Homework for 4/9 Due 4/16

Transcription

1 Name: ID: Homework for 4/9 Due 4/16 1. [ 13-6] It is covetioal wisdom i military squadros that pilots ted to father more girls tha boys. Syder 1961 gathered data for military fighter pilots. The sex of the pilots offsprig were tabulated for three kids of flight duty durig the moth of coceptio, as show i the followig table. Father s Activity Female Offsprig Male Offsprig Flyig Fighters Flyig Trasports Not Flyig a. Is there ay sigificat differece betwee the three groups? Use α b. I the Uited States i 1950, males were bor for every 100 females. Are the data cosistet with this sex ratio? Use α Hit: this is similar to the authorship example. We are comparig pilots with geeral males. Thus we eed to combie all pilots i order to make a compariso. a. First, we ca fid the totals: Father s Activity Female Offsprig Male Offsprig Total Flyig Fighters Flyig Trasports Not Flyig Total Let the ull hypothesis be that there is o differece betwee these groups. The if H 0 is true, the expected couts for these groups would be Father s Activity Female Offsprig Male Offsprig Flyig Fighters Flyig Trasports Not Flyig Thus the Pearso s χ 2 test statistic is X

2 Moreover, the distributio of X 2 is approximately χ 2 2 df Sice α 0.05, the rejectio regio is R {X 2 > 5.99} 5.99 χ ,2. Sice < 5.99, we do ot reject H 0. I other words, there is o sigificace differece betwee the groups. b. I this part, we compare pilots with geeral males. Thus we combie the data for pilots. Father Female Offsprig Male Offsprig Total Pilot Geeral male Total Let the ull hypothesis be that there is o differece betwee pilots ad geeral males. The if H 0 is true, the expected couts for pilots ad geeral males would be Father Female Offsprig Male Offsprig Pilot Geeral male Ad the Pearso s χ 2 test statistic is X The distributio of the test statistic is approximately χ 2 1 df Sice α 0.05, the rejectio regio is R {X 2 > 3.84} 3.84 χ ,1. Sice < 3.84, we do ot reject H 0. I other words, there is o sigificace differece betwee pilots ad geeral males, that is, the data is cosistet with this sex ratio. 2

3 2. [ 13-16] A market research team coducted a survey to ivestigate the relatioship of persoality to attitude toward small cars. A sample of adults i a metropolita area were asked to fill out a 16-item selfperceptio questioaire, o the basis of which they were classified ito three types: cautious coservative, middle-of-the-roader, ad cofidet explorer. They were the asked to give their overall opiio of small cars: favorable, eutral, or ufavorable. Is there a relatioship betwee persoality type ad attitude toward small cars? Use α Persoality Type Attitude Cautious Midroad Explorer Favorable Neutral Ufavorable We first fid the totals. Persoality Type Attitude Cautious Midroad Explorer Total Favorable Neutral Ufavorable Total Let the ull hypothesis be that there is o relatioship, that is, Persoality ad Attitude are idepedet. The if H 0 is true, the expected couts for pilots ad geeral males would be Persoality Type Attitude Cautious Midroad Explorer Favorable Neutral Ufavorable Ad the Pearso s χ 2 test statistic is X The distributio of the test statistic is approximately χ 2 4 df Sice α 0.05, the rejectio regio is R {X 2 > 9.49} 9.49 χ ,4. Sice > 9.49, we reject H 0. I other words, there is some relatioship betwee persoality type ad attitude toward small cars. 3

4 3. [ 14-2] For the followig data: x y a. Fit a lie y a + bx by the method of least squares. b. Fit a lie x c + dy by the method of least squares. a. We have x i 0.46, yi , x 2 i , x i y i 9.452, y i 0.75 ad 10. Thus, by the method of least squares y i x i x i y i a x 2 i , x i y i x i y i b Thus the lie is y 0.904x

5 b. We iterchage the role of x ad y. x i y i y i x i c y 2 i 2 yi 2 y i , y i x i y i x i b 2 yi 2 y i Thus the lie is x 1.055y

6 4. [ 14-10] Show that the least squares estimates of the slope ad itercept of a lie may be expressed as ad ˆβ 1 ˆβ 0 ȳ ˆβ 1 x x i xy i ȳ. x i x 2 Hit: begi with ˆβ 1 ad expad x i xy i ȳ ad x i x 2. We will use the followig idetities several times: x i x ad y i ȳ. First we have x i xy i ȳ x i y i x i ȳ xy i + xȳ x i y i xȳ xȳ + xȳ x i y i ȳ x i x y i + xȳ x i y i xȳ 1 x i y i x ȳ [ 1 ] x i y i x i y i. 6

7 Similarly, x i x 2 x 2 i 2x i x + x 2 x 2 i 2 x 2 + x 2 x 2 i 2 x x i + x 2 x 2 i x 2 1 x 2 i x I fact, we ca save the calculatio by usig the first result ad replace y with x. Therefore, x i xy i ȳ x i x 2 [ 1 1 ] x i y i x i y i 2 x i y i x i y i 2 ˆβ1. 7

