THE ROYAL STATISTICAL SOCIETY HIGHER CERTIFICATE

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Bethany Sibyl Lucas
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1 THE ROYAL STATISTICAL SOCIETY 00 EXAMINATIONS SOLUTIONS HIGHER CERTIFICATE PAPER I STATISTICAL THEORY The Socety provdes these solutos to assst caddates preparg for the examatos future years ad for the formato of ay other persos usg the examatos The solutos should NOT be see as "model aswers" Rather, they have bee wrtte out cosderable detal ad are teded as learg ads Users of the solutos should always be aware that may cases there are vald alteratve methods Also, the may cases where dscusso s called for, there may be other vald pots that could be made Whle every care has bee take wth the preparato of these solutos, the Socety wll ot be resposble for ay errors or omssos The Socety wll ot eter to ay correspodece respect of these solutos RSS 00

2 Hgher Certfcate, Paper I, 00 Questo () A ad B are depedet So P( A B) = P( A) P( B) PA = PA = ; PB = ; so PA ( B) = 3 6 () A B = [( A B) C] [( A B) C], wth the two evets [( A B) C] ad [( A B) C] beg dsjot Hece PA ( B C) = PA ( B) PA ( B C) = = () B C = ( A B C) ( A B C), these beg dsjot Further, ( A B C) ( A B C) = ( A B) Hece P( B C) = PA ( B C) + PA ( B) PA ( B C) = + P( A C) P( A B C) 4 3 sce ABare, depedet But A, C are also depedet ad so are AC, Therefore 5 63 PB ( C) = + + = = = 0 (v) (v) PB ( C) 7 / 0 7 PB ( C) = = = PC 3/5 PA ( B C) /4 5 PAB ( C) = = = PB ( C) 7 / 0 7 PA ( B C) 5 (v) PA ( B A C) = = 4 = PA ( C) (A, C are depedet)

3 Hgher Certfcate, Paper I, 00 Questo (a) Fx the posto of M (suppose hm to be the host) M Label the postos clockwse () (6) () M, M 3, W, W, W 3 may be arraged 5! = 0 ways (5) (3) (4) () M, M 3 must occupy (3) ad (5); W, W, W 3 may occupy the other places 3! ways, makg 3! arragemets The probablty s the = 0 0 () M, M 3 must occupy () ad (6) or () ad (3) or (5) ad (6) I each case, M, M 3 ca be placed two orders, makg 6 postos altogether for the three me The wome may aga fll the remag places 3! ways The probablty s = () EITHER =, because ths s the oly other arragemet possble besdes () ad (); OR by havg M (), M 3 (4) or (5); M (6), M 3 (3) or (4); M (3), M 3 (4); M (4), M 3 (5); or ay of these wth M, M 3 terchaged, gvg postogs of the me There are aga 3! orders for the wome, so the probablty s 6 = (b) () Evet D s "has dsease", T s "tests postve" PD ( T) PT ( DPD ) PT ( DPD ) PD ( T) = = = P( T) P( T) P( T D) P( D) + P( T D) P( D) pp 0 = p p + ( p )( p ) 0 0 () = = 0087 ( ) + ( ) The error rates the clcal tests are large compared to the chace of havg the dsease, so the calculated probablty s very small

4 Hgher Certfcate, Paper I, 00 Questo 3 (a) () PS ( 300) = PS ( < 300) = Φ 300 = Φ () = 0843 = PH ( 300) = Φ = Φ( 6) 5 = = 0945 () P( S H) P( S H 0) > = >, where (S H) s N( 500, ) e N( 500, 35 ) Hece P( S > H) =Φ =Φ( 5385) = 0060 (b) () The lfetme X s S wth probablty 06 ad H wth probablty 04 Hece E[ X ] = = 00 hrs ( > 600 ) = ( > 600 ) + ( > 600 ) P X P X S P S P X H P H =Φ 06 + Φ (usg the approprate tal areas from Normal tables) = 06 Φ( ) + 04 Φ( 08) = = = () P( H ) ( > H) P( H) P( X > ) P X > 600 = = = () X wll have mea 00 It s ot Normally dstrbuted but we way apply the Cetral Lmt Theorem f we kow ts varace I large samples we may take X as approxmately N(00, 3468 ), so that P( X > 300) = Φ = Φ =Φ( 8835) = /

