MS exam problems Fall 01 (From: Rya Mart) 1. (Stat 401) Cosder the followg game wth a box that cotas te balls two red, three blue, ad fve gree. A player selects two balls from the box at radom, wthout replacemet. The player ws $5 for each red ball selected, $1 for each blue, ad $0 for each gree. Let deote the player s total wgs. (a) Fd P( 4). (b) Fd P( 4 > 0) (c) Fd E().. (Stat 411) You have a co whch you would lke to test for faress. Let θ (0, 1) deote the probablty of the co ladg o heads, ad cosder testg H 0 : θ = 0.5 versus H 1 : θ < 0.5. You toss the co utl you see the frst head. Let deote the umber of tosses. (a) Show that the uformly most powerful test rejects H 0 f c, where c > 1 s some costat to be determed. (b) Wthout loss of geeralty, the costat c ca be a teger. Fd c such that the sgfcace level s o more tha a specfed α (0, 1). Solutos 1. Start by wrtg the PMF table: x 0 1 5 6 10 p (x) c ( ) 5 c ( 5 3 ) 1)( c ( ) 3 c ( 5 ) 1 1)( c ( 3 ) 1 1)( c ( ) 1 where c = ( 10 ) 1. After evaluatg the bomal coeffcets, the table looks lke: Now the rest s easy. x 0 1 5 6 10 p (x) 10/45 15/45 3/45 10/45 6/45 1/45 (a) P( 4) = p (5) + p (6) + p (10) = 17/45. (b) Usg defto of codtoal probablty: P( 4 > 0) = P( 4) P( > 0) = 17/45 1 10/45 = 17 35. (c) E() = 1 (15 + 6 + 50 + 36 + 10) = 117/45 =.6. 45 1
. (a) Fx θ 1 < 0.5 ad cosder the smple hypothess testg problem H 0 : θ = 0.5 versus H 1 : θ = θ 1. The Neyma Pearso lemma states that the most powerful test s oe that rejects whe the lkelhood rato L(0.5)/L(θ 1 ) s too small. But the lkelhood rato ca be wrtte as L(0.5) L(θ 1 ) = (1/) ( 1/ ). θ 1 (1 θ 1 ) = θ 1(1 θ 1 1 ) 1 θ 1 1/ Sce θ 1 < 0.5, t follows that 1 θ 1 < 1; therefore, the lkelhood rato s mootoe decreasg. Ths mootocty mples that the lkelhood rato s too small f ad oly f s too large. Cosequetly, the most powerful test for the smple hypotheses rejects f c for some costat c. Sce the crtcal rego s depedet of θ 1, ths test must be most powerful for all θ 1 < 0.5 ad, hece, uformly most powerful for H 1 : θ < 0.5. (b) To fd the crtcal value c, we must solve the followg equalty: P 0.5 ( c) α. The left-had sde ca be smplfed usg the fact that Geo(0.5) uder H 0. Usg some geometrc seres trcks, oe gets (for teger c): P 0.5 ( c) = 0.5 c 1. The the smallest c such that 0.5 c 1 α s the soluto we wat: 0.5 c 1 α c 1 + log α/ log 0.5, or, equvaletly, c = 1 + log α/ log 0.5. For example, f α = 0.05, the c = 5.
