M1 M1 A1 M1 A1 M1 A1 A1 A1 11 A1 2 B1 B1. B1 M1 Relative efficiency (y) = M1 A1 BEWARE PRINTED ANSWER. 5

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1 4769 Mark Scheme Jue 006 Q (i) L e σ π ( W μ) σ e σ π ( W μ) σ M M A Product form. Two Normal terms. Fully correct. (ii) l L cost ( W μ) ( W σ σ d l L ( W μ) + ( W dμ σ σ 0 σ W σ μ + σ W σ W + σ W ˆ μ σ + σ Check this is a maximum. d l L E.g. < 0 dμ σ σ σ μ σ μ μ + E( ˆ) μ σ + σ ubiased. μ) μ) σ μ 0 M A M A A A M Differetiate w.r.t. μ. BEWARE PRINTED ANSWER. A M A (iii) Var( ˆ) μ σ σ σ σ + σ + σ ( σ σ 4 + σ σ ) 4 B B First factor. Secod factor. Simplificatio ot required at this poit. (iv) T W + W ) ( ( σ 4 + σ Var(T) ) B Var( ˆ) μ M Relative efficiecy (y) Var( T ) M σ σ + σ σ + σ σ ( σ σ 4 ( σ 4σ + 4 ) ) 4 σ + σ A Ay attempt to compare variaces. If correct. A BEWARE PRINTED ANSWER. 5 (v) E.g. cosider σ + σ σ σ ( σ σ ) 0 M Deomiator umerator, fractio E [Both μˆ ad T are ubiased,] μˆ has smaller variace tha T ad is therefore better. E E 4 4

2 4769 Mark Scheme Jue 006 Q k f ( x) + k x e k! x, [ x > 0 -u Give: u m e du m! 0 ( > 0, k iteger 0)] (i) M X ( ) E[e 0 x ] k! k + x k k!( ) e k + ( ) x k + k + 0 dx Put ( )x u u k e u du M M M A A A A For obtaiig this expressio after substitutio. Take out costats. (Dep o subst.) Apply give : itegral k! (Dep o subst.) BEWARE PRINTED ANSWER. 7 (ii) (iii) Y X + X + + X By covolutio theorem:- mgf of Y is {M X ()} k + B i.e. μ M (0) k + k M ( ) ( k )( ) ( ) M A k + μ A σ M (0) μ k + k M ( ) ( k + ) ( k )( ) ( ) M M (0) ( k + )( k + + ) / A k k σ k + k ( + )( + + ) ( + ) [Note that M Y (t) is of the same fuctioal form as M X (t) with k + replaced by k +, i.e. k replaced by k +. This must also be true of the pdf.] M A 8 Pdf of Y is [for y > 0] k + k + y y ( k + )! e B B B Oe mark for each factor of the expressio. Mark for third factor show here depeds o at least oe of the other two eared. 3 (iv), k, 5, Exact P(Y > 0) Use of N(5, 5) M M Mea. ft (ii). Variace. ft (ii).

3 4769 Mark Scheme Jue P(this > 0) P N(0, ) > Reasoably good agreemet CLT workig for oly small. A c.a.o. A c.a.o. E (E, E) [Or other sesible commets.] 6 4

4 4769 Mark Scheme Jue 006 Q3 (i) x y 45.5 s s 4.89 s s B If all correct. [No marks for use of s which are ad 4.83 respectively.] Assumptios: Normality of both populatios equal variaces B B H 0 : μ A μ B H : μ A μ B B Do NOT accept X Y or similar. Where μ A, μ B are the populatio meas. B Pooled + s Test statistic is B (.7444) M A Refer to t 0. M No ft from here if wrog. Double tailed 5% poit is 086. A No ft from here if wrog. Not sigificat. A ft oly c s test statistic. No evidece that populatio mea times differ. A ft oly c s test statistic. (ii) Assumptio: Normality of uderlyig B populatio of differeces. H 0 : μ D 0 H : μ D > 0 B Do NOT accept D 0 or similar. Where μ D is the populatio mea of before after differeces. B The directio of D must be CLEAR. Allow μ A μ B etc. Differeces are 6.4, 4.4, 3.9, -.0, 5.6, 8.8, -.8, M. ( x 4.8 s ) [A ca be awarded here if NOT awarded i part (i)]. Use of s (4.3396) is NOT acceptable, Test statistic is M.9(64) A eve i a deomiator of Refer to t 7. M No ft from here if wrog. Sigle tailed 5% poit is.895. A No ft from here if wrog. Sigificat. A ft oly c s test statistic. Seems mea is lowered. A ft oly c s test statistic. 0 s (iii) The paired compariso i part (ii) elimiates the variability betwee workers. E (E, E) 4

