HSA-6752 Statistic in Health Care Management

HSA-6752 Statistic in Health Care Management

HSA-6752 Statistic in Health Care Management

Florida National University HSA-6752 Statistic in Health Care Management Week 2 Critical Reflection Paper: Chapters 3 & 4 Objective: To critically reflect your understanding of the readings and your ability to apply them to your Health care Setting. ASSIGNMENT GUIDELINES (10%): Students will critically analyze the readings from Chapter 3 and 4 in your textbook. This assignment is designed to help you review, critique, and apply the readings to your Health Care setting as well as become the foundation for all of your remaining assignments. You need to read the chapters assigned for week 2 and develop a 2-3-page paper reflecting your understanding and ability to apply the readings to your Health Care Setting. 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 and articles you read, in your own words. 2. Your Critique (50%) What is your reaction to the content of the chapters? What did you learn about Prevalence, Incidence and the relationships between them? What did you learn and how you can apply Ordinal and Categorical Variables? Did these Chapters change your thoughts about comparing the extent of Disease between groups? If so, how? If not, what remained the same? 3. Conclusion (15%) Briefly summarize your thoughts & conclusion to your critique of the Chapter you read. How did these Chapters impact your thoughts on quantifying the extent of disease and summarizing Data Collected in the sample? Evaluation will be based on how clearly you respond to the above, in particular: a) The clarity with which you critique the chapters. b) The depth, scope, and organization of your paper; and, c) Your conclusions, including a description of the impact of these Chapters on any Health Care Setting. ASSIGNMENT RUBRICS Assignments Guidelines Introduction Your Critique Conclusion Total 10 Points 25 Points 50 Points 15 Points 100 points ASSIGNMENT GRADING SYSTEM A B+ B C+ C D F Dr. Gisela LLamas 90% – 100% 85% – 89% 80% – 84% 75% – 79% 70% – 74% 60% – 69% 50% – 59% Or less. 10% 25% 50% 15% 100% Chapter 3 Quantifying the Extent of Disease Learning Objectives • Define and differentiate prevalence and incidence • Select, compute, and interpret the appropriate measure to compare the extent of disease between groups • Compare and contrast relative risks, risk differences, and odds ratios • Compute and interpret relative risks, risk differences, and odds ratios Critical Components of RCT • Randomization • Control group – ethical issues • Monitoring – Interim analysis – Data and safety monitoring board • Data management • Reporting Prevalence • Proportion of participants with disease at a particular point in time Number of persons with disease Point Prevalence = Number of persons examined at baseline Example 3.1. Computing Prevalence Prevalence of CVD = 379/3799 = 0.0998 = 9.98% Prevalence of CVD in Men = 244/1792 = 0.1362 = 13.62% Prevalence of CVD in Women = 135/2007 = 0.0673 = 6.73% Example: H1N1 Outbreak • H1N1 outbreak first noticed in Mexico. • Large outbreak early on in La Gloria—a small village outside of Mexico City • Studied extensively in the first report on H1N1 (Fraser, Donelly, et al. “Pandemic potential of a strain of Influenza (H1N1): Early findings,” Science Express, 11 May 2009.) • Important questions – Who is most likely to be impacted? – What are characteristics of people commonly impacted? Computing Prevalence (1 of 2) Data on H1N1 outbreak in La Gloria, Mexico: n = 1575 villagers (out of 2155) were surveyed to determine if they had influenza-like illness (ILI) between 2/15/09 and 4/27/09. Age No ILI ILI Total ≤ 44 years 703 522 1225 > 44 years 256 94 350 959 616 1575 Total Computing Prevalence (2 of 2) Age No ILI ILI Total ≤ 44 years 703 522 1225 > 44 years 256 94 350 959 616 1575 Total Prevalence of ILI = 616/1575 = 0.3911 = 39.11% Prevalence of ILI in ≤ 44 = 522/1225 = 0.4261 = 42.61% Prevalence of ILI in > 44 = 94/350 = 0.2686 = 26.86% Incidence • Likelihood of developing disease among persons free of disease who are at risk of developing disease Cumulative Incidence = Incidence Rate = Number of persons who develop disease during a specified period Number of persons at risk at baseline Number of persons who develop disease during a specified period Sum of the lengths of time during which persons are disease-free Computing Incidence • Cumulative incidence requires complete follow-up on all participants. • Person-time data is used to take full advantage of available information in incidence rate. • Incidence rate often expressed as an integer per multiple of participants over a specified time. Incidence of CVD? Incidence Rate Incidence Rate = Number of persons who develop disease during a specified period Sum of the lengths of time during which persons are disease-free Incidence of CVD Incidence = 2/(10 + 9 + 3 + 10 + 5) = 2/37 = 0.054 5.4 per 100 person-years Example 3.2. Computing Incidence Men Women Total Develop CVD 190 119 309 Total Follow-Up Time (years) 9984 12153 22137 Incidence Rate of CVD in Men = 190/9984 = 0.01903 = 190 per 10,000 person-years Incidence Rate of CVD in Women = 119/12153 = 0.00979 = 98 per 10,000 person-years Computing Incidence Developed ILI Total Follow-Up Time (years) ≤ 44 years 522 20,064 > 44 years 94 3,514 Total 616 23,578 Incidence Rate of ILI in ≤ 44 = 522/20,064 = 0.0260 = 260 per 10,000 person-years Incidence Rate of ILI in > 44 = 94/3514 = 0.0268 = 268 per 10,000 person-years Comparing Extent of Disease Between Groups (1 of 2) • Risk difference (excess risk) = Prevalence exposed − Prevalence unexposed = Cumulative Incidence exposed − Cumulative Incidence unexposed = Incidence Rate exposed − Incidence Rate unexposed Comparing Extent of Disease Between Groups (2 of 2) • Risk difference of prevalent CVD in smokers versus nonsmokers = Prevalencesmokers – Prevalencenonsmokers = 81/744 – 298/3055 = 0.1089 – 0.0975 = 0.0114 Population Attributable Risk of CVD in Smokers vs. Nonsmokers Prevalenceoverall – Prevalencenonsmokers = Prevalenceoverall = (0.0998 – 0.0975) / 0.0998 = 0.023 = 2.3% Comparing Extent of Disease Between Groups (1 of 7) • Risk difference of history of ILI in males and females in La Gloria = Prevalence Females − Prevalence Males No ILI ILI Total Males 517 260 777 Females 442 356 798 Total 959 616 1575 = 356/798 – 260/777 = 0.4461 – 0.3346 = 0.1115 Comparing Extent of Disease Between Groups (2 of 7) • Relative risk = Prevalence exposed Prevalence unexposed Comparing Extent of Disease Between Groups (3 of 7) • Relative risk of CVD in smokers versus nonsmokers Prevalencesmokers = Prevalence nonsmokers = 0.1089/0.0975 = 1.12 Comparing Extent of Disease Between Groups (4 of 7) • Relative risk of ILI in females versus males Prevalence = females Prevalence males No ILI ILI Total Males 517 260 777 Females 442 356 798 959 616 1575 = 0.4461/0.3346 = 1.33 Comparing Extent of Disease Between Groups (5 of 7) • Odds ratio Prevalence exposed = Prevalence unexposed (1 − Prevalence exposed ) (1 − Prevalence unexposed ) Comparing Extent of Disease Between Groups (6 of 7) • Odds ratio of CVD in hypertensives versus hypertensives 181/840 = 188/2942 (1 − 181 / 840) (1 − 188 / 2942) = 0.275 / 0.725 = 4.04 0.068 / 0.932 Comparing Extent of Disease Between Groups (7 of 7) • Odds ratio of ILI in younger group versus older group Age No ILI ILI Total ≤ 44 years 703 522 1225 > 44 years 256 94 350 Total 959 616 1575 522/1225 = 94/350 (1 − 522 / 1225) (1 − 94 / 350) = 0.426 / 0.574 = 2.02 0.269 / 0.731 Relative Risks and Odds Ratios • Not possible to estimate relative risk in case-control studies • Possible to estimate odds ratio because of its invariance property Invariance Property of Odds Ratio (1 of 2) • Case-control study to assess association between smoking and cancer Invariance Property of Odds Ratio (2 of 2) Odds ratio for cancer in smokers versus nonsmokers = (40/29) / (10/21) = 2.90 Odds of smoking in patients with cancer versus not = (40/10) / (29/21) = 2.90(!) Chapter 4 Summarizing Data Collected in the Sample Learning Objectives (1 of 3) • Distinguish between dichotomous, ordinal, categorical, and continuous variables • Identify appropriate numerical and graphical summaries for each variable type • Compute a mean, median, standard deviation, quartiles and range for a continuous variable Learning Objectives (2 of 3) • Construct a frequency distribution table for dichotomous, categorical, and ordinal variables • Provide an example of when the mean is a better measure of location than the median • Interpret the standard deviation of a continuous variable Learning Objectives (3 of 3) • Generate and interpret a box plot for a continuous variable • Produce and interpret side-by-side box plots • Differentiate between a histogram and a bar chart Variable Types • Dichotomous variables have two possible responses (e.g., yes/no). • Ordinal and categorical variables have more than two responses, and responses are ordered and unordered, respectively. • Continuous (or measurement) variables assume in theory any values between a theoretical minimum and maximum. Biostatistics • Two areas of applied biostatistics – Descriptive statistics—summarize a sample selected from a population – Inferential statistics—make inferences about population parameters based on sample statistics. Vocabulary • Data elements/data points • Subjects/units of measurement • Population versus sample Sample vs. Population • Any summary measure computed on a sample is a statistic. • Any summary measure computed on a population is a parameter. n = Sample Size N = Population Size Example 4.1. Dichotomous Variable Frequency Distribution Table Relative Frequency Bar Chart for Dichotomous Variable Categorical Outcome (1 of 2) Sample: n = 50 Population: Patients at health center Variable: Marital status Marital Status Number of Patients Married 24 Separated 5 Divorced 8 Widowed 2 Never married 11 Total 50 Categorical Outcome (2 of 2) Frequency Distribution Table Marital Status Number of Patients (f) Relative Frequency (f/n) Married 24 0.48 Separated 5 0.10 Divorced 8 0.16 Widowed 2 0.04 Never married 11 0.22 Total 50 1.00 Frequency Bar Chart Ordinal Outcome (1 of 2) Sample: n =50 Population: Patients at health center Variable: Self-reported current health status Health Status Number of Patients Excellent 19 Very good 12 Good 9 Fair 6 Poor 4 Total 50 Ordinal Outcome (2 of 2) Frequency Distribution Table Heath Status Freq. Rel. Freq. Cumulative Freq. Cumulative Rel. Freq. Excellent 19 38% 19 38% Very good 12 24% 31 62% Good 9 18% 40 80% Fair 6 12% 46 92% Poor 4 8% 50 100% 50 100% % Relative Frequency Histogram 40 35 30 25 20 15 10 5 0 Poor Fair Good Very Good Health Status Excellent Example 4.2. Ordinal Variable Frequency Distribution Table Relative Frequency Histogram for Ordinal Variable Continuous Variable (1 of 9) • Assume, in theory, any value between a theoretical minimum and maximum • Quantitative, measurement variables Continuous Variable (2 of 9) • Population: Patients 50 years of age with coronary artery disease • Sample: n = 7 patients • Outcome: (mmHg) Systolic blood pressure Continuous Variable (3 of 9) Sample data X 100 110 114 121 130 130 160 Continuous Variable (4 of 9) X  Sample mean = X = n X 865  X= = = 123.6 n 7 X 100 110 114 121 130 130 160 865 Continuous Variable (5 of 9) Consider a second sample from the same population. We record SBP on each subject in the second sample: 120 121 122 124 125 126 127 X n =7 = 865 / 7 = 123.6. What is different between the two samples? Continuous Variable (6 of 9) • Dispersion X (X – X) 100 –23.6 110 –13.6 114 –9.6 121 –2.6 130 6.4 130 6.4 160 36.4 865 0 Continuous Variable (7 of 9) • Dispersion X (X – X ) 100 –23.6 110 –13.6 114 –9.6 121 –2.6 130 6.4 130 6.4 160 36.4 865 0 Mean absolute deviation (MAD): Σ|X -X| MAD = n Continuous Variable (8 of 9) • Sample variance X (X – X ) (X –X )2 100 –23.6 556.96 110 –13.6 184.96 114 –9.6 92.16 121 –2.6 6.76 130 6.4 40.96 130 6.4 40.96 160 36.4 1324.96 865 0 2247.72 2 Σ(X − X ) s2 = n −1 2247.72 s = = 374.6 6 2 Continuous Variable (9 of 9) • Sample standard deviation s= s 2 s = 374.6 = 19.4 • Standard summary n = 7, X = 123.6, s = 19.4 Median • Median—holds 50% of values above and 50% of values below • Order data — For n odd—median is middle value — For n even—median is mean of two middle values 100 110 114 121 130 130 160 Median Quartiles • Q1 = first quartile holds approximately 25% of the scores at or below it. • Q3 = third quartile holds approximately 