Std Dev 128.2 126.5 20.1 Assess Equality of Variances • Ratio of sample variances: 17.52/20.12 = 0.76 Sp = Sp = (n 1 − 1)s 12 + (n 2 − 1)s 22 n1 + n 2 − 2 (1623 − 1)17.52 + (1911 − 1)20.12 1623 + 1911 − 2 = 359.12 = 19.0 Confidence Intervals for (m1 − m2) 1 1 (X1 – X 2 ) ± ZSp + n1 n 2 1 1 (128.2 – 126.5) ± 1.96 (19.0) + 1623 1911 1.7 ± 1.26 (0.44, 2.96) New Scenario • Outcome is continuous – SBP, weight, cholesterol • Two matched study samples • Data – On each participant, measure outcome under each experimental condition. – Compute differences (D = X1 – X2) – n, X , s d d Two Dependent/Matched Samples Subject ID 1 2 . . Measure 1 55 42 Measure 2 70 60 Measures taken serially in time or under different experimental conditions Crossover Trial Eligible Participants Treatment Treatment Placebo Placebo R Each participant measured on treatment and placebo Confidence Intervals for md • Continuous outcome • Two matched/paired samples n ≥ 30 n < 30 sd Xd Z n sd Xd t n (Find Z in Table 1B) (Find t in Table 2, df = n – 1) Example 6.8. Confidence Interval for md (1 of 3) • In a crossover trial to evaluate a new medication for depressive symptoms, patients’ depressive symptoms were measured after taking new drug and after taking placebo. • Depressive symptoms were measured on a scale of 0 to100 with higher scores indicative of more symptoms. Example 6.8. Confidence Interval for md (2 of 3) • Construct a 95% confidence interval for the mean difference in depressive symptoms between drug and placebo. • The mean difference in the sample (n = 100) is –12.7 with a standard deviation of 8.9. Example 6.8. Confidence Interval for md (3 of 3) sd Xd Z n 8.9 –12.7 ±1.96 100 –12.7 ± 1.74 (–14.1, –10.7) New Scenario • Outcome is dichotomous – Result of surgery (success, failure) – Cancer remission (yes/no) • Two independent study samples • Data – On each participant, identify group and measure outcome (yes/no) – n1 , p̂1 , n 2 , p̂2 Confidence Intervals for (p1 − p2) • Dichotomous outcome • Two independent samples min[n 1p̂1 , n1 (1 − p̂1 ), n 2 p̂2 , n 2 (1 − p̂2 )] 5 p̂1 (1 – p̂1 ) p̂ 2 (1 – p̂ 2 ) (p̂1 – p̂ 2 ) ± Z + n1 n2 (Find Z in Table 1B) Example 6.10. Confidence Interval for (p1 – p2) (1 of 3) • A clinical trial compares a new pain reliever to that considered standard care in patients undergoing joint replacement surgery; the outcome of interest is reduction in pain by 3+ scale points. • Construct a 95% confidence interval for the difference in proportions of patients reporting a reduction between treatments. Example 6.10. Confidence Interval for (p1 – p2) (2 of 3) Reduction of 3+ Points Treatment n Number Proportion New 50 23 0.46 Standard 50 11 0.22 Example 6.10. Confidence Interval for (p1 – p2) (3 of 3) p̂1 (1 – p̂1 ) p̂ 2 (1 – p̂ 2 ) (p̂1 – p̂ 2 ) ± Z + n1 n2 0.46(1 – 0.46) 0.22(1 – 0.22) (0.46 – 0.22) ±1.96 + 50 50 0.24 ± 0.18 (0.06, 0.42) Confidence Intervals for Relative Risk (RR) • Dichotomous outcome • Two independent samples (n1 – x1 )/x1 (n 2 – x 2 )/x 2 ln(R̂R) ± Z + n1 n2 exp(lower limit), exp(upper limit) (Find Z in Table 1B) Example 6.12. Confidence Interval for RR (1 of 2) Reduction of 3+ Points Treatment n Number Proportion New 50 23 0.46 Standard 50 11 0.22 Construct a 95% CI for the relative risk. Example 6.12. Confidence Interval for RR (2 of 2) p̂1 0.46 R̂R = = = 2.09 p̂ 2 0.22 27/23 39/11 ln(2.09) 1.96 + 50 50 0.737 ± 0.602 (0.135, 1.339) exp(0.135), exp(1.339) (1.14, 3.82) Confidence Intervals for Odds Ratio (OR) • Dichotomous outcome • Two independent samples 1 1 1 1 ln( ÔR) Z + + + x1 (n1 − x1 ) n 2 (n 2 − x 2 ) exp(lower limit), exp(upper limit) (Find Z in Table 1B) Example 6.14. Confidence Interval for OR (1 of 2) Reduction of 3+ Points Treatment n Number Proportion New 50 23 0.46 Standard 50 11 0.22 Construct a 95% CI