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# Introduction to Mathematical Statistics APPM 5520

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This 9 page Study Guide was uploaded by Dr. Filomena Hegmann on Thursday October 29, 2015. The Study Guide belongs to APPM 5520 at University of Colorado at Boulder taught by Jem Corcoran in Fall. Since its upload, it has received 55 views. For similar materials see /class/231868/appm-5520-university-of-colorado-at-boulder in Applied Math at University of Colorado at Boulder.

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Date Created: 10/29/15

H 10 r gt 01 905 Review Problems for Exam II Sections 7172 fw secac2 7 5dw 4t37t216t f WC 1 f 1Cos z dw 02 178205 I dw f 1029 0190 far4 1 i2nzz dw 3 sin 26016 f sin1n w dw d f 82 APPM 45520 fsirf1 zdz ffwg lnwdw fcsc wdw f Cot 1n 12 dw z 1444 v1 tan2wdw fe z coswdw fw barf1 wdw f 33 3x dw ff 1n wdw 12 dw APPM 45520 Mome Review Problems for Exam III For questions 35 38 determine whether the series is conditionally convergent absolutely convergent or divergent 35 Eilem ln lg 36 Eggnnilw 37 yaw 38 co VVT 711 ln n For questions 39 42 nd the sum of the series 22n1 39 231 5T 40 371 We 41 f1tan 1n 1 7 tan 1 n 42 220 43 Express the repeating decimal 12345345345 as a fraction 44 For What values of w does the series Zill n w converge 45 Find the sum of the series 221 7 73 correct to four decimal places 46 a Show that the series 211 is convergent b Deduce that limnnm 0 4 K1 Prove that if the series 211 an is absolutely convergent then the series Willi n1 is also absolutely convergent For questions 48 49 nd the radius of convergence and interval of convergence of the given series 7 n 2 48 220 00 n 49 2711 gins 50 Find the radius of convergence of the series 51 Find the Taylor series expansion of ne sin x at c 7r6 52 Find the Taylor series expansion of ne cos x at c 7r3 For questions 53 607 nd the Maclarin series for f and its radius of convergence Use either the direct method or the manipulation of known series Note You might nd it easiest in some cases to use the binomial series7 even though it is the subject of Section 811 53 M 54 far xliwz 55 f w ln17w 56 far wezz 58 f 59 far 1316753 gt gt gt 57 fwsinw4 gt gt 60 fw173w5 1 I 9quot r gt APPM 45520 Final Exam Review Problems 971 7 Consider a random sample X1 Xn from a distribution with pdf fw 61 7 w 0 lt w lt 1 a gt 0 Give the form of the GLRT for testing H7 6 1 against H1 6 7 1 Suppose that X N binn1p1 and Y N binn2p2 and that we Wish to test whether the proportions are equal H02p1p2p versus 11111213175132 where p is unknown a Give the GLR statistic b Since your GLR does not simplify very well give an approximate GLRT of size 04 based on large sample sizes Let X1 X2 Xn be a random sample from the Nu02 distribution where 72 is known a Derive the GLRT of size 04 for testing H0 u p0 versus Ha u 7 MD b How does your GLRT compare to the standard common sense two tailed test we had for this problem towards the begining of the course c Show that the asymptotic approximation for this test based on the chi squared distri bution is exact in this case Let X1X2 Xn be a random sample from the two parameter exponential distribution with pdf 1 7 M M 7 Xe I 9 lime for A gt O and 700 lt 6 lt 00 Note that l have changed the role of A in order to simplify some upcoming computations Let A and be the MLEs of A and 6 One can show that A and are independent This is a result of a theorem known as Basu s Theorem77 which can be found in most any