Statstcs 401&411 MS Exam Fall Semester 01 1. (STAT401) Suppose that,..., 1..d. cotuous radom varables wth c.d.f. F( x). Let (1) ( ) be the order statstcs. () What s the jot dstrbuto of F( 1),, F( )? () Fd PF ( ( ) 0.5). (1) () Fd the jot dstrbuto of F( (1)) ad F( ( ) ). Soluto: () Frst, F( 1),, F( ) are..d. because they are fuctos of..d.,..., 1. Next, sce F s the c.d.f. of cotuous r.v., ts verse fucto F 1 exsts, ad 1 1 PF ( ( ) y) P ( F = ( y)) = FF ( ( y)) = y. Therefore, () F F ) ( 1),, ( are..d. Uf(0,1). PF ( ( ) 0.5) = P(at least oe F ( ) 0.5) (1) = 1 P(all F( ) > 0.5) = 1 P( F( ) > 0.5) = 1 0.5. () Frst, the jot c.d.f. of F( (1)) ad F( ( ) ) s F1, ( y, z) = P( F( (1) ) y, F( ( ) ) z) = P( F( ( ) ) z) P( F( (1) ) y, F( ( ) ) z) for ay 0 y z 1. Next, the jot p.d.f of F( ) ad F ) s = PF ( ( ) z) Py ( F( ) z) = z ( z y), (1) ( ( ) for ay 0 y z 1. F ( y, z) = ( 1)( z y) yz 1,,
. (STAT411) Suppose that,..., 1 f( x λ) are..d. dscrete radom varables wth p.m.f. λ λx = (1 e ) e, x = 0,1,, where λ > 0 s a ukow parameter. () Fd the MLE of λ, ad the expected Fsher formato,..., 1 cocerg λ. () Costruct a asymptotc 100(1 α)% cofdece terval for λ. Soluto: () The log-lkelhood fucto s l( λ) = log(1 e λ ) x. The score fucto s the λ λ dl( λ) e = x λ. Settg t equal to zero ad solvg for λ yelds the MLE dλ 1 e ˆ 1 λ = log 1+ x, where x s the sample mea. Note that the MLE does NOT exst whe x = 0. Next, s the expected Fsher formato of λ. λ ( ) dl( λ) e I λ = E d = λ (1 e ) λ () The MLE s asymptotcally cofdece terval for λ s 1 N( λ, I( λ) ), therefore a asymptotc 100(1 α)% ˆ 1/ 1/ λ ˆ λ), ˆ z I( ˆ ( z I ( λ+ λ) ). α/ α/
3. (STAT 411) Suppose that,..., 1 are..d. from e xp(1/ θ ) wth c.d.f. x/ θ F( x θ ) = P( x) = 1 e, f x > 0, where θ > 0 s a ukow parameter. () Fd a compete ad suffcet statstc for θ. () Fd a MVUE of g( θ ) = 1 F( t θ), where t s a fxed costat. Soluto: () The p.d.f. s f x = e I x, ad the lkelhood s 1 x/ θ ( θ) θ (0, ) ( ) (0, ) { } (0, ). x / ( ) θ L θ = θ e I ( x ) = exp log θ x / θ I ( x ) Clearly, t s a member of expoetal famly, ad hece that a complete ad suffcet statstc for θ s x. () Note that I(, t )( x1) s a ubased estmate of g( θ ), therefore a MVUE of g( θ ) s ( (, t ) 1 ) ( 1 = ) T( x) = E I ( ) = x = P t x t t = P = x = P x 1 1 x 1 The last equalty follows from Basu's theorem ad Beta(1, 1) s a acllary statstc, whch s due to the fact that 1 Gamma(1, θ ), Gamma( 1, θ ) ad they are depedet. Applyg the c.d.f. formula of Beta dstrbuto, 1 ( ) 1 T x = t t = P = x x 1.
Stat401, Problem, Fall 01: Let Y 1 < Y < < Y be the order statstcs of a radom sample from a dstrbuto wth pdf Let Z = Y log. (a) Fd the cdf of Z. (b) Determe the lmtg dstrbuto of Z as goes to. Stat401, Soluto, Fall 01:
Stat 481, Problem, Fall 01: A egeer a textle mll studes the effect of temperature ad tme o the brghtess of a sythetc fabrc a process volvg dye. Several small radomly selected fabrc specmes were dyed uder each temperature ad tme combato. The brghtess of the dyed fabrc was measured o a 50-pot scale, ad the results of the vestgato are as follows: Temperature (degrees Fahrehet) Tme (cycles) 350 375 400 40 38,3,30 37,35,40 36,39,43 50 40,45,36 39,4,46 39,48,47 Deote the observatos by Y jk, where =1, (factor A, tme), j=1,,3 (factor B, temperature), k=1,,3 (replcatos). Some relevat summary statstcs are = 8584; = 71; = 330, = 38; = 1, = 39, = 5. We also fd that SS Error =186.0. (a) Obta approprate ANOVA table ad test whether the row (factor A), colum (factor B), ad teracto (AB) effects are sgfcat. You may eed these crtcal F-values: F(0.05; 1,1)=4.75, F(0.05;,1)=3.89. (b) Summarze your fdgs ad tell the egeer that how the tme factor ad the temperature factor affect the brghtess of the dyed fabrc. Stat 481, Soluto, Fall 01, Je Yag: (a) SSTO = 8584 (71) /18 = 40.4; SS A = (330 + 38 )/9 - (71) /18 = 150.; SS B = (1 + 39 + 5 )/6 - (71) /18 = 80.8; SS AB = 40.4-150. - 80.8-186.0 = 3.4 The ANOVA table s the obtaed as follows: Source SS df MS F A (tme) 150. 1 150. 9.69 B (temperature) 80.8 40.4.61
AB 3.4 1.7 0.11 Error 186.0 1 15.5 Total 40.4 17 The teracto s sgfcat sce F AB = 0.11 < F(0.05;,1) = 3.89. Tme s a mportat factor sce F A = 9.69 > F(0.05; 1, 1)=4.75. Temperature s ot qute sgfcat sce F B =.61 < F(0.05;, 1) = 3.89. (b) The data aalyss shows the absece of teracto ad the strog ma effect of tme wth loger tmes creasg the brghtess. The crease temperature creases brghtess too. However, because the large varablty of dvdual measuremets, the temperature effect s oly border-le sgfcat. Addtoal observatos would help stregthe the evdece for a temperature effect.
