Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
10/13/2016 Normal distribution Statistics in medicine • Properties: – Bell shape – Symmetric about the mean – Mean=median=mode – The area under the curve = 1 • 68% within mean + 1SD • 95% within mead + 2SD • 99% within mean+ 3SD Workshop 2: Describing variation, and probabilities October 6, 2016 10:00 AM to 10:50 AM Hope 110 Fatma Shebl, MD, MS, MPH, PhD Assistant Professor Chronic Disease Epidemiology Department Yale School of Public Health [email protected] SLIDE 0 SLIDE 1 Readiness assessment questions Readiness assessment questions Q1: A study of serum triglycerides levels among patients with history of myocardial infarction revealed that serum triglycerides is normally distributed with a mean of 200 mg/dL and standard deviation of 10 mg/dL. Within the study population 68% of the serum triglycerides values will lie within the limits of? Q1: A study of serum triglycerides levels among patients with history of myocardial infarction revealed that serum triglycerides is normally distributed with a mean of 200 mg/dL and standard deviation of 10 mg/dL. Within the study population 68% of the serum triglycerides values will lie within the limits of? A. B. C. D. 180 mg/dL to 210 mg/dL 170 mg/dL to 230 mg/dL 180 mg/dL to 220 mg/dL 190 mg/dL to 210 mg/dL Since 68% lie within mean + 1SD then the limits will be 200 + 10 = 190 mg/dL to 210 mg/dL SLIDE 2 Probabilities rules SLIDE 3 Probabilities rules • Probability – The probability (P) of an event A is written as P(A) –P(A)= Number of times A occurs / Number of times A can occur • Product(multiplication) rule – Used to estimate probability of two or more events being true – Depends on the assumption of independence or not • One probability is not influenced by the outcome of another probability i.e. P(A+|B+)= P(A+|B-) • Independence assumed – P(A and B)= P(A) x P(B) • Independence NOT assumed (used in Bayes theorem) – P(A and B)= P(A) x P(B|A) = P(B) x P(A|B) SLIDE 4 SLIDE 5 1 10/13/2016 Probabilities rules Probabilities rules • Conditional probability –P(B|A) = P(A and B)/ P(A) x P(B) • Addition rule – All possible different probabilities in situation must add to one – Mutually exclusive events • P(A or B)= P(A) + P(B) – Mutually inclusive events (non-mutually exclusive) • Modified addition rule • P(A or B or both)= P(A) + P(B) – P(A and B) SLIDE 6 Readiness assessment questions SLIDE 7 Readiness assessment questions Q2-3 : An investigator measured body mass index(BMI) of fitness club members. He reported that 15% were under-weight, 30% had healthy weight, and 15% are overweight, 25% are obese, and 15% extremely obese. The members were mostly men (75%). Q2-3 : An investigator measured body mass index(BMI) of fitness club members. He reported that 15% were under-weight, 30% had healthy weight, and 15% are overweight, 25% are obese, and 15% extremely obese. The members were mostly men (75%). Q2: What is the probability that a randomly selected member of the club members will be healthy or overweight? Explanation: Because are being healthy and being overweight are mutually exclusive events therefore we can use the addition rule for mutually exclusive events P(A or B)= P(A) + P(B)= 30%+15%=45% Q2: What is the probability that a randomly selected member of the club members will be healthy or overweight? A. 20% B. 25% C. 45% D. 50% SLIDE 8 Readiness assessment questions SLIDE 9 Readiness assessment questions Q2-3 : An investigator measured body mass index(BMI) of fitness club members. He reported that 15% were under-weight, 30% had healthy weight, and 15% are overweight, 25% are obese, and 15% extremely obese. The members were mostly men (75%). Q2-3 : An investigator measured body mass index(BMI) of fitness club members. He reported that 15% were under-weight, 30% had healthy weight, and 15% are overweight, 25% are obese, and 15% extremely obese. The members were mostly men (75%). Q3: The probability that a randomly selected member will be male and of healthy weight? Q3: The probability that a randomly selected member will be male and of healthy weight? A. .0009 B. 009 C. .09 D. . 225 Explanation: Because the probability of being male is independent of the weight, therefore we can use the multiplication rule to calculate the probability of the two independent events. P(A, and B)= P(A) x P(B)= .75 x .3=.225 S L I D E 10 S L I D E 11 2 10/13/2016 Variable: is any characteristic of a subject that can be measured. Scales 1. Nominal: Used for characteristics that have no numerical values, no measurement scales and no rank order. 