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Statistics for clinicians

Statistics for clinicians.

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Statistics for clinicians

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  1. Statistics for clinicians • Biostatistics course by Kevin E. Kip, Ph.D., FAHAProfessor and Executive Director, Research CenterUniversity of South Florida, College of NursingProfessor, College of Public HealthDepartment of Epidemiology and BiostatisticsAssociate Member, Byrd Alzheimer’s InstituteMorsani College of MedicineTampa, FL, USA

  2. SECTION 2.1 Module Overview and Introduction Probability theory and discrete and continuous sampling distributions

  3. SECTION 2.4 Bayes Theorem

  4. Bayes Theorem • Procedure for updating a probability based on new information. • Rule can be used to compute a conditional probability based on specific, available information (i.e. links degree of belief in a proposition before and after accounting for evidence). • Can represent a subjective degree of belief that changes over time to account for new evidence. • Often used in meta analyses and synthesis of evidence • P(B|A) P(A) • P(A|B) = --------------- • P(B)

  5. Bayesian Methods (example) Question: What is the probability that Daphne’s next male child will be affected by the X-linked disorder? Aaron Bart Barbara Naïve: 1/2 Carlos Cathy 1/4 Desmond Daphne 1/8 Earl Joseph 1/16

  6. Question: What is the probability that Daphne’s next male child will be affected by the X-linked disorder? Aaron Bart Barbara Cathy Cathy carrier non-carrier 1/2 1/2 1/3 2/3 Carlos Prior: Posterior: Cathy Daphne Daphne carrier non-carrier 1/6 5/6 1/21 20/21 Desmond Daphne Prior: Posterior: Earl Joseph Thus, the probability that Daphne’s next male child will be affected is: 1/2 x 1/21 = 1/42.

  7. SECTION 2.5 Binomial Distribution Model

  8. Probability Model: • Mathematical equation or formula used to generate probabilities based on certain assumptions about the process. Very important for statistical inference. • Binomial Model: • Two possible outcomes – often labeled as “success” or “failure”, or as “disease” or “no disease”. • Allows computation of observing a specified number of responses (e.g. successes) when the process is repeated a specific number of times (e.g. among a set of patients). • For the binomial model with a set number of trials: • p = probability of success, and • q = 1 – p

  9. Binomial distribution model n = number of times process is repeated (e.g. # of patients) x = number of successes (outcomes) p = probability of outcome for any individual (i.e. independent) ! = factorial n! P(x outcomes) = ----------- px(1-p)n-x x!(n-x)! Example: Assume that a medication is effective 80% (0.80) of the time (i.e. p=0.8) Assume that the medication will be given to 10 patients (i.e. n=10) What is the probability the medication will be effective in exactly 7 patients? (x=7) (i.e. if we had to guess, we would think it was most likely that the medication would be effective in 8 patients) 10! P(7 successes) = ----------- 0.807(1-0.80)10-7 7!(10-7)!

  10. Binomial distribution model n! P(x outcomes) = ----------- px(1-p)n-x x!(n-x)! 10! P(7 successes) = ----------- 0.807(1-0.80)10-7 7!(10-7)! 10! 10(9)(8)(7)(6)(5)(4)(3)(2)1 --------- = -------------------------------------- 7!(10-7)! [7(6)(5)(4)(3)(2)(1)][(3)(2)(1)] 10(9)(8) = ---------- = 120 3(2) P(7 successes) = (120)(0.807))(1-0.80)10-7) P(7 successes) = (120)(0.2097)(0.008) = 0.2013

  11. Binomial distribution model Many times we do not want to know a probability of an outcome in an exact number of persons, but rather a certain number or more persons. For example, using our previous scenario, what is the probability that the medication will be effective at least 7 of the 10 patients? (i.e. P(> 7 successes)) To do this, we need to compute individual probabilities for all the combinations. P(7 successes) = 0.2013 P(8 successes) = 0.3020 P(9 successes) = 0.2684 P(10 successes) = 0.1074 P(>7 successes) = 0.2013 + 0.3020 + 0.2684 + 0.1074 = 0.8791