25% of the scores at or above it. • Q2 = ?? Continuous Variable Order data 100 110 114 121 130 130 160 Q1 Median Q3 Box and Whisker Plot Min Q1 Median Q3 100 110 120 130 140 150 160 Max Comparing Samples with Box and Whisker Plots 100 110 120 130 140 150 160 Summarizing Location and Variability • • When there are no outliers, the sample mean and standard deviation summarize location and variability. When there are outliers, the median and interquartile range (IQR) summarize location and variability, where IQR = Q3 – Q 1. Example (1 of 2) Sample: n = 51 participants in a study of cardiovascular risk factors. Variable: age (years) 60 62 63 64 64 65 65 66 66 66 66 66 67 67 70 70 70 71 71 72 72 73 73 75 75 75 76 76 7779 82 83 85 85 87 65 67 73 77 65 68 73 77 65 68 73 77 65 68 73 77 Example (2 of 2) Sample mean: ΣX 3637 X= = = 71.3 n 51 Sample variance: 2 2 2 Σ (Σ /n ) 261,439 (3637 /51 ) X X 2 = = 41.4 s = n -1 50 Sample standard deviation: s = 41.4 = 6.4 Standard summary: n = 51, X = 71.3, s = 6.4 Outliers IQR = Interquartile Range = Q3 – Q1 = Range of middle half of the data • Outliers are values that either: – Exceed Q3 + 1.5 IQR – Fall below Q1 – 1.5 IQR – Or, are outside ± 3s X Check for Outliers in Example • Q1 = 66, Q3 = 76, IQR = 10 – Lower = 66 – 1.5(10) = 51 – Upper = 76 + 1.5(10) = 91 • X ± 3s = 52.1 to 90.5 Presenting Data (1 of 2) • Suppose we collapse ages into five mutually exclusive and exhaustive categories Age Class 60–64 65–69 70–74 75–79 80–84 Number of Individuals (freq.) 5 17 12 12 2 85–89 3 Presenting Data (2 of 2) Age Class 60-64 65-69 70-74 75-79 80-84 85-89 Total Cumulative Freq. Rel. Freq. Freq. Rel. Freq. 5 0.10 5 0.10 17 0.33 22 0.43 12 0.24 34 0.67 12 0.24 46 0.91 2 0.04 48 0.95 3 0.06 51 1.00 51 1.00 Frequency Frequency Histogram 18 16 14 12 10 8 6 4 2 0 6064 6569 7074 7579 Age Class 8084 8589 Example 4.3. Summarizing Continuous Variables • Diastolic blood pressures in n = 10 randomly selected participants attending the seventh examination of the Framingham Offspring Study 76 72 64 81 62 63 81 67 70 77 Summarizing Location • What is a typical diastolic blood pressure? Sample mean: = Sum of diastolic blood pressures/n = 713/10 = 71.3 Notation • Let X represent the outcome of interest (e.g., X = diastolic blood pressure) X  Sample mean = X = n Summarizing Variability • Sample range: = maximum – minimum = 81 – 62 = 19 • Sample variance: (x − x)  s = 2 n −1 2 Sample Variance (1 of 2) DBP 76 64 62 81 70 72 81 63 67 77 S X = 71.3 Deviation from Mean (76 – 71.3) = 4.7 (64 – 71.3) = –7.3 (62 – 71.3) = –9.3 9.7 –1.3 0.7 9.7 –8.3 –4.3 5.7 S Deviations from Mean = 0 Sample Variance (2 of 2) DBP 76 64 62 81 70 72 81 63 67 77 S X = 71.3 Deviation from Mean Squared Deviations (76 – 71.3) = 4.7 22.09 (64 – 71.3) = –7.3 53.29 (62 – 71.3) = –9.3 86.49 9.7 94.09 –1.3 1.69 0.7 0.49 9.7 94.09 –8.3 68.89 –4.3 18.49 5.7 32.49 S Deviations = 0 S Deviations2 = 472.10 Sample Variance and Sample Standard Deviation (x − x) 472.10  s = = = 52.46 2 2 n −1 9 (x − x)  s= = 52.46 = 7.2 2 n −1 Median • Median holds 50% of values above and 50% of values below • Order data – For n odd—median is middle value – For n even—median is mean of two middle values Median = 71 62 63 64 64 70 | 72 76 77 81 81 Quartiles • Q1 = first quartile holds 25% of values below it • Q3 = third quartile holds 25% of values above it Median = 71 62 63 64 64 70 | 72 76 77 81 81 Q1 Q3 Determining Outliers • Outliers—values below Q1 – 1.5(Q3 – Q1) or above Q3 + 1.5(Q3 – Q1) • In Example 4.3: lower limit = 64 – 1.5(77 – 64) = 44.5 and upper limit = 77 + 1.5(77 – 64) = 96.5 • Outliers? • Mean or median? • s or IQR? Box Plot for Continuous Variable 80 dbp 75 70 65 60 Numerical and Graphical Summaries (1 of 2) • Dichotomous and categorical – Frequencies and relative frequencies – Bar charts (freq. or relative freq.) • Ordinal – Frequencies, relative frequencies, cumulative frequencies, and cumulative relative frequencies – Histograms (freq. or relative freq.) Numerical and Graphical Summaries (2 of 2) • Continuous – Mean, standard deviation, minimum, maximum, range, median, quartiles, interquartile range – Box plot

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