5 Hgher Certfcate, Paper I, 00 Questo 4 A easy method s to cosder X as X, where X are a set of Beroull varables wth P( X ) p, P( X 0) ( p) X = p, so E X = p = = = = The E[ ] [ ] X = p, so Var X = p p ad Var X = p p = pq Also E ALTERNATIVELY: [ ] x x x x! p q E X = x p q = x= 0 x x= ( x! ) ( x)! = p p q p x= x x ( ) ( x ) = Smlarly, Var ( X ) = E X ( X ) + E[ X] -( E[ X] ), ad we have E X ( X ) = x( x ) p q = x( x ) p q x= 0 x x= x x x x x = ( ) p p q = p x= x x ( ) ( x ), ad hece Var X = p + p p = p p = pq PGFs or MGFs could also be used () (a) = 0000 approx () 6 (b) P(o oe gets all 6 rght), probablty s { 075 } = { } = 039 P(B A > 0) ca be studed usg a Normal approxmato to the dfferece ( { }) B A, e N 6{ }, 6 ( 05 05) + ( ), e N( 4,7) The probablty s foud as takes dscrete values P B A> usg a cotuty correcto sce B A

6 Hece t s Φ =Φ =Φ( 7008) [Note: ths would be wthout the cotuty correcto] E X = E X = p = = A () [ ] set Smlarly, E Y = 6 05 = 8 set B There are studets A ad 5 B, so that Var Var ( X ) ( Y ) = = set A = = set B 5 5

7 λ x+ ( + ) λ Hgher Certfcate, Paper I, 00 Questo 5 p x e x! λ = = for x = 0,,, λ x p x x+! e λ x+ λ =, ; so 03037, 00758, p = p = p = p = For Graph of p(x) for λ = ½ For λ =, p(0) = 03534; so p = = p, p 3 = 08045, p 4 = 0090, p 5 = Graph of p(x) for λ =

8 () x xt λ x t t Xt e e λ λe λ λ λe t exp( λ{ } ) M X t = E e = = e = e e = e x! x! x= 0 x= 0 t M t λ( e ) = λee ; put t= 0 ad ths s λ, whch s therefore E[ X] t t M t t λ( e ) = ( λ e + λe ) e ; put t = 0 ad ths s λ + λ, but t s also t E X Hece ( [ ]) Var X = E X E X = λ + λ λ = λ 3 t M 3 3t t t λ( e ) 3 3 = 3 ( λ e + 3 λ e + λe ) e = λ + 3 λ + λ at t = 0 Ths s E X t Now, E ( X ) E X 3 E X 3 E[ X] λ = λ + λ λ = = 3 3 λ 3λ λ 3λ λ λ 3 λ λ λ λ For Y = X + X + + X we have = t X λ Yt Xt E e E e M ( t) exp e = = = Ths s the mgf of a Posso dstrbuto wth parameter λ ad so E[Y] = Var(Y) = λ PY ( 40) = PY ( 39) Φ, 5 usg cotuty correcto, ad µ = λ = 5 Ths s Φ 9 = Wth a postvely skew dstrbuto, the Normal approxmato s lkely to uderestmate the probablty the rght had tal ad so we expect ths aswer to be less tha the true value

9 Hgher Certfcate, Paper I, 00 Questo 6 p L p q p p q p = p x x = = = ( ) l L = l p l p + xl( p) x + x ( l L) = + = 0 whe = whch gves pˆ = as mlestmate p p p p pˆ pˆ x x ( l L) ( x ) = p p ( p) whch s < 0, cofrmg the maxmum l L = + E ( X) = + = + p p p p p p p E ( p) ( p) ( ) Hece Var ( pˆ ) ( p) p = p ( p) 56 fx = 448, f = 56, pˆ = = Var ( pˆ) = = , SE ( pˆ) = Approxmate 95% cofdece terval for p s p ˆ ± 96SE ( pˆ ), whch s 05 ± , e 05 ± or (00944, 0556) 5 Whe p =, we have ~ N,, e N(0667, ); therefore 6 X the probablty of obtag pˆ 05 s approxmately Φ = Φ( 0496) = The cofdece terval for p dd ot clude ; also ow the probablty beg very 6 small s cosstet wth rejectg a ull hypothess that p =, e that the de s far 6

