18.05 Problem Set 9, Spring 2014 Solutions

Transcription

1 18.05 Problem Set 9, Sprig 2014 Solutio Problem 1. (10 pt.) (a) We have x biomial(, θ), o E(X) =θ ad Var(X) = θ(1 θ). The rule-of-thumb variace i jut 4. So the ditributio beig plotted are biomial(250, θ), N(250θ, 250θ(1 θ)), N(250θ, 250/4). Note, the whole rage i from 0 to 250, but we oly plotted the part where the graph were ot all 0. We otice that for each θ the blue dot lie very cloe to the red curve. So the N(θ, θ(1 θ)) ditributio i quite cloe to the biomial(, θ) ditributio for each of the value of θ coidered. I fact, thi i true for all θ by the Cetral Limit Theorem. For θ = 0.5 the rule-of-thumb give the exact variace. For θ =0.3 the rule-of-thumb approximatio i very good: it ha maller peak ad lightly fatter tail. For θ =0.1 the rule-of-thumb approximatio break dow ad i ot very good. I ummary we ca ay two thig about the rule-of-thumb approximatio: 1. It i good for θ ear 0.5 ad break dow for extreme value of θ. 2. Sice the rule-ofthumb overetimate the variace (the rule-of-thumb graph are horter ad wider) it give u a cofidece iterval that i larger tha i rictly eceary. That i a 95% rule-of-thumb iterval actually ha a greater tha 95% cofidece level. (b) Uig the rule-of-thumb approximatio, we kow that x i approximately N(θ, 1/4). For a 80% cofidece iterval, we have α =0.2 o z α/2 = qorm(0.9,0,1) = So the 80% cofidece iterval for θ i give by z 1 x 0. 2, x + z =0.5195,

2 18.05 Problem Set 9, Sprig 2014 Solutio For the 95% cofidece iterval, we ue the rule-of-thumb that z So the cofidece iterval i 1 1 x, x + =0.497, It okay to have ued the exact value of z Thi give a cofidece iterval: 1.96 x, x =0.498, (c) With prior Beta(1, 1), if oberve x ad the the poterior i Beta(x +1, x). I our cae x = 140. So, uig R we get the 80% poterior probability iterval: prob iterval = qbeta(0.1, 141, 111), qbeta(0.9, 141, 111) = , Thi i quite cloe to the 80% cofidece iterval. Though the two iterval have very differet techical meaig, we ee that they are coitet (ad umerically cloe). Both give a type of etimate of θ. Problem 2. (10 pt.) (a) We have =20adα =0.1 o t α/2 = qt(0.05,19) = Thu the 90% t-cofidece iterval i give by x t α/2, x + t α/2 = , Give that the ample mea ad variace are oly reported to 2 decimal place the extra digit are a puriou preciio. It i worth otig that to the give preciio the 90% cofidece iterval i 68.08, (The problem did ot ak you to do thi.) (b) We have z α/2 = qorm(0.05) = So the 90% z-cofidece iterval i give by x z α/2, x + z α/2 = , A i part (a) takig the preciio of the mea ito accout we get the iterval 68.16, (c) We eed uch that 2 z 0.05 / =1. So =(2 z 0.05 ) 2 = Sice you eed a whole umber of people the awer i = 154. (d) We eed to fid o that 2 t 0.05 / = 1. Becaue the critical value t 0.05 deped o the oly way to fid the right i by ytematically checkig differet value of. = 157 t05 = qt(0.95,-1) = width = (2*qrt(2)*t05/qrt()) = (very cloe to 1). (Our actual code ued a for loop to ru through the value = 130 to = 180 ad prit the width to the cree for each.) 2

3 18.05 Problem Set 9, Sprig 2014 Solutio We fid = 157 i the firt value of where the width 90% iterval i le tha 1. Thi i ot guarateed. I a actual experimet the value of 2 wo t ecearily equal If it happe to be maller the the the 90% t cofidece iterval will have width le tha 1. Problem 3. (10 pt.) (a) Theamplemeaix = 356. Sice z =1.96, = 3 ad = 9, the 95% cofidece iterval i x z 0.025, x + z = , (b) We have z 0.01 = qorm(0.99) =2.33. So the 98% cofidece iterval i x z 0.01, x + z 0.01 = , (c) The ample variace i 2 = (xi x ) 2 = var( ) = Sice = 9 the umber of degree of freedom for the t-tatitic i 8. Redo (a): t 8,0.025 = qt(0.975, 8) = So the 95% cofidece iterval i x t 8,0.025, x + t 8, , Redo (b): t 8,0.01 = qt(0.99, 8) = So the 98% cofidece iterval i x t 8,0.01, x + t 8, , Thee iterval are larger tha the correpodig iterval from part (a) ad (b). The ew ucertaily regardig the value of meaweeedlargeritervaltoachievetheame level of cofidece. Thi i reflected i the fact that the t ditributio ha fatter tail tha the ormal ditributio). Problem 4. (10 pt.) (a) Thi i imilar to problem 3c. We aume the data i ormally ditributed with ukow mea μ ad variace 2. We have the umber of data poit = 12. Uig Matlab we fid data = 6.0, 6.4, 7.0, 5.8, 6.0, 5.8, 5.9, 6.7, 6.1, 6.5, 6.3, 5.8; x i x = = mea(data) = (xi x ) 2 = = var(data) = c = qchiq(0.975, 11) = c = qchiq(0.025, 11) =

4 18.05 Problem Set 9, Sprig 2014 Solutio So the 95% cofidece iterval i ( 1) 2 c 0.025, ( 1) 2 = , c i our poit etimate for 2 ad the cofidece iterval i our rage etimate with 95% cofidece. (b) We have aumed that the plama choleterol level are idepedet ad ormally ditributed. Thi might ot be a good aumptio becaue choleterol for me ad wome might follow differet ditributio. We d have to do further exploratio to udertad thi. Problem 5. (10 pt.) (a) We have =10ad 2 =4.2 Aumig that the weight are ( 1) ormally ditributed with mea μ =52advariace 2, we kow that 2 χ We have c = qchiq(0.975, 9) = c = qchiq(0.025, 9) = The 95% cofidece iterval for i give by 2 ( 1), 2 ( 1) = , c c (b) I order to ue a χ 2 cofidece iterval we aumed that the weight of the pack of cady are idepedet ad ormally ditributed with mea 52 ad variace 2. 4

5 MIT OpeCoureWare Itroductio to Probability ad Statitic Sprig 2014 For iformatio about citig thee material or our Term of Ue, viit: