Exam 2. Instructor: Cynthia Rudin TA: Dimitrios Bisias. October 25, 2011
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1 Exam 2 Istructor: Cythia Rudi TA: Dimitrios Bisias October 25, 2011 Gradig is based o demostratio of coceptual uderstadig, so you eed to show all of your work. Problem 1 You are i charge of a study that compares how two weight-loss techiques (Diet ad Exercise) affect the weight loss of overweight patiets. The table below shows the umber of obese ad severely obese people who lost sigificat weight (successes) ad the umber of people who did t lose ay weight (failures) for both of the weight-loss techiques. Obese Severely Obese Diet successes Diet failures Exercise successes Exercise failures Withi each category of obesity, compare success rates for the weight-loss techiques. What do these rates idicate? 2. Compare the total success rates for the two techiques. Explai ay differeces from (1). Solutio 1. For the obese category the success rate for diet is 10/40 = 25% while the success rate for exercise is 22/80 = 27.5%. For the severely obese category, the success rate for diet is 60/80 = 75% while the success rate for exercise is 35/40 = 87.5%.These figures show that exercise has higher success rates for both of the categories. 2. The total success rate for diet is 70/120 = 58.3% while the total success rate for exercise is 57/120 = 47.5%. This cotrasts with the result of part (1). This is a example of Simpso s paradox. The lurkig variable is type of obesity. There are far less obese people i case of diet tha exercise, who have the lower success rates. 1
2 Problem 2 You are the ower of a bakery sellig cheesecakes. Some days you sell a lot of cheesecakes ad some other days you do t, idepedetly of other days. I fact, the distributio of the umber of cheesecakes you sell each day is show below. If we measure cheesecakes sales for 100 days, what is the approximate probability that the average cheesecakes sales (over the 100 days) will be more tha 102 cheesecakes? Solutio Let X be the umber of cheesecakes sold i oe day. The due to symmetry E[X] = 100 ad var(x) = = 175 We are asked to fid P ( X > 102). From CLT we kow that X is approximately ormally distributed with mea 100 ad variace 175 = 1.75, therefore we have: P ( X > 102) P (Z > 1 ) = 1 Φ(1.51) = /100 2
3 Problem 3 Suppose that the observatios X 1, X 2,, X are iid with commo mea θ ad kow variace σ 2. Cosider the followig estimator of the mea θ: X 1 + +X Θˆ = What is the bias of this estimator? What happes to the bias as we icrease the size of the sample? 2. What is the MSE of the estimator? Hit: Remember to use either what you kow about V ar(x ), or what you kow about the variace of the sum of idepedet radom variables. Solutio ˆ 1. It is: Θ = +1 X. Therefore, ad Thus, we have: As the bias goes to 0. E[Θˆ ] = E[X ] = θ var[θˆ ] = var[x ] = σ 2 ( + 1) 2 ( + 1) 2 1 Bias = E[Θˆ ] θ = θ θ = θ It is: MSE(Θˆ ) = Bias 2 + var(θˆ ) = + σ 2 ( + 1) 2 ( + 1) 2 θ 2 3
4 Problem 4 The weight of a object is measured eight times usig a electroic scale that reports the true weight plus a radom error that is ormally distributed with zero mea ad variace σ 2 = 4. Assume that the errors i the observatios are idepedet. The followig results are obtaied: {4.45, 4.02, 5.51, 1.10, 2.62, 2.38, 5.94, 7.64} Compute a 95% cofidece iterval for the true weight. Solutio A 95% cofidece iterval for the true weight is σ σ 2 2 [X 1.96, X ] = [ , ] = [2.82, 5.59] 8 8 4
5 Problem 5 The returs of a asset maagemet firm i differet years are idepedet ad ormally distributed with ukow mea ad variace. The asset maagemet firm claims that the stadard deviatio of the returs is as low as σ = 2% ad the mea of the returs is µ = 22%. You believe that this is too good to be true. To verify your suspicio, you take the returs from the last 10 years. These are: r = {20.6, 19.2, 17, 19.1, 18.7, 22.5, 27.2, 17.9, 22.5, 21.3} (To help you with the calculatios, we give you that: i=1 r(i) = 206 ad (r(i) r )2 = 79.34) 1. Calculate the sample mea ad stadard deviatio of the above radom sample. 2. What is the probability that whe we draw a ew radom sample of size 10, its sample mea will be below the oe you calculated i part 1? Assume that the compay s claims are correct. 3. What is the probability that whe we draw a ew radom sample of size 10, its sample stadard deviatio will be larger tha the oe you calculated i part 1, assumig that the compay s claims are correct? Solutio = i=1 i=1 r(i) i =1(r(i) r ) The sample mea is: r = = 20.6 ad the sample stadard deviatio is s = = 2. Uder the compay s claims the sample mea is ormally distributed with mea 22% ad stadard deviatio 2/ 10%. It is: P ( X < 20.6) = P (Z < ) = P (Z < 2.21) = / Uder the compay s claims the statistic: ( 1)S 2 is distributed as a chi-square with -1=9 degrees of freedom. Therefore: σ 2 ( 1)S P (S > 2.97) = P ( > ) = P (χ 9 2 > 19.85) = σ 2 4 5
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