5 4769 Mark Scheme Jue 006 Q4 (i) Lati square. Layout such as: B Locatios I A B C D E Surf II B C D E A -aces III C D E A B IV D E A B C V E A B C D B B (letters paits) Correct rows ad colums. A correct arragemet of letters. SC. For a descriptio istead of a example allow max out of. 3 (ii) X ij μ + α i + e ij B μ populatio grad mea for whole experimet. α i populatio mea amout by which the i th treatmet differs from μ. e ij are experimetal errors ~ id N(0, σ ). (iii) Totals are: 3, 35, 307, 355, 9 (each from sample of size 5) Grad total: Correctio factor CF Total SS CF Betwee paits SS CF CF 6.96 Residual SS (by subtractio) Source of SS df MS variatio Betwee paits Residual Total MS ratio B B B B B B B B M M A B B M M A Allow ucorrelated. Mea. Variace. 9 For correct methods for ay two SS. If each calculated SS is correct. Degrees of freedom betwee paits. Degrees of freedom residual. MS colum. Idepedet of previous M. Dep oly o this M.

6 4769 Mark Scheme Jue 006 Refer to F 4, 0 M No ft if wrog. But allow ft of wrog d.o.f. above. Upper 5% poit is.87 A No ft if wrog. Sigificat. A ft oly c s test statistic ad d.o.f. s. Seems performaces of paits are ot all the same. A ft oly c s test statistic ad d.o.f. s. 4

7 4769 Mark Scheme Jue 007 ) f (x) 0 x (i) (ii) (iii) B Write-dow, or by symmetry, or by E[ X ] itegratio. E[ X ] E[ X ] E[ X ] M A E 4 ubiased x.3.3 x 0.46 x B But we kow E estimator ca give osese aswers, E (E, E) i.e. essetially useless 4 y Y max{x i }, g(y) 0 y MSE (ky) E[( ky ) ] M E[ k Y E[ k Y + ] ] E[ ] + k Y k Y BEWARE PRINTED ANSWER dmse M dk ke[ Y ] E[ Y] 0 M E[ Y] A for k E[ Y ] d MSE E[ Y ] > 0 dk this is a miimum M M E[ Y ] dy A E 0 [ Y ] 0 y y dy miimisig k (iv) With this k, ky is always greater tha the sample maximum So it does ot suffer from the disadvatage i part (ii) M A M A E E (E E) (E E) 4 89

8 4769 Mark Scheme Jue 007 (i) (ii) (iii) (iv) X x x G( t) E[ t ] ( pt) ( p) M x 0 x [( p ) + pt] Available as B for write-dow or as + for algebra ( q + pt) 4 μ G () G ( t) p( q + pt) G () p p σ G () + μ μ G ( t) ( ) p ( q + pt) G' '() ( ) p σ p p + p p M p + p pq 6 μ Z X σ Mea 0, Variace B For BOTH M( ) G( e ) ( q + pe ) Z ax + b with: μ p a ad b σ pq σ q M Z ( ) e M Z b ( ) e M X p q ( a ) q + pe pq M (v) qe M Z p p pq pq + pe ( ) ( q qp qp + pq pq + BEWARE PRINTED ANSWER M For expasio of expoetial terms 5 terms i 3,,.. + pq pq p + + +) pq pq M For idicatio that these ca be eglected as. Use of result give i questio ( + + ) e 4 90