for the odds ratio. Example 6.14. Confidence Interval for OR (2 of 2) x1 /(n1 – x1 ) 23/27 ÔR = = = 3.02 x 2 /(n 2 – x 2 ) 11/39 1 1 1 1 ln(3.02) 1.96 + + + 23 27 11 39 1.105 ± 0.870 (0.235, 1.975) exp(0.235), exp(1.975) (1.26, 7.21) Chapter 5 The Role of Probability Learning Objectives (1 of 3) • Define the terms “equally likely” and “at random” • Compute and interpret unconditional and conditional probabilities • Evaluate and interpret independence of events • Explain the key features of the binomial distribution model Learning Objectives (2 of 3) • Calculate probabilities using the binomial formula • Explain the key features of the normal distribution model • Calculate probabilities using the standard normal distribution table • Compute and interpret percentiles of the normal distribution Learning Objectives (3 of 3) • Define and interpret the standard error • Explain sampling variability • Apply and interpret the results of the Central Limit Theorem Two Areas of Biostatistics Goal: Statistical Inference POPULATION SAMPLE =? n, X Descriptive Statistics Sampling from a Population SAMPLES n n n n Population N n n n n n n Sampling: Population Size = N, Sample Size = n (1 of 2) • Simple random sample – Enumerate all members of population N (sampling frame), select n individuals at random (each has same probability of being selected). • Systematic sample – Start with sampling frame; determine sampling interval (N/n); select first person at random from first (N/n) and every (N/n) thereafter. Sampling: Population Size = N, Sample Size = n (2 of 2) • Stratified sample – Organize population into mutually exclusive strata; select individuals at random within each stratum. • Convenience sample – Non-probability sample (not for inference) • Quota sample – Select a predetermined number of individuals into sample from groups of interest. Basics • Probability reflects the likelihood that outcome will occur. • 0 ≤ Probability ≤ 1 Number with outcome Probability = N Example 5.1. Basic Probability (1 of 2) P(Select any child) = 1/5290 = 0.0002 Example 5.1. Basic Probability (2 of 2) P(Select a boy) = 2560/5290 = 0.484 P(Select boy age 10) = 418/5290 = 0.079 P(Select child at least 8 years of age) = (846 + 881 + 918)/5290 = 2645/5290 = 0.500 Conditional Probability • Probability of outcome in a specific subpopulation • Example 5.1. P(Select 9-year-old from among girls) = P(Select 9-year-old | girl) = 461/2730 = 0.169 P(Select boy | 6 years of age) = 379/892=0.425 Example 5.2. Conditional Probability (1 of 2) Example 5.2. Conditional Probability (2 of 2) P(Prostate cancer | Low PSA) = 3/64 = 0.047 P(Prostate cancer | Moderate PSA) = 13/41 = 0.317 P(Prostate cancer | High PSA) = 12/15 = 0.80 Sensitivity and Specificity Sensitivity = True positive fraction = P(test+ | disease) Specificity = True negative fraction = P(test– | disease free) False negative fraction = P(test– | disease) False positive fraction = P(test+ | disease free) Example 5.4. Sensitivity and Specificity Sensitivity and Specificity Sensitivity = P(test+ | disease) = 9/10 = 0.90 Specificity = P(test– | disease free) = 4449/4800 = 0.927 False negative fraction = P(test– | disease) = 1/10 = 0.10 False positive fraction = P(test+ | disease free) = 351/4800 = 0.073 Independence • Two events, A and B, are independent if P(A | B) = P(A) or if P(B | A) = P(B) Example 5.2. • Is screening test independent of prostate cancer diagnosis? – – – – P(Prostate cancer) = 28/120 = 0.023 P(Prostate cancer | Low PSA) = 0.047 P(Prostate cancer | Moderate PSA) = 0.317 P(Prostate cancer | High PSA) = 0.80 Bayes’ Theorem (1 of 2) • Using Bayes’ Theorem we revise or update a probability based on additional information. – Prior probability is an initial probability. – Posterior probability is a probability