Questio 1 (i) EITHER: 1 S xy = xy x y = 198.56 1 19.8 140.4 =.44 x x = 1411.66 1 19.8 = 15.657 1 S xx = y y = 1417.88 1 140.4 = 9.869 14 Sxy -.44 r = = SxxSyy 15.6579.