9 4769 Mark Scheme Jue 007 (vi) N(0,) Because e is the mgf of N(0,) E ad the relatioship betwee distributios ad their mgfs is uique E 3 (vii) Ustadardisig, N( μ, σ ) ie N( p, pq) Parameters eed to be give. 9

10 4769 Mark Scheme Jue 007 3(i) H H 0 : μ μ A : μ μ A B B Do NOT allow X Y or similar Where μ, μ are the populatio meas A B Test statistic M M M Accept absece of populatio if correct otatio μ is used. Hypotheses stated verbally must iclude the word populatio. Numerator Deomiator two separate terms correct A Refer to N(0,) Double-tailed 5% poit is.96 Not sigificat No evidece that the populatio meas differ No FT if wrog No FT if wrog 0 (ii) CI ( for μ A μ B ) is.0 ± ± ( 0.856,.356) (iii) (iv) H 0 is accepted if.96< test statistic <.96 i.e. if x y.96 < < i.e. if.556 < x y <. 556 I fact, X Y ~ N(, ) So we wat P(.556 < N(, ) <.556) P < N(0,) < P ( 4.48 < N(0,) < ) Wilcoxo would give protectio if assumptio of Normality is wrog. Wilcoxo could ot really be applied if uderlyig variaces are ideed differet. Wilcoxo would be less powerful (worse Type II error behaviour) with such small samples if Normality is correct. M B M A cao M M A M M M A cao E E E Zero out of 4 if ot N(0,) 4 SC Same wrog test ca get M,M,A0. SC Use of.645 gets out of 3. BEWARE PRINTED ANSWER Stadardisig 7 3 9

11 4769 Mark Scheme Jue (i) There might be some cosistet source of plotto-plot variatio that has iflated the residual ad which the desig has failed to cater for. (ii) Variatio betwee the fertilisers should be compared with experimetal error. E E E Some referece to extra variatio. E Some idicatio of a reaso. (iii) If the residual is iflated so that it measures more tha experimetal error, the compariso of betwee - fertilisers variatio with it is less likely to reach sigificace. Radomised blocks E (E, E) 3 C B A D E... SPECIAL CASE: Lati Square 4 (, E)... E E E Blocks (strips) clearly correctly orieted w.r.t. fertiliser gradiet. All fertilisers appear i a block. Differet (radom) arragemets i the blocks. 4 (iv) Totals are: (each from sample of size 4) Grad total 50.6 Correctio factor CF Total SS 360. CF Betwee fertilisers SS CF CF M M For correct method for ay two Residual SS (by subtractio) A If each calculated SS is correct Source of variatio SS df MS MS Ratio Betwee fertiliser Residual M M,A Total Refer to F 4, 5 No FT if wrog -upper 5% poit is 3.06 No FT if wrog Sigificat - seems effects of fertilisers are ot all the same (vii) Idepedet N (0, σ [costat]) 3 93

12 4769 Mark Scheme Jue Statistics 4 Q (i) x x e e L x! x! xi e x! x! x! M A product form fully correct l L cost + d l L xi + d ˆ xi ( x) 0 Check this is a maximum x i d l L e.g. < 0 d x i l M A M A A M A CAO 9 (ii) P ( X 0) e B (iii) We have R ~ B(, e ), so E (R) e Var( R ) e ( e ) M B B ~ R M ~ E ( ) e i.e. ubiased ~ e ( e ) Var( ) A A A BEWARE PRINTED ANSWER 7 77

13 4769 Mark Scheme Jue 008 (iv) Relative efficiecy of ~ wrt ML est Var(ML Est) ~ Var( ) e e ( e ) e M M A ay attempt to compare variaces if correct BEWARE PRINTED ANSWER Eg:- Expressio is + +! M always < ad this is if is small 0 if is large E E E Allow statemet that 0 as e 7 78

