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fall_11_midterm_2_review

# fall_11_midterm_2_review - Concepts you need to know for...

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Unformatted text preview: Concepts you need to know for midterm two Fall 2 011 / Professor Esfandiari Chapter 12 Kaiser Foundation has 100 hospitals in four states. There are a total of 10,000 doctors and 50,000 nurses in the four states. Number of Number of hos - itals doctors nurses _ Two 2500 12,500 Suppose that hospital number 10 in state one has 1500 nurses. Picking a random sample of 150 nurses from the 1500 is a simple random sample. Picking a one percent random sample from the doctors and nurses in the four in the four states is a stratified random sample. We will be randomly picking 30 doctors and 150 nurses from the ten hospitals in state one, etc. Randomly picking one of the ten hospitals in each state and having all of the doctors in that hospital participate in the study is a cluster random sample. Chapter 13 Let us say that we use the data given in the example under chapter 12 Observational study: Comparing the attitude of a random sample of male and female doctors towards the work ethic of nurses is an observational study (no intervention or treatment), no control group, no causal inference is possible. Experimental study: Randomly assign a group of nurses from state one into two groups. Have group one learn a new procedure through participation in a hands-on workshop offered by an expert and the other group through participation in the same workshop in addition to watching procedure done online two times. At the end of the training, observe the nurses do the procedure and rank their effectiveness on a scale of one to ten. There is random assignment , there is a control and experimental group. It is possible to draw causal conclusions. Randomized block design: In state three, first block the pediatricians by gender, then randomly assign them to participate in two trainings (reading a chapter on normal distribution, vs. reading the chapter and participation in a one-hour talk by a statistician) on how to use the statistical tables and norms to communicate the weight and height of the newborns to the parents. Then ask the parents to tell you on a scale of one to ten how well they understood what the doctors communicated to them. Chapter 14 Disioint events: Suppose in one of the hospitals you have 100 doctors 30 of whom are internists and 20 are heart specialist and 50 are other specialties. These are disjoint or mutually exclusive events. Overlapping events: Suppose you have 20 doctors, 10 are internists, 10 are pediatrics and two are both. Chapter 15 Suppose you are given the following information about the doctors in one of the above hospitals Is there a relationship between gender and type of specialization? Visual answer given by looking at row or column percentages or segmented bar charts. If the two events are independent, then P(A| B) = P(A) and this should be true for all the cells. Example: P (male I orthopedics) = P (male) In the above example : P(male lorthopedic) # P(male) P(80|100) # 130/200 Also if two events are independent P(Aﬂ B) = P(A) * P(B) For the above P(malenheart) = P(male) * P(heart) 50/200 # 130/200*100/200 1/: # 13/40 if the events were inde - endent — P(maleﬂ heart) = P(male) * P(heart) 50/200 = 100/200*100/200 1/; = 1/4 P(male lorthopedic) =P(male) P(50/100) =(100/200) Chapter 16 Suppose the discrete random event you are interested in is getting the right answer to two multiple—choice questions about which you know nothing. The probability model would be as follows: (x—M)A2* P(x) riht 0.25 %*2= 0.50 1/2=0.50 riht RW riht 0.25 1/2=0.50 E(X) or M = M2 = (X—u)"2* P(x) = 2( X * p(x)) = 0.50 1.00 Check calculation: Mean = n*p = 2*1/2 = 1 (n = number of questions) Variance = n*P*(1—p) = 2 1/21/2= 2/4 = 0.50 Chapter 17 We will use the following table to discuss probability tree £(heart surgery) = 0.50 ~__; p(Mn HS) = 0.5*0.5 = 0.25 P (Male) = 0.5 x \\.\ ‘3. ll “P(orthopedic surgery) = 0.50...) p(Mn OS) = 0.5*0.5 = 0.25 P(heart surgery) = 0.50 9 P(Fﬂ HS) = 0.5*0.5 = 0.25 . ,7? f P (Female) = 0.50 ﬁg ll V} P(orthopedic surgery) = 0.50 .7) p(Fﬂ HS) = 0.5*0.5 = 0.25 We will use this example to show binomial events Suppose the discrete random event you are interested in is getting the right answer to two multiple—choice questions about which you know nothing. What is the probability that you will get: ' Exactly one right answer. ' At least one right answer. - At most one right answer. ° N = number of questions 2 2 ° K = number of correct answers at random Possible Probability of Probability of number of correct the binomial correct answer event answers 2c 0 = 1 1/1*1/2 = 1*4: 1/4 ww 1 4 2C1 =2 1/2*1/2=14 2*1/4: 1/2 WR RW 2C2 1/2*1/2—1/4 1*%=1/4 RR _ W= wrong R: right P® = 1/2 P(W) = 1/2 N c k = n!/[k! * (n-k)!] ° Exactly one right answer 2 p(k = 1) = 14 ° At least one right answer = p[k = 1) + p(k = 2) = 14 + 1/2 0 At most one right answer = p(k = 0) + p(k = 1) = 1/2 + 1/4 ...
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