Math 10B: Mock Mid II. April 13, 2016

Transcription

1 Name: Soluions Mah 10B: Mock Mid II April 13, ( poins) Sae, wih jusificaion, wheher he following saemens are rue or false. (a) If a 3 3 marix A saisfies A 3 A = 0, hen i canno be inverible. True. Noe ha A 3 = A de(a 3 ) = de( A) [de(a)] 3 = de(a) [de(a)] 3 de(a) = 0 de(a)([de(a)] 1) = 0 de(a) = 0. (b) A compued 95% confidence inerval for a populaion parameer has a 95% probabiliy of conaining he rue value of he parameer. False. A common misconcepion. We can only say ha if a large number of confidence inervals are consruced using samples of he same size, 95% of hem will conain he rue value. (c) If he produc of wo marices A and B is he ideniy marix, hen A and B are inverses of one anoher. False. The marices A and B could be non-square and hence no inverible. (d) The populaion sandard deviaion is never larger han he sample sandard deviaion. True. The sample SD s is relaed o he populaion SD s by s = s n n 1 so s s. (The equaliy comes ino play when boh are zero) (e) If he produc of wo marices A and B is he zero marix, hen a leas one of A and B is no inverible. True. If eiher of A or B is non-square, i canno be inverible. If boh are square, hen AB = 0 implies ha de(ab) = de(0) de(a) de(b) = 0 de(a) = 0 or de(b) = 0. ( ) 1 (f) The marix A = is inverible for all real values of. 1 False. We have de(a) = ( 1)( 1) = 1 = ( 1) which is zero for = 1. 1

2 . (4 6 poins) Find he general soluions of he following differenial equaions. (a) y () 8y () 16y() = 0. The corresponding auxiliary equaion is r 8r 16 = 0. This can be facorized o yield (r 4) = 0 r = 4, 4. Since he auxiliary equaion has a repeaed real soluion, we infer ha for some consans C 1 and C. y() = C 1 e 4 C e 4 (b) ( ) y () y() =. Noe ha since his equaion is nonlinear, we canno apply he inegraing facor mehod. Rearranging his equaion however gives ( ) y () = (y() 1) y () = ( ) (y() 1). This is a separable equaion. Rearranging and inegraing boh sides yields 1 dy = y 1 d ( ) 1 y 1 dy = d. (1) ( ) Recall ha 1 dy = arcan(y). For he oher inegral, we need o use parial y 1 fracions; we have ( ) = A ( ) B ( ) = A B( ). Plug in = o infer ha A = and hence B = 1. I follows ha ( ) d = ( ) 1 d = ln C where C is some consan. Plug hese in (1) o ge arcan(y) = ( ) ln C y() = an ln C.

3 3. (3 4 poins) In a random sample of 00 individuals, he handedness and eye-color was noed. The resuls are displayed below. Lef Righ Blue 18 6 Brown 98 (a) From he resuls above, consruc a 95% confidence inerval for he proporion of lef-handed individuals. Noe ha he proporion of lef-handed individuals in he given sample is ˆp = = 0. so a 95% confidence inerval for he rue proporion is [ ] (0.)(0.8) (0.)(0.8) 0., [0.1434, 0.566]. (b) Consruc a able showing he expeced frequencies under he null hypohesis H 0 ha handedness and eye-color are independen. Lef Righ Toal Blue Brown Toal (c) Explain how o use he ables of observed and expeced frequencies o carry ou a χ es. Under wha circumsances will we rejec he null hypohesis? We firs compue he value of he goodness-of-fi saisic R ha measure he deviaion beween expeced and observed frequencies: r = (O E) E = (18 16) 16 ( 4) 4 (6 64) 64 (98 96). 96 Nex, we need o compue he probabiliy ha he goodness-of-fi saisic R is a leas r. This probabiliy is ermed he p-value. The p-value can be found by using he χ disribuion wih degree(s) of freedom ν = ( 1)( 1) = 1. The p-value is equal o he area beneah he χ 1 curve and o he righ of r. The compued p-value now needs o be compared o he significance level of he es. Typically, his level is se a If he p-value is below he significance level, he null hypohesis is rejeced in favor of he alernaive hypohesis, namely, ha handedness and eye-color are no independen. On he oher hand, if he p-value is above he significance level, we do no rejec he null hypohesis. 3

4 4. (7 poins) A fair dice is rolled en imes. Find he expeced number of imes an even number is followed by an odd number. Define he random variables X i by { 1 he ih roll is even and he (i 1)s roll is odd X i = 0 oherwise for i = 1,,..., 9. Observe hen ha P(X i = 1) = (3/6)(3/6) = 1/4 so ha E[X i ] = 1(1/4) 0(3/4) = 1/4. Le Y = X 1 X... X 9. Noe ha Y couns precisely he occurrences of an even number followed by an odd number. Using he lineariy of expecaion, we have E[Y ] = E[X 1 X... X 9 ] = E[X 1 ] E[X ]... E[X 9 ] = 9E[X 1 ] =

5 5. (4 6 poins) A can is placed in a refrigeraor. The can is iniially a 33 C while he refrigeraor s emperaure is held consan a 3 C. (a) Le T n be he emperaure of he can n minues afer i was refrigeraed. Assume ha he T n s obey T n1 = 0.98T n 0.06 for n 0. Find he emperaure afer wo hours. Noe ha his is an inhomogeneous linear recursive relaion; i has he soluion ( ) n T n = (0.06) (0.98) n T ( ) The emperaure afer wo hours (= 10 minues) is T 10 = 10 (0.06) (0.98) 10 = ( )( 33(0.98) 10 = 3 30(0.98) C (b) Wrie a differenial equaion for he can s emperaure, assuming ha is rae of change is proporional o is difference from ha of he refrigeraor, Le C() be he can s emperaure minues afer refrigeraion. We hen have C () = k(c() where k is some consan. We also have C(0) = 33. (c) Afer an hour inside, he can s emperaure is down o 13 C. Solve he equaion from (b) o find he emperaure afer wo hours. We have C () kc() = 3k. The inegraing facor is herefore I() = e k d = e k so muliplying I() in he differenial equaion, we ge C ()e k kc()e k = 3ke k d d (C()e k ) = 3ke k C()e k = 3ke k d = 3k e k ( k) A = 3e k A C() = 3 Ae k. Plug in C(0) = 33 o ge 33 = 3 A A = 30. Nex, use C(60) = 13 o ge 13 = 3 30e 60k e 60k = 1 3 ek = ( 1 1/60. We herefore have C() = 3 30(e k ) = 3 30 ( 1 /60. The emperaure afer wo hours hen is C(10) = 3 30 ( 1 10/60 ( = ) 3 = 3 30 = C. 5