14 4769 Mark Scheme Jue 008 Q (i) P( X x) q x p B FT ito pgf oly X Pgf G( t ) E( t ) pt pt ( + qt + q t pt ( qt) μ G' () σ x + ) G' '() x q x + μ μ G'( t) pt( )( qt) ( q) + p( qt) pqt( qt) + p( qt) M A A M A BEWARE PRINTED ANSWER [cosideratio of qt < ot required] for attempt to fid G'(t) ad/or G"(t) G' () pq( q) + p( q) q + p p 3 G' '( t) pqt( )( qt) ( q) + pq( qt) p( )( qt) ( q) G''() pq ( q) q q + p p 3 + pq( q) q q q σ + + p p p p q q (q + p ) p p + pq( q) + pq + p p + A A A M A BEWARE PRINTED ANSWER For isertig their values BEWARE PRINTED ANSWER 79

15 4769 Mark Scheme Jue 008 (ii) X umber of trials to first success X ext YX +X +..X.. total o of trials. to the th success. X th E E pgf of Y (pgf of X ) p t ( qt) μ μ Y X p P(N(0,)>-.343) q σ Y σ X p 5 (iii) N(cadidate s μ Y, cadidate s σ Y ) (iv) Y o of tickets to be sold ~ radom variable as i (ii) with 40 ad p 0.8 E ~ Approx N ( , ) 0.8 (0.8) Do ot award if cty corr P ( Y 60) P(N(75,43.75) > 59 ) M abset or wrog, but FT if 60 used -.68, For ay sesible discussio i cotext (eg groups of passegers ot idep.) Q3 X amout of salt ~ N( μ [750], σ [0 ]) Sample of 9 (i) Type I error: rejectig ull hypothesis whe it is true. A A E E B B CAO Allow B for P(rej H 0 whe true) 7 Type II error: acceptig ull hypothesis whe it is false. B B Allow B for P(acc H 0 whe false) OC: P (acceptig ull hypothesis as a fuctio of the parameter uder ivestigatio) (ii) Reject if x < 735 or x > α P ( X < 735 or X > 765 X ~ N(750, )) 9 ( )3 P(Z <.5 0 ( )3 or Z >.5) 0 B B M A A [ P(type II error the true value of the parameter) scores B+B] Might be implicit 6 ( ) A CAO This is the probability of rejectig good output ad uecessarily re-calibratig the machie seems small [but ot very small?] E E Accept ay sesible commets 6 80

16 4769 Mark Scheme Jue 008 (iii) Accept if 735 < x < 765, ad ow μ 75. β P(735 < X < 765 X P(.5 < Z< 6) ~ N(75, 0 )) This is the probability of acceptig output ad carryig o whe i fact μ has slipped to 75 small[-ish?] M A A A E E might be implicit CAO If upper limit 765 ot cosidered, maximum of these 4 marks. If Ф(6) ot cosidered, maximum 3 out of 4. accept sesible commets 6 (iv) OC P( 735 < X < 765 X ~ N( μ, 0 )) 9 Ф 765 )3 735 ( μ Ф ( μ )3 0 0 M Ф Ф M A both correct μ70:ф(6.75) Ф(.5) : : ( ) : similarly or by write-dow from part (ii) [ FT ] : if ay two correct 760, 770, 780 by symmetry [FT]: , 0.66, 0.0 Q4 (i) x ij μ + α i + eij μ populatio.. grad mea for whole experimet α i populatio.. mea by which i th treatmet differs from μ e ij are experimetal errors ~ id N (0, σ ) 3 Allow ucorrelated for id N; for 0; for σ. 9 6 (ii) Totals are 40, 46, 54, 64, 96 each from sample of size 5 Grad total Correctio factor CF Total SS CF

17 4769 Mark Scheme Jue 008 Betwee cotractors SS CF CF Residual SS ( by subtractio) M M A For correct methods for ay two, if each calculated SS is correct. Source of SS df MS MS ratio Variatio Betwee Cotractors Residual Total M M A CAO Refer to F 3,6 Upper 5% poit is 3.4 Sigificat Seems performaces of cotractors are ot all the same NO FT IF WRONG NO FT IF WRONG (iii) Radomised blocks B Descriptio E E Take the subject areas as blocks, esure each cotractor is used at least oce i each block 3 8