8 Fially, we have x 2 i y i x i x i y i ˆβ 0 2 x 2 i ȳ ȳ ˆβ 1 x. y i 1 2 x i y i x i y i x i x i y i x 2 i x 2 i [ x i y i x i y i 1 ] x i y i x i 1 2 x i ȳ 2 [ x i y i 1 ] x i y i x 2 x i y i x i y i 2 Oe ca also go backwards. x 8

9 Name: ID: Homework for 4/11 Due 4/16 1. [ 13-17] Let X ad Y be radom variables with E[X] µ x Var[X] σ 2 x E[Y ] µ y Var[Y ] σ 2 y Cov[X, Y ] σ xy Cosider predictig Y from X as Ŷ α + βx, where α ad β are chose to miimize E[Y Ŷ 2 ], the expected squared predictio error. a. Show that the miimizig values of α ad β are β σ xy σ 2 x α µ y βµ x Hit: E[Y Ŷ 2 ] E[Y ] E[Ŷ ]2 + Var[Y Ŷ ]. Sectio 4.3 may be helpful. Especially, Theorem A, Corollary A, ad Corollary B. b. Show that for this choice of α ad β Var[Y ] Var[Y Ŷ ] Var[Y ] where r xy is correlatio betwee X ad Y : r xy r 2 xy, Cov[X, Y ] Var[X]Var[Y ]. a. First we have E[Ŷ ] E[α + βx] α + βe[x] α + βµ x, Var[Ŷ ] Var[α + βx] β2 Var[X] β 2 σx, 2 ad Cov[Y, Ŷ ] Cov[Y, α + βx] βcov[y, X] βσ xy. Thus Var[Y Ŷ ] Var[Y ] + Var[Ŷ ] 2Cov[Y, Ŷ ] σ2 y + β 2 σx 2 2βσ xy. I order to miimize E[Y Ŷ 2 ], we otice that E[Y Ŷ 2 ] E[Y ] E[Ŷ ]2 + Var[Y Ŷ ] µ y α βµ x 2 σy 2 + β 2 σx 2 2βσ xy, fα, β.

10 Furthermore, f α 2µ y α βµ x, f β 2µ xµ y α βµ x 2σ 2 xβ 2σ xy α 2 2, α β 2 f β α 2µ x. β 2 2µ2 x 2σ 2 x, ad The solutio to { 2µ y α βµ x 0 2µ x µ y α βµ x 2σ 2 xβ 2σ xy 0 is Sice α 2 β α α µ y βµ x α β β 2 β σ xy σx µ x 2µ x 2µ 2 x 2σx 2 4σ2 x > 0, we see that fα, β achieve its miimum at α µ y βµ x β σ xy σx 2. b. From a, we immediately have Var[Y ] Var[Y Ŷ ] Var[Y ] σ2 y σ 2 y + β 2 σ 2 x 2βσ xy σ 2 y β 2 σ2 x σy 2 + 2β σ xy σy 2 r 2 xy. σxy σ 2 x 2 σ2 x σ 2 y + 2 σxy σ 2 x σxy σ 2 y σ2 xy σ 2 x σ 2 y 10

11 2. [ 13-18] Suppose that Y i β 0 + β 1 x i + e i, i 1,..., where the e i are idepedet ad ormally distributed with mea zero ad variace σ 2. Fid the mle s of β 0 ad β 1 ad verify that they are the least squares estimates. Hit: Uder these assumptios, the Y i are idepedet ad ormally distributed with meas β 0 + β 1 x i ad variace σ 2. Write the joit desity fuctio of the Y i ad thus the likelihood. Sice e i s are i.i.d. N0, 1 radom variables, we have Y i Nβ 0 + β 1 x 1, σ 2 ad Y i s are idepedet. Let f i y i be the pdf of Y i. We have { 1 f i y i β 0, β 1 exp 1 } 2πσ 2 2σ 2 [y i β 0 + β 1 x i ] 2. Sice Y i s are idepedet, the joit pdf of Y 1,..., Y is fy 1,..., y 2 β 0, β 1 f i y i β 0, β 1. Correspodigly, the likelihood fuctio ad log-likelihood fuctio are likβ 0, β 1 where Sβ 0, β 1 f i β 0, β 1 Y i, lβ 0, β 1 log likβ 0, β 1 ad log f i β 0, β 1 Y i log 2πσ 2 1 2σ 2 [Y i β 0 + β 1 x i ] 2 log 2πσ 2 1 2σ 2 [Y i β 0 + β 1 x i ] 2 log 2πσ 2 1 2σ 2 Sβ 0, β 1, [Y i β 0 + β 1 x i ] 2. It follows that the miimizer of Sβ 0, β 1 is the maximizer of lβ 0, β 1. Therefore, the mle for β 0 ad β 1 11

12 are the least square estimates: x 2 i Y i x i x i Y i ˆβ 0 2 x 2 i x i x i Y i x i Y i ˆβ