STAT 416 - August 01 Two populatos, Y are of same dstrbuto forms but dfferet measures of cetral tedecy (mea). A radom sample of sze 5 s draw from the two populatos respectvely, ad data s recorded the followg table: : 1.6, 11., 13., 9.4, 1; Y : 16.1, 13.4, 15.4, 11.3, 14. (1). State the hypotheses to test f populato Y has a larger mea tha populato. (). Choose a approprate test statstc to make decso o the hypotheses gve α = 0.05. x Soluto: (1). Hypotheses: H 0 : θ = µ Y µ = 0 vs H 1 : θ = µ Y µ > 0 (). Pooled sequece: 9.4, 11., 11.3, 1, 1.6, 13., 13.4, 14, 15.4, 16. Wlcoxo rak sum test statstc for rak sum of sample: W N = 1 + + 4 + 5 + 6 = 18. The p-value of the test s P (W N 18) = 0.08 < 0.05. Table for 1 = = 5. So the ull hypothess s rejected favor of H 1 : θ > 0 at sgfcace level α = 0.05. 1
STAT 481 - August 01 A lear regresso model wth two covarates was ft to =0 observatos {(x 1, x, Y ), = 1,..., 0}. (1). Wrte dow the lear regresso model wth coeffcets β 0, β 1, β ad ecessary model assumptos. (). It s gve that the sum squares of regresso SSR = 66, sum squares of total SST = 00. Calculate ad terpret determato of the coeffcet. (3). Costruct ANOVA table ad test H 0 : β 1 = β = 0 at sgfcace level 0.05. [F 0.05 (1, 17) = 4.45; F 0.05 ( (, ) 17) = 3.59.] ( ) (4). Provded that studetzed t-statstcs t ˆβ0 = 3.3, t ˆβ1 = ( ) 4.6, t ˆβ = 1., could the bvarate model be smplfed at level α = 0.05? [t 0.05 (17) =.11; t 0.05 (, 17) = 1.74.] Soluto: (1). Bvarate lear regresso model Y = β 0 +β 1 x 1 +β x +ε, where..d. errors ε N (0, σ ). (). Determato of the coeffcet: R = 66/00 = 0.33. 33% of total varablty the respose s explaed by the ftted model. (3). ANOVA table Source SS DF M S F Reg 66 33 4.19 Error 134 17 7.88 Total 00 19 F statstc: F = 4.19 > F 0.05 (, 17) = 3.59. We reject H 0 : β 1 = β = 0. (4). Crtcal rego C = { t o > t 0.05 (17)} = { t o >.11}. It mples that β s a coeffcet ot sgfcatly from 0. Hece we ca smplfy the model to a smple lear regresso model: Y = β 0 + β 1 x 1 + ε, = 1,...,.