  12. Binomial distribution model (practice) n! P(x outcomes) = ----------- px(1-p)n-x x!(n-x)! Assume that a drug is effective 90% of the time Assume that the medication will be given to 12 patients What is the probability the drug will be effective in exactly 10 patients? Complete the formula below: P(10 successes) = ------------------

  13. Binomial distribution model (practice) n! P(x outcomes) = ----------- px(1-p)n-x x!(n-x)! Assume that a drug is effective 90% of the time Assume that the medication will be given to 12 patients What is the probability the drug will be effective in exactly 10 patients? Complete the formula below: 12! P(10 successes) = ----------- 0.9010(1-0.90)12-10 10!(12-10)!

  14. Binomial distribution model (practice) Complete the formula below: 12! P(10 successes) = ----------- 0.9010(1-0.90)12-10 10!(12-10)! 12! 12(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)1 --------- = -------------------------------------- 10!(12-10)! [10(9)(8)7(6)(5)(4)(3)(2)(1)][(2)(1)] 12(11) = ---------- = 66 (2) P(10 successes) = (66)(0.9010))(1-0.90)12-10) P(10 successes) = (66)(0.3487)(0.01) = 0.2301

  15. Binomial distribution model (calculating standard deviation) Our Example: Assume a medication is effective 80% (0.80) of the time (i.e. p=0.8) Assume the medication will be given to 10 patients (i.e. n=10) What is the probability the medication will be effective in exactly 7 patients? (i.e. x=7) The expected number of outcomes of a binomial population is: µ = np So, in our example, µ = (10 x 0.8) = 8 The standard deviation (σ) = sqrt[(n(p))(1-p)] So, in our example, σ = sqrt[(10 x 0.8) x (1-0.8)] σ = sqrt[(8 x 0.2)] σ = sqrt[1.6] σ = 1.265

  16. Binomial distribution model (practice) (calculating standard deviation) Example: Assume that a medication is effective 90% of the time Assume that the medication will be given to 20 patients What is the expected number of outcomes: µ = np So, in this example, µ = ______________ Calculate the standard deviation (σ) = sqrt[(n(p))(1-p)] So, in this example, σ = _______________________

  17. Binomial distribution model (practice) (calculating standard deviation) Example: Assume that a medication is effective 90% of the time Assume that the medication will be given to 20 patients What is the expected number of outcomes: µ = np So, in this example, µ = (20 x 0.9) = 18 Calculate the standard deviation (σ) = sqrt[(n(p))(1-p)] So, in this example, σ = sqrt[(20 x 0.9) x (1-0.9)] σ = sqrt[(18 x 0.1)] σ = sqrt[1.8] σ = 1.34

  18. SECTION 2.6 Poisson Distribution Model

  19. Poisson Distribution Model • A discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time (i.e. X = 0,1,2,3, ….) • Usually associated with rare events • Approximates the binomial distribution when N is large (i.e. >100) and p is small (i.e. <0.01) • Requirements for the Poisson Distribution • Length of time period is fixed in advance; • Events occur at a constant average rate; • Events can be counted in whole numbers • Number of events occurring in disjoint intervals are statistically independent.

  20. Poisson Distribution Model Illustration: Assume deaths from typhoid fever in a given population are Poisson distributed with a mean of 2.3 deaths per year. What is the probability distribution of deaths in this population? Pr(X = k) k

  21. Poisson Distribution Model Poisson Formula Suppose we conduct a Poisson study in which the average number of health events within a given time period is μ. Then, the Poisson probability is: P(x; μ) = (e-μ) (μx) / x! where x is the actual number of events that result from the study, and e is approximately equal to 2.71828.

  22. Poisson Formula Example: The average number of colds for toddlers in day care is 2 per year. Knowing this, what is the estimated probability that a new toddler to day care will have exactly 3 colds during the following year? P(x; μ) = (e-μ) (μx) / x! μ = 2; since 2 colds per year, on average. x = 3; since we want to find the likelihood that 3 colds will occur in the next year. e = 2.71828; since e is a constant equal to approximately 2.71828. Plug these values into the Poisson formula: P(3; 2) = (2.71828-2) (23) / 3! P(3; 2) = (0.13534) (8) / 6 P(3; 2) = 0.180 Thus, the probability of a toddler having 3 colds in the next year is 0.180.