mathematical statistics textbook a Let W1 27137 6A W2 271637 6A and W3 27AM Show that W1 N van W2 N X22 and W3 N X22n 7 2 Hint Note that W1 W2 W3 b If V1 N X2011 and V2 N X2n2 are independent random variables the random variable Vii711 V27l2 is said to have an F distribution with 711 and 712 degrees of freedom We write F N F011 712 What is the distribution of it 7 1 7 6A7 c Find the GLR for a test of H0 6 60 versus 6 gt 60 d Show that the GRLT is equivalent to rejecting H0 if n 7 7 60gtS is larger than an F critial value Give a notation for that critical value 5 Let X1X2X3X4 N0 1 Show that 4 2 2 i X 7 X X 7 X X 2 X 7 X X X 3 2XiiXgt21 2l3 12ll4 12 3lr 7 2 32 43 271 Argue that the three terms on the righthand side are independent and nd the distribution of each H 10 U1 9 T 90 J APPM 45520 Review Problems for Exam II Sections 7374 f dw f dy f d9 f 7 dz Solve x2 4 3L 3 ya 0 14 Zdz 112 4z f t3 dt 1t24 fxinzdw APPM 45520 Review Problems for Exam II Sections 5657 Note Sections 56 and 57 are about surface area and centers of massi the resulting integrals in this review may use techniques that appeared later in Chapter 7 H 10 9 r gt U1 9 T 00 Find the area of the surface generated by revolving the curve 3 sinw for O S x S 27r about the w aXis Find the area of the surface generated by revolving the curve 3 32 for O S x S 1 about the w aXis Find the area of the surface generated by revolving the curve 3 lnw for 1 S x S 6 about the line 3 71 Find the area of the surface generated by revolving the curve w M431 7 32 for 1 S y S 2 about the y aXis Find the center of mass of a thin plate of constant density 6 bounded by y 2sin2w and the line 3 O on the interval 0 S x S 7r2 Find the center of mass of a thin plate of constant density 6 bounded by the curves 3 3 andy Find the center of mass of a thin plate bounded by the curves 3 4Zand y 74 and the lines w 1 and w 4 if the plate s density at the point 5331 is 690 lw Find the center of mass of a thin plate bounded by the curves 3 and y i and the line 3 sic2 if the plate s density at the point 5331 is 6y y 1 APPM 45520 Solutions for Review Problems for Exam II Sections 7374 6 Let x sec 6 Then 32 7 1 sec2 6 7 1 tan2 6 and dye sec 6 tan Ode The integral becomes 6 t 6 6 sec an sec d6 d6 tan4 6 tan3 6 Yuchl Could we play around with trig identities until we make this work Yup But we could try partial fractions too7 don t forget about this option 1 1 A B C D wlw7l2 w12w712 mwl2 w7l2 The solution is A 147 B 14C 7147 and D 14 So the integral becomes 1 1 1 1 1 7 7d 7d 7 d 7d 4lx1 wwl2 7 v71 ww712 7 Now each one is a simple u substitution The answer is dw l l l W1lnwll7wl 71nw717w71l 0 7 Let u cos 6 Then du 7 sin GdO and the integral becomes 61 712 u 7 2 Now it s a partial fractions problem The answer is 1 u 2 1 cos 6 2 gm u71i07 1n cos 7li0 8 Let u 2E Then du l and the integral becomes 1 2 du 1N 1 u2 7 71 1 7 31 if an lwg 2lt6 4gt 6 3 Let t 2tan 6 for 77r2 lt 67r2 Then dt 2sec2 GdO and t2 4 4tan2 64 4sec2 6 The integral becomes 8tan3 6 2sec2 GdO 3 gsece 78tan OseCOdO sin3 6 sin23 6 sin 6 1 7 C082 6 sin 6 TSCos46d678 C0846 deis C0846 d6 Let u cos 6 Then du 7 sin GdO and the integral becomes 17712 71271 1 1 78 u4 du8 u4 du8u727gdu 1 1 1 1 1 1 ST f C8H c8 mm0 12 2 32 87secesec3 6 C87tT4 C 10 Let x sin 6 for 77r2 S 6 S 7r2 Then dw COS 6 and 17 932 C082 6 The integral becomes 1 1 VC082 gCOSGd9COSG gCOS since 77r2 S 6 S 7r2 1 2 1 1Cos2 1 gCOS Odeigfdei lcos26d 1 1 1 1 E6 sm26 CE sin C0s6C sin13x3xw179x2 0

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