STAT 431 : Samplg Techques for Fall 01 Problem: The purpose s to estmate the proporto of UIC studets of a partcular morty commuty studyg o bak loa of a amout exceedg US$ 10,000.00 per year. It s kow that there are altogether N = 753 studets belogg to ths specfc commuty. It s desred to take a smple radom sample wthout replacemet [SRS (N, )] of a certa umber, say, of these studets ad ask them about ther facal stuato as to f aybody s gog for stpulated amout of bak loa or ot. Formulate the problem as oe of determato of sample sze () ad provde reasoable solutos, wth or wthout ay assumpto o the ature of true proporto. Expla your solutos wth llustratve examples. ------------------------------------------------- Soluto: As we see, there s a specfc commuty-based populato of studets of sze N ad we wat to study a specfc feature of the populato as a whole, vz., proporto of studets amog them studyg o bak loa exceedg US$ 10,000.00 per year. Let P deote the true [ukow] proporto of such studets ths commuty. We wat to determe adequate sample sze uder SRS(N, ) so that the sample proporto p wll esure a tolerable ad acceptable devato from the true ukow P. We deote by d the acceptable devato. Iterpretato of Acceptable Devato: If p = 0.73 ad d = 0.05, the we wsh P to belog to the terval [0.68, 0.78], Aga, f p = 0.38 ad d = 0.01, the we wsh P to be cluded the terval [0.37, 0.39].Theoretcally, t s kow that for ay choce of d, there s a value of, sce = N would lead to what s kow as cesus whch correspods to d=0. Clearly, ths arrowg dow of acceptable devato calls for creasg sample sze. Techcally, we eed -d < p P < d,.e., p d < P < p + d where p s based o SRS (N, ). We must ote that we caot gve ths assurace 100% cases; we wll deftely mss out at tmes; so we would lke to keep hgh chace of meetg the acceptable devato level. Wth ths motvato, we formulate the problem as oe of satsfyg the requremet Pr.[p d < P < p + d] = 1- α (1)
where α s a small fracto close to 0. We ow use the fact that uder SRS(N, ), () E(p) = P; () V(p) = (1/ 1/N) P(1-P) So, defg Z = (p - P) / [V(p)] whch s approx. N(0, 1), we have Z_α/ = d / (V(p) whch yelds, usg z for Z_α/, d / z = V (p) = P(1-P)[1/ 1/N]..() From (), assumg a large populato sze, we have (approx.) = P(1-P) z / d..(3) I the absece of ay kowledge about the ature of the true proporto P, we may replace P by 1/.e., P (1-P) by 1/4. So, = z /4d (4) Ad for α = 0.05; we kow z = 1.96 =.00 approx., so that a tal estmate of s o = 1/ d...(5) Now we ca apply fte populato correcto ad coclude that the fal sample sze s gve by f = o tmes N /[N + o ].. (6) Illustratve Examples: () d = 0.05; α = 0.05 so that o = 1/ d [approx.] = 400 ad hece f = 400 x 753/[753+400] = 6. () d = 0.05, α = 0.01 so that z =.58 ad hece o = z / 4d = (.58) /4x(0.01) =16641.
So fally f = 16641x753/[753+16641] = 71. Remark : Note that the basc formula for sample sze computato s gve by (3). At tmes, we may have some guess values about P such as P < 0.35 or, P > 0.6 ad the lke. I such cases, we ca use ths kd of vague formato to mprove o our tal estmate of. The dea s ot to replace P by 0.5 as was doe arrvg at (4). Rather, replace P by the upper or lower lmt as the case may be. However, f the vague formato terms of a terval cludes the value 0.5, the we eed to use ½ place of P. The followg examples llustrate the procedure to be followed : (I) P < 0.35.replace P by 0.35; (II) P > 0.65.replace P by 0.65; (III) 0.30 < P < 0.45.replace P by 0.45 (IV) 0.57 < P < 0.77.replace P by 0.57 (V) 0.43 < P < 0.67.replace P by 0.5.
MS Exam (Fall 01) Soluto for STAT 461 problems Problem 1. I roll a sx-sded de ad observe the umber N o the uppermost face. I the toss a far co N tmes ad observe, the total umber of heads to appear. What s the probablty that N=3 ad =1? What s the probablty that =4?
Probelm. Let {_} be a Markov cha wth the state space {0, 1, }. The trasto probablty matrx s gve by P= 1 0 0 0.1 0.5 0.4 0. 0.4 0.4 If the cha starts from state 1, fd the probablty that t does ot vst state pror ts absorpto.