  23. Poisson Formula (Practice) Assume that the average number of cases of tuberculosis within a nursing home is 4 per year. Knowing this, what is the estimated probability that exactly 4 new cases of TB will occur during the following 6-months? P(x; μ) = (e-μ) (μx) / x! μ = ________ x = ________ e = ________ Plug these values into the Poisson formula:

  24. Poisson Formula (Practice) Assume that the average number of cases of tuberculosis within a nursing home is 4 per year. Knowing this, what is the estimated probability that exactly 4 new cases of TB will occur during the following 6-months? P(x; μ) = (e-μ) (μx) / x! μ = 2; since 4 cases of TB per year = 2 cases of TB per 6 months, on average. x = 4; since we want to find the likelihood that 4 cases will occur in next 6 months. e = 2.71828; since e is a constant equal to approximately 2.71828. Plug these values into the Poisson formula: P(4; 2) = (2.71828-2) (24) / 4! P(4; 2) = (0.13534) (16) / 24 P(4; 2) = 0.0902 Thus, the probability of exactly 4 new cases of TB in the next 6-months is 0.09.

  25. http://statpages.org/ctab2x2.html

  26. SECTION 2.7 Properties of the Normal Distribution

  27. Normal Distribution • Appropriate for a continuous outcome if: • Mean = median = mode • Symmetric around the mean • P(x > µ) = p(x < µ) where x is continuous variable and µ is mean • ~68% of all values fall between the mean and one 1 SD • (i.e. P(µ - σ < x < P(µ + σ) = 0.68) • ~95% of all values fall between the mean and 2 SD • (i.e. P(µ - 2σ< x < P(µ + 2σ) = 0.95) • ~99% of all values fall between the mean and 3 SD • (i.e. P(µ - 3σ< x < P(µ + 3σ) = 0.99)

  28. Normal Distribution

  29. Normal Distribution ~68% of values between mean and one 1 SD ~95% - ~68% = 27% / 2 = ~13.6% of all values between -1 to -2 SD and +1 to +2SD ~99% - ~95% = 4% / 2 = ~2.0% of all values between -2 to -3 SD and +2 to +3SD

  30. Normal Distribution (Practice) Assume that BMI is normally distributed with µ = 29.4 and σ = 4.6 1. Put in the appropriate values on the normal distribution curve for BMI values plus or minus 1, 2, and 3 SD from the mean 2. What is the median BMI value? _______ 3. Approximately what percentage of the population has a BMI > 34? _________ 4. Approximately what percentage of the population has a BMI between 15.6 and 20.2? __________ 5. What is the approximate minimum and maximum BMI in the population? Min: _________ Max: __________ Body Mass Index (BMI) 29.4 µ ? ? ? ? ? ?

  31. Normal Distribution (Practice) Assume that BMI is normally distributed with µ = 29.4 and σ = 4.6 1. Put in the appropriate values on the normal distribution curve for BMI values plus or minus 1, 2, and 3 SD from the mean 2. What is the median BMI value? _29.4__ 3. Approximately what percentage of the population has a BMI > 34? __16.0%___ (i.e. 13.6% + 2.2% + 0.2%) 4. Approximately what percentage of the population has a BMI between 15.6 and 20.2? ____2.2%______ 5. What is the approximate minimum and maximum BMI in the population? Min: ____~15_____ Max: ____~44____ Body Mass Index (BMI) 13.6% 34% 34% 13.6% 0.2% 0.2% 2.2% 2.2% 29.4 15.6 20.2 24.8 34.0 38.6 43.2 µ+2σ µ+1σ µ-3σ µ-2σ µ-1σ µ+3σ µ