2. Ordinal: Used for characteristics that have an underlying order to their values 3. Interval: It is used for characteristics that can be given numerical values that are equally spaced and has no true zero point. 4. Ratio: It is used for characteristics that can be given numerical values that are equally spaced and HAS true zero point. Types 1. Qualitative •Nominal •Ordinal?? 2. Quantitative •Ordinal •Interval •Ratio Readiness assessment questions Q4: A patient had undergone a cholecystectomy operation. Post-operatively, he complained of pain. The attending surgeon asked him to rate his pain on a scale of 0 (no pain) to 10 (the worst pain). The patient reported that his pain is a 9. After the administration of 100mg ketoprofen, the patient reported that the pain is now a 5 on the same scale. After the administration of morphine sulfate, given as an intravenous push, the pain is 0. This pain scale is a: A. B. C. D. Nominal scale Continuous scale Qualitative scale Ratio scale S L I D E 12 Readiness assessment questions S L I D E 13 Application Questions Q4: The pain scale is subjective and it might be arguable that a pain score of 6 is not quit as twice as a score of 3, therefore sometime pain scale is considered as ordinal scale. However, the pain scale has a true 0, indicating the absence of pain, therefore it could be considered as a ratio scale (i.e., for a continuous variable with a true 0 point). Although the pain scale is indeed continuous, it is a special and more specific case of continuous (i.e., ratio). Dichotomous scales are binary (only two options) and are a special case of nominal scales. Neither binary nor nominal applies to the 0-10 pain scale discussed in this question. Qualitative measures are completely devoid of objective scales by definition. Q1- The table below lists the results of blood type evaluation of 50 men and 50 women? Q1 Blood type Gender Male Female Totals O 20 20 40 A 17 18 35 B 8 7 15 AB 5 5 10 50 50 100 Totals S L I D E 14 S L I D E 15 Application Questions Application Questions Q1- What is the probability that an individual selected at random from the group has type O blood or is a man? Q1- What is the probability that an individual selected at random from the group has type O blood or is a man? A. B. C. D. Explanation: Sex and blood group are not mutually exclusive, therefore we will use the general addition rule 0.2 0.35 0.5 0.7 P(A or B)= P(A) + P(B) – P (A and B) =(40/100) + (50/100) – (20/100)=(70/100)=0.7 S L I D E 16 S L I D E 17 3 10/13/2016 Standard normal distribution (z) Standard normal distribution (z) • Properties: – Bell shape – Symmetric about the mean – Mean=median=mode – Mean=0 – Standard deviation=1 – The area under the curve = 1 • 68% within µ + 1σ • 95% within µ + 2σ • 99% within µ + 3σ • The normal distribution with mean 0 and standard deviation 1 • If the mean#0 and SD#1do z transformation allow using the standard normal table 𝑥−𝜇 – 𝑧= , where x is the value of the 𝜎 variable, µ is the mean, σ is the SD • A positive z means the value is above the mean • A negative z means the value is below the mean • If the z is known you can get the x – x= µ + zσ Graph generated by R Graph generated by R S L I D E 18 Application Questions Standard normal distribution (z) tables Q2- Data on RBC cholinesterase were collected from a healthy population. The data revealed that RBC cholinesterase had a mean of 11 µmol/min/ml and SD of 2 µmol/min/ml. What is the probability that an individual randomly selected from the population will have an RBC cholinesterase value between 11.95 and 13.95 µmol/min/ml? Z score Area under the curve to the left i.e. below z Areas under the standard normal curve (z scores) • Could be used to find proportion above ,below , or between any z scores • The first column includes the stem of the z value • The top row includes the second and third digit of the z value S L I D E 19 A. B. C. D. Negative z .9306 .6844 .2462 .3156 Positive z Source: http://image.slidesharecdn.com/copyofz-table-130515110049-phpapp02/95/copy-of-ztable-1 -638.jpg?cb=1368615687 S L I D E 20 Application Questions Application Questions Q2- It means we want the area between 11.95 and 13.95 RBC cholinesterase 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 S L I D E 21 18 19 20 21 22 S L I D E 22 • Solution: – Find the z score equivalent to 11.95 𝑥−𝜇 𝑧= = (11.95-11)/2= 𝜎 .95/2=.475=.48 – Find the z score equivalent to 13.95 𝑥−𝜇 𝑧= = (13.95-11)/2= 𝜎 2.95/2=1.475=1.48 – Find the probability that z < 0.48 0.6844 Find the probability that z < 1.48 0.9306 Then the area between 11.95 and 13.95 = P (.48 < z < 1.48) = 0.9306-0.6844=0.2462 S L I D E 23 4