10 Hgher Certfcate, Paper I, 00 Questo 7 P X x = F x (a) for =,,, () F ( x) = P( X x, X x,, X x) = P( X x) Xmax = = F x () For Xm x, we requre every X to be x Now, ( ) = ( ) = P X x F x, so P X x F x, for all (,, ) F x = P X x = P X x = P X x X x X m m m = F x (b) () Pdfs are dervatves of cdfs: f X = F x f x max F x f X = F( x) f ( x), sce = f ( x) m x x

11 F x x α = du = = 0 + ( + u) ( + u) ( + x) Meda M s such that α α α 0 α ( + M) ( + M) F M = So we have α ( / α ) = or = or = ( + M) or M = α Usg (a)(), by α F X = = m α ( + x) ( + x) x α, also Pareto but wth α replaced The meda of X m s the / < 0 or / <, e l < l, gvg Hece 8 α, whch s / f α = We requre l 0693 > = = 77 l 00953

12 Hgher Certfcate, Paper I, 00 Questo 8 y = α + β x + ε where y s the respose (observato) ad x the value of the explaatory varable o that ut; { ε } s a set of depedet, detcally dstrbuted radom varables wth mea 0 ad the same varaceσ Usually they are assumed Normal as a bass for ferece x s assumed "fxed", ot "radom" () (a) There s a creasg tred, ad the relatoshp betwee y ad x appears curvlear (b) I smple regresso R s the square of the correlato, r, betwee x ad y I geeral t s the proporto of the varace of y whch ca be explaed by the depedece of y o all explaatory varables { x } the model; hece t s the square of the correlato betwee ŷ ad y I the ANOVA, regresso SS = 6933 = 0803, or 803% total SS (c) As x (% operatg capacty) creases by ut so y (proft) creases by 0356 uts A 95% cofdece terval s 0356 ± t , whch s , 0457 ± or (d) Values of proft have bee predcted for capacty 5%, 50% ad 75% The cofdece tervals for these predctos are those gve; but ote that 5% s far outsde the rage of avalable data (hece the remark about extreme x values) A 95% cofdece terval s a terval whch should cover the true y at a gve x wth probablty 095, based o the ftted lear regresso () (a) The logarthmc plot shows that a lear regresso these uts s a much better ft There s stll a creasg tred (b) e log 0 (profts) = (capacty) profts capacty = (c) For capacty = 5, the 95% lmts are 073 ad +00, ad the actual predcto s 00760, log 0 uts At-loggg these (e rasg 0 to these powers) we fd the 95% lmts are 0533 ad 33 The 'predcto' s 08395

13 These lmts do ot overlap the lmts o the prevous model The predcto ow s for a small proft, compared wth a loss o the prevous model () The scatter plots dcate that the logarthmc model s preferred, ad so do the plots of resduals whch show a radom patter (as compared wth a systematc, curved oe for the prevous model) R also hgher (94%) o the log model But a log model caot predct egatve profts e losses whch are qute possble geeral though ot for these data f used wth the rage of x values gve Extrapolato dow to 5% s well outsde the data ad so s ot relable o ay model

THE ROYAL STATISTICAL SOCIETY 2016 EXAMINATIONS SOLUTIONS HIGHER CERTIFICATE MODULE 5

THE ROYAL STATISTICAL SOCIETY 06 EAMINATIONS SOLUTIONS HIGHER CERTIFICATE MODULE 5 The Socety s provdg these solutos to assst cadtes preparg for the examatos 07. The solutos are teded as learg ads ad should