  32. Normal Distribution Question: What do we if we want to calculate a probability when the value of interest is not the mean or a multiple of the standard deviation? Answer: Compute a z-score and use a table of probabilities for a “standard” normal distribution with mean of 0 and standard deviation of 1. Standard Normal Distribution; µ=0, σ=1 µ 2 3 -3 -2 1 -1

  33. Normal Distribution Standardized Score (z-score) x - µ Z = ------------- σ Where: x is the value of interest µ is the mean σ is the standard deviation Example: Body Mass Index x is 35 µ is 29.4 σ is 4.6 35 – 29.4 Z = ------------- = 1.17 4.6

  34. Body Mass Index µ is 29.4 σ is 4.6 Body Mass Index (BMI) If x = 34, then z = 1 i.e. 34 – 29.4 ----------- = 1 4.6 Thus, P(z > 1 = (0.136 + 0.022 + 0.002 = 0.16) 13.6% 34% 34% 13.6% 0.2% 0.2% 2.2% 2.2% 20.2 24.8 15.6 38.6 43.2 29.4 34.0 If x = 35, then z = 1.17 i.e. 35 – 29.4 ----------- = 1.17 4.6 Thus, P(z > 1.17 = (?????) Refer to Appendix Table 1 P(z > 1.17 = 0.121) Body Mass Index (BMI) Area under the curve P 35.0

  35. Body Mass Index (BMI) In Appendix Table 1: The probability value of 0.879 (and 1 – 0.879 = 0.121) is determined by first looking at the left column for z to 1 decimal place, and then across the top row for z to the second decimal place. Area under the curve P 35.0 Can also get the exact probability for z = 1.17 http://stattrek.com/online-calculator/normal.aspx Cumulative probability: P(Z < 1.17) = 0.879 Thus, probability: P(Z > 1.17) = (1 - 0.879) = 0.121

  36. Standard Normal Distribution (Practice) • A. What is the probability that a person in the population • will have a BMI > 37.0? • z = _______________ P(Z > ???) = __________ • B. What is the probability that a person in the population • will have a BMI < 27.0? • z = _______________ P(Z < ???) = __________ Body Mass Index (BMI) x - µ Z = ------------- σ After calculating z for questions A and B, refer to Appendix Table 1 Body Mass Index µ is 29.4 σ is 4.6 Area under the curve Area under the curve P P 27.0 37.0

  37. Standard Normal Distribution (Practice) • A. What is the probability that a person in the population • will have a BMI > 37.0? • z = (37 – 29.4) / 4.6 = 1.65 P(Z > 1.65) = 0.0495 • B. What is the probability that a person in the population • will have a BMI < 27.0? • z = (27-29.4) / 4.6 = -0.52 P(Z < -0.52) = 0.3015 Body Mass Index (BMI) x - µ Z = ------------- σ After calculating z for questions A and B, refer to Appendix Table 1 Body Mass Index µ is 29.4 σ is 4.6 Area under the curve Area under the curve P P 27.0 37.0

  38. Percentiles from Standard Normal Distribution Standard normal distribution can also be used to compute percentiles. x = µ + zσ Example: Calculate the 90th percentile for BMI: In Appendix Table 1: The z value for the 90th percentile is found by looking in the body of the table for a value of 0.90, or the nearest value. In this case, it is 0.8997 which corresponds to a z value of 1.28 (the actual z value is 1.282) So, x = 29.4 + (1.282 x 4.6) = 35.3 Body Mass Index (BMI) Body Mass Index µ is 29.4 σ is 4.6 µ 29.4 90th percentile

  39. Percentiles from Standard Normal Distribution (Practice) x = µ + zσ Calculate the 33rd percentile for BMI: z = _________ so, x = _________________________ Body Mass Index (BMI) Body Mass Index µ is 29.4 σ is 4.6 µ 33rd percentile 29.4

  40. Percentiles from Standard Normal Distribution (Practice) x = µ + zσ Calculate the 33rd percentile for BMI: z = -0.44 so, x = 29.4 + (-0.44 x 4.6) = 27.4 Body Mass Index (BMI) Body Mass Index µ is 29.4 σ is 4.6 µ 33rd percentile 29.4