18 4769 Mark Scheme Jue Statistics 4 Jue 009 Q Follow-through all itermediate results i this questio, uless obvious osese. (i) P(X ) ( ) ( ) [o.e.] 0 L [] [( )] ( ) 0 [(-) ] M A M A A Product form Fully correct BEWARE PRINTED ANSWER 5 (ii) l L + ) l + 0 ) l ( ) ( 0 ( M A (iii) d l L d ( - ˆ ) ( 0 + ) ˆ ( 0 ) 0 + ˆ 0 E(X) x x ( ) x 0 {0 + ( ) + ( ) 3 + 3( ) + } M A M A M A Divisible, for algebra; e.g. by GP of GPs BEWARE PRINTED ANSWER 6 (iv) So could sesibly use (method of momets) ~ ~ give by ~ X ~ + X To use this, we eed to kow the exact umbers of faults for compoets with two or more. 4 x ~ Also, from expressio give i questio, ~ ( 0 877) Var( ) M A E B B B BEWARE PRINTED ANSWER 6 CI is give by ± 96 x (0 87, 0 937) M B M A For For.96 For

19 4769 Mark Scheme Jue 009 Q (i) z tz e tz Mgf of Z E (e ) π Complete the square z tz ( z t) + t t ( z t) e e dt π Pdf of N(t,) t e dz M M A A M M M A t For takig out factor e For use of pdf of N(t,) For pdf t For fial aswer e 8 (ii) Y has mgf M Y (t) t( b) Mgf of ay + b is [e ay + E ] bt ( at) Y bt e E[e ] e M ( at) (iii) (iv) μ Z X, so X σ Z + μ σ M X ( t) e μt.e ( σt ) Y e σ t μt + X W e k X k kx E( W ) E[(e ) ] E(e ) M ( k) X M M M A A bt For factor e ( at) Y For factor E [e ] For fial aswer For factor μt e ( σt) For factor e For fial aswer 4 X k For E [(e ) ] For E ( e kx ) For M X (k) 4 E ( W ) M X () e σ μ+ M A μ+ σ E ( W ) M X () e μ σ Var( W ) e e + μ+ σ μ + σ σ [ e (e )] M A A 8 8

20 4769 Mark Scheme Jue 009 Q3 (i) x 6 s s y 33 9 s s H 0 : μ A μ B H 0 : μ A μ B Where μ A, μ B are the populatio meas. Pooled s [ ] Test statistic is Refer to t A B M A A if all correct. [No mark for use of s, which are ad respectively.] Do ot accept X Y or similar. No FT if wrog (ii) Double-tailed 0% poit is 753 Not sigificat No evidece that populatio mea cocetratios differ. There may be cosistet differeces betwee days (days of week, types of rubbish, ambiet coditios, ) which should be allowed for. E E No FT if wrog 0 Assumptio: Normality of populatio of differeces. Differeces are [ d 4, s 3 86 ( s 4 95)] Use of s ( 3 64) is ot acceptable, eve i a deomiator of s / ] 4 0 Test statistic is / 9 Refer tot 8 Double-tailed 5% poit is 306 Sigificat Seems populatio meas differ M M A A Ca be awarded here if NOT awarded i part (i) No FT if wrog No FT if wrog 0 8

21 4769 Mark Scheme Jue 009 (iii) Wilcoxo rak sum test Wilcoxo siged rak test H 0 : media A media B H : media A media B B B [Or more formal statemets] 4 Q4 (i) (ii) (iii) Descriptio must be i cotext. If o cotext give, mark accordig to scheme ad the give half-marks, rouded dow. Clear descriptio of rows. Ad colums As extraeous factors to be take accout of i the desig, with treatmets to be compared. Need same umbers of each Clear cotrast with situatios for completely radomised desig ad radomised treds. e ~ id N (0, σ ) ij α i is populatio mea effect by which ith treatmet differs from overall mea Source of Variatio Betwee Treatmets SS df MS MS ratio Residual Total E E E E E E E E E 9 Allow ucorrelated For 0 For σ 5 Refer to F 4, 5 No FT if wrog No FT if wrog Upper % poit is 4 89 Sigificat, seems treatmets are ot all the same 0 83