that is revised or updated based on additional information. Bayes’ Theorem (2 of 2) P(B | A)P(A) P(A | B) = P(B) P(B | A)P(A) P(A | B) = P(A)P(B | A) + P(A’ )P(B | A’ ) Example (1 of 2) • In Boston, 51% of adults are male. • One adult is randomly selected to participate in a study. Prior probability of selecting a male = 0.51 Example (2 of 2) • Selected participant is a smoker. • 9.5% of males in Boston smoke as compared to 1.7% of females. • Find the probability that we selected a male given he is a smoker. Example: Find P(M | S) • P(M) = 0.51 P(M’) = 0.49 P(S | M) = 0.095 P(S | M’) = 0.017 • Bayes’ Theorem P(S | M)P(M) P(M | S) = P(M)P(S | M) + P(M’ )P(S | M’ ) 0.095(0.51) P(M|S) = = 0.853 0.51(0.095) + 0.49(0.017) • Knowing the participant smokes—increases P(M) Example 5.8. Bayes’ Theorem (1 of 3) P(disease) = 0.002 Sensitivity = 0.85 = P(test+ | disease) P(test+) = 0.08 and P(test–) = 0.92 What is P(disease | test+)? Example 5.8. Bayes’ Theorem (2 of 3) What is P(disease | test+)? P(disease) = 0.002 Sensitivity = 0.85 = P(test+ | disease) P(test+) = 0.08 and P(test–) = 0.92 P(test+ | disease)P(disease) P(disease | test+) = P(test+) Example 5.8. Bayes’ Theorem (3 of 3) P(disease) = 0.002 Sensitivity = 0.85 = P(test+ | disease) P(test+) = 0.08 and P(test–) = 0.92 P(test+ | disease)P(disease) = P(test+) Binomial Distribution (1 of 2) • Model for discrete outcome • Process or experiment has two possible outcomes: success and failure. • Replications of process are independent. • P(success) is constant for each replication. Binomial Distribution (2 of 2) • Notation n = number of times process is replicated p = P(success) x = number of successes of interest 0≤x≤n n! P(x successes) = p x (1 − p) n − x x! (n − x)! Example 5.9. Binomial Distribution • Medication for allergies is effective in reducing symptoms in 80% of patients. If medication is given to 10 patients, what is the probability it is effective in 7? 10! 7 10-7 P(7 successes) = 0.8 (1- 0.8) 7!(10 – 7)! = 120(0.2097)(0.008) = 0.2013 Binomial Distribution (1 of 4) • Antibiotic is claimed to be effective in 70% of the patients. If antibiotic is given to five patients, what is the probability it is effective on exactly three? Success = Antibiotic is effective: n = 5, p = 0.7, x = 3 5! P(X = 3) = 0.73 (1- 0.7)5-3 3!(5 – 3)! = 10(0.343)(0.09) = 0.3087 Binomial Distribution (2 of 4) • What is the probability that the antibiotic is effective on all five? 5! 5 5-5 P(X = 5) = 0.7 (1- 0.7) 5!(5 – 5)! =1(0.1681)(1) = 0.1681 Binomial Distribution (3 of 4) • What is the probability that the antibiotic is effective on at least three? P(X ≥ 3) = P(3) + P(4) + P(5) = 0.3087 + 0.3601 + 0.1681 = 0.8369 Binomial Distribution (4 of 4) • Mean and variance of the binomial distribution m = np s2 = np (1 – p) For example, the mean (or expected) number of patients in whom the antibiotic is effective is 5*0.7 = 3.5 Normal Distribution (1 of 3) • Model for continuous outcome • Mean = median = mode Normal Distribution (2 of 3) Notation: = mean and s = standard deviation −3s −2s −s +s +2s +3s Normal Distribution (3 of 3) • Properties of normal distribution I) The normal distribution is symmetric about the mean (i.e., P(X > ) = P(X < ) = 0.5). ii) The mean and variance, and s2, completely characterize the normal distribution. iii) The mean = the median = the mode. P( – s < X < + s) = 0.68 P( – 2s < X < + 2s) = 0.95 P( – 3s < X < + 3s) = 0.99 iv) P(a < X < b) = the area under the normal curve from a to b. Example 5.11. Normal Distribution (1 of 10) • Body mass index (BMI) for men age 60 is normally distributed with a mean of 29 and standard deviation of 6. • What is the probability that a male has BMI less than 29? Example 5.11. Normal Distribution (2 of 10) Example 5.11. Normal Distribution (3 of 10) P(X