  41. Standard Normal Distribution Z values for common percentiles: PercentileZ 1st -2.326 2.5th -1.960 5th -1.645 10th -1.282 25th -0.675 50th 0.0 75th 0.675 90th 1.282 95th 1.645 97.5th 1.960 99th 2.326 x - µ Z = ------------- σ

  42. SECTION 2.8 Sampling Distributions and Central Limit Theorem

  43. Sampling Distributions • In estimating the mean of a continuous variable in a population, the mean of a representative sample is a good estimate of the unknown population mean, but it is only an estimate. • When making estimates about population parameters based on sample statistics, it is very important to quantify the precision of the parameter estimates (e.g. standard error of the mean). • Illustration: • Assume a population of 6 measurements of self-reported pain (on a 0 to 100 scale) after total hip replacement with scores as follows: • 25 50 80 85 90 100 • The population mean (μ) is: ∑X/ N = 71.7 and • the standard deviation is: sqrt[∑(X-μ)2 / N = 25.9

  44. Sampling Distributions 25 50 80 85 90 100 The population mean (μ) is: ∑X / N = 71.7 and the standard deviation is: sqrt[∑(X- μ)2 / N = 25.9 Suppose we did not have population data and wanted to estimate the mean from a sample, taking a sample size of 4. There are 15 different possible samples with n=4 when sampling without replacement is used (i.e. each individual can only be sampled once in a given sample). The probability of selecting any one of the 15 possible samples is 1/15 – see next slide.

  45. The table represents the sampling distribution of the sample means So, with the original population sample of N=6, the population mean is: µ = ∑X / N = 71.7 with SD of 25.9. However, the mean of the sample means, denoted as µX is 71.7 and a standard deviation of σX = 8.5. Note that µ = µX (71.7), yet σ (25.9) is much smaller than σX = 8.5 This is because the range of the population data (25 to 100) is much larger than the range of the sample means (60 to 88.8). These properties are formally stated in the Central Limit Theorem.

  46. Central Limit Theorem If we take simple random samples of size n from the population with replacement, then for large samples (n > 30), the sample distribution of the sample means is approximately normally distributed with: µX = µ and σX = σ / sqrt(n) Because the distribution of the sample means is approximately normal, the normal probability model can be used to make inferences about a population mean. The parameter σX = σ / sqrt(n), as noted above, is the standard error (meaning the standard deviation of the sample means) For a dichotomous outcome, the theorem holds for samples that: Minimum[np, n(1-p)] > 5, where n is the sample size and p is probability of the outcome in the population.

  47. Central Limit Theorem (Practice) Sample Population Sample Means µX = µ σX = σ / sqrt(n)

  48. Central Limit Theorem (Practice) Sample Population Sample Means µX = µ σX = σ / sqrt(n) Note that µX = µ because the sample population mean is an unbiased estimate of the true mean WhereasσX = σ are different quantities: σ is an estimate of variability in the sample σX is an estimate of precision of the mean estimate Thus, as n increases, σX is smaller, but σ may be > or <

  49. Central Limit Theorem -- Application Assume that in adults over age 50 that HDL cholesterol has: µ = 54 and σ = 17. Suppose a physician has 40 patients (older than 50) and wants to determine the probability that their mean HDL is >60 i.e. P(X>60) Intuitively, this should appear very unlikely….. X - µ 60 – 54 6 Z = ------------ Z = ---------------- = ---- = 2.22 σ / sqrt(n) 17 / sqrt(40) 2.7 From appendix Table 1, P(z > 2.22) = (1 – 0.9868) = 0.0132 (i.e. very unlikely)

  50. Central Limit Theorem – Application (Practice) Assume that in adults over age 50 that HDL cholesterol has: µ = 44 and σ = 16. Suppose a physician has 50 patients (older than 50) and wants to determine the probability that their mean HDL is <40 i.e. P(X<40) X - µ Z = ------------ Z = ---------------- σ / sqrt(n) X = ____ µ = ____ σ = ____ n = ____ From appendix Table 1, P(z < ????) =

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