22 4769 Mark Scheme Jue 00 Questio (i) (ii) (iii) (iv) x / xe f( x) > 0 / E( ) e x X x dx 0 ( x ) { } / x x x e.xe / dx {[ 0 0] } +.. ( ) ( ) ( ) ˆ [ ] E X E X E X ˆ is ubiased. Var ( ˆ) Var ( X ) E( ) e x / X x dx 0 Var ( X ) { } 3 / x e 3 x x x e / dx {[ 0 0 ]} + 3E( X ) 6. ( ) ( ) { ( )} Var X E X E X 6 4. ( ˆ ) Var. Variace of ˆ becomes very small as icreases. It is ubiased ad so becomes icreasigly cocetrated at the correct value. E + +. is ubiased. ( ) ( 8 4 8) ( ) ( ) Var relative efficiecy of to ˆ is So ˆ is preferred. /6 3 / Special case. If doe as Var( ) / Var( ˆ ), award out of for the secod M ad the A i the scheme. M for itegral for E(X) M for attempt to itegrate by parts For secod term: M for use of itegral of pdf or for itegr'g by parts agai A M A E M M for use of E(X ) By parts M M for use of E(X) A for 6 A [7] A [7] E E [] E ( ) M A : B; "ubiased": E M ay compariso of variaces M correct compariso A for 8/9 [Note. This MMA is allowable i full as FT if everythig is plausible.] E (FT from above) [8]

23 4769 Mark Scheme Jue 00 Questio ( t) X (i) G() t E( t ) [M] x 0 e x! x t e + t+ +...! [A] ( ) e e t e t [A] [Allow omissio of previous A step ad write-dow of this for A provided opeig M has bee eared (NB aswer is give)] [3] ( ) (ii) Mea G'() [M] G'() [A] G' () t e t Variace G''() + mea mea ( ) G'' e t t [M] G''() [A] () (iii) (iv) variace + [A] X μ Z : mea 0 [B] variace [B] σ (e ) Mgf of X is M G e e [B] ( ) ( ) Liear trasformatio result is M ( ) e b M ( ) ax + b X a [B if fully correct, ay equivalet form. Allow B if either factor correct.] [5] [] Use with a σ ad b μ [M] σ e/ e/ M ( ) e e e Z [A] [A] [A] [NB aswer is give] [7] (v) Cosider 3 / e /! 3! [M] / 3/ + terms i,,,... [A] as [M] [some explaatio required] ( ) / M e as e / Z (vi) is the mgf of N(0, ) [E], [A] [aswer give] [4] ad the relatioship betwee distributios ad their mgfs is uique [E]. "Ustadardisig", X teds to N(μ, σ ) i.e. N(, ) [B, parameters must be give]. [3]

24 4769 Mark Scheme Jue 00 Questio 3 (i) H 0 is accepted if.96 < value of test statistic <.96 i.e. if ( x x ) ( 0) < < i.e. if < x x < i.e. if.0(8) < x x <.0(8) Note. Use of μ μ istead of x x ca score M B M0 M A0 A0. (ii) x x.4 which is outside the acceptace regio so H 0 is rejected. M double iequality B.96 M um r of test statistic M de r of test statistic A A Special case. Allow out of of the A marks if.645 used provided all 3 M marks have bee eared. B FT if wrog M [FT ca's acceptace regio if reasoable] E [6] (iii) CI for μ μ :.4 ± ( ), i.e..4 ±.5796, i.e. ( 0.8 [ 0.796],.97[96] ) Wilcoxo rak sum test (or Ma-Whitey form of test) M for.4 B for.576 M for 0.63 A cao for iterval M [7] Raks are: First Secod W [or if M-W used] M Combied rakig A Correct [allow up to errors; FT provided M eared] B Refer to W 6,8 [or MW 6,8 ] tables. Lower ½% critical poit is 9 [or 8 if M-W used]. M No FT if wrog A Cosideratio of upper ½% poit is also eeded. Eg: by usig symmetry about mea of ( ) ( , critical poit is 6. [For M-W: mea is 6 8 4, hece critical poit is 40.] Result is sigificat. Seems (populatio) medias may ot be assumed equal. ) Special case. If M for W 6,8 has ot bee awarded (likely to be because rak sum 43 has bee used, which should be referred to W 8.6 ), the ext two M marks ca be eared but othig beyod them. M M for ay correct method A if 6 correct E, E Special case (does ot apply if Special Case has bee ivoked). These marks may be eared eve if oly or of the precedig 3 have bee eared. [] 3

25 4769 Mark Scheme Jue 00 Questio 4 (i) Radomised blocks B Eg:- WEST D C D EAST A B C C A A B D B (ii) Plots i strips (blocks) correctly aliged w.r.t. fertility tred. Each letter occurs at least oce i each block i a radom arragemet. μ populatio [B] grad mea for whole experimet [B] α i populatio [B] mea amout by which the ith treatmet differs from μ [B] M E M E 4 marks, as show [5] e ij ~ id N [B, accept "ucorrelated"] (0 [B], σ [B] ) (ii) Totals are all from samples of size 5 3 marks, as show [7] Grad total 65. "Correctio factor" CF 65. / Total SS CF Betwee varieties SS CF CF Residual SS (by subtractio) Source of variatio SS df MS [M] MS ratio [M] Betwee varieties [B] [A cao] Residual [B] Total Refer MS ratio to F 3,6. Upper 5% poit is 3.4. Sigificat. Seems the mea yields of the varieties are ot all the same. M for attempt to form three sums of squares. M for correct method for ay two. A if each calculated SS is correct. 5 marks withi the table, as show M No FT if wrog A No FT if wrog E E [] 4

26 4769 Mark Scheme Jue 0 Questio Aswer Marks Guidace (i) P X x F ( x). F ( x). M use of cdfs (ii) B G x B x G x f x f ( x) f ( x) ie cdf of X is F F ( ) F ( ) A ie (by differetiatig) pdf of X is A Aswer give E X xf ( x)dx xf ( x)dx B G B G [] (iii) E X x f( x)dx M x f B( x)dx x f G( x)dx M Use of "E X " B G B M A Var( X ) E( X ) E( X) B G 4 B 4 G B G 4 B G (iv) [Cetral Limit Theorem] Approx dist of N, B G 4 B G X is G [3] M [aswer give; eeds some idicatio of method] M A A Aswer give [7] B B B B B4 4 marks as show [4] (v) X X X X MM X X 4 st B G E st B G Var either B ad Var st B [4] 5

27 4769 Mark Scheme Jue 0 Questio Aswer Marks Guidace (vi) E EX ( ) EX ( st ) ( B G ) EX ( ) ie they are ubiased. E Clearly Var X Var X st, M for ay attempt to compare variaces Cadidates are ot required to ote that the variaces are equal i M the case μb μg. Var X Var X st [FT c X st is the more efficiet. E More efficiet [5] (i) Mea of X 3.5 (immediate by symmetry) B E X M 6 9 A 7 35 Var X 6 [3] 6 t t X t 6 t 6 t 6 6 t t 6 6 t t... t (ii) G E M 6t A [] Aswer give Aswer give (iii) [P(N 0) ½, P(N ) (½)(½),] P(N r) (½) r.(½) B aswer give; must be covicig [] 0 t t N t t 4 t 8 (iv) H E M t t t A Aswer give 6

28 4769 Mark Scheme Jue 0 Questio Aswer Marks Guidace (v) Mea H'(), variace H''() + mea mea. M for use of st derivative H' t t t mea A t t t 3 3 H'' M for use of d derivative variace + A For variace [4] (vi) K(t) H{G(t)} { G(t)} M (vii) 6 5 t t t t t tt... t 6 t 6 t 3 6 tt t... t 6 6 tt... t 6 6 t tt t t t t 3 t 4 t 5 K' t tt t t t t t t 6 t t t t t t 3 t 4 3 K'' M M A [4] M M M isertig G(t) use of hit give Aswer give reasoable attempt to differetiate K(t) reasoable attempt at d derivative for use of derivatives mea K '() 6( 6) () /6 7/ A Substitutio show K''() ( 6 3 ) + (6 6 70) (49/) + (70/6) A variace A 6 4 [6] Ft c s K () ad/or K () provided variace positive Exact. 7

29 4769 Mark Scheme Jue 0 Questio Aswer Marks Guidace (viii) We have: 7/ M for correct use of cadidate's values for meas ad variaces X X N N Q 35/ 39 / Isertig i the quoted formula gives A aswer hoestly obtaied (commo deomiator show) A0 if differet from (vii) as required. [] 3 (i) H 0 : populatio medias are equal B [Note: "populatio" must be explicit] ) Explicit statemet re shapes of distributios. H : populatio media for A < populatio media for B B (eg that they are the same shape) is ot required. ) More formal statemets of hypotheses gai both marks [eg cdfs are F(x) ad F(x Δ), H 0 is Δ 0 etc).] Wilcoxo rak sum test (or Ma-Whitey form of test) Raks are: A M Combied rakig B A Correct [allow up to errors; FT provided M eared] W B [or if M-W used] Refer to W 6,8 [or MW 6,8 ] tables M No FT if wrog Lower 5% critical poit is 3 [or 0 if M-W used] A No FT if wrog Result is ot sigificat A Seems media yields may be assumed equal A [9] 8

30 4769 Mark Scheme Jue 0 Questio Aswer Marks Guidace 3 (ii) H 0 : populatio meas are equal B H : populatio mea for A < populatio mea for B B "populatio" must be explicit, either i words or by otatio For A: x.4, s.9 [ s.3875] B For all. Use of s scores B0 For B: y.575, s.05[ s.05] Pooled s (5.9) (7.05) Test statistic (5) M A M A for ay reasoable attempt at poolig (but ot if s used) If correct Ft if icorrect Refer to t M No FT if wrog Lower sigle-tailed 5% critical poit is.78 A No FT if wrog must compare.83 with.78 uless it is clear ad explicit that absolute values are beig used Sigificat A Seems mea yield for A is less tha that for B A [] 3 (iii) t test is "more sesitive" if Normality is correct. E No-rejectio of Normality supports t. E But Wilcoxo is more reliable if ot Normal E ad we do ot have proof of Normality. E [4] 4 (i) Lati square B 4 4 layout, with rows clearly represetig castig techiques ad colums clearly represetig chemical compositios [or vice B,,0 B if completely correct, B if oe poit omitted versa] Labels each appearig exactly oce i each row ad i each B for correct structure colum represetig maufacturers. B withi the square [5] 9

31 4769 Mark Scheme Jue 0 Questio Aswer Marks Guidace 4 (ii) Totals are all from samples of size 4 Grad total 8.7 "Correctio factor" CF 8.7 / Total SS CF Betwee maufacturers SS M for attempt to form three sums of squares CF CF M for correct method for ay two Residual SS (by subtractio) A if each calculated SS is correct. Source of variatio SS df MS MS ratio Betwee treatmets Residual (6) Total B B M M A Betwee treatmets df Residual df Method for calculatig either mea square MS ratio cao at least 3sf, codoe up to 6 sf 4 (iii) Refer MS ratio to F 3,. M No FT if wrog quotatio of a extreme poit (ot just the 5%poit); eg % poit, A 5.95, or 0.% poit, Must be some referece to very highly sigificat A ad very strog/overwhelmig evidece that "the maufacturers are ot all the same. A [] y e B B if ay two RHS terms correct ij i ij populatio B Populatio ; award here or for i grad mea for whole exp't B i populatio mea amout by which the ith treatmet differs B from e ij is experimetal error does ot eed to be stated explicitly here, is subsumed i error assumptios below. e ij ~ id N [*] ( accept "ucorrelated") (0 [*], [*] ) B if all three * compoets correct, B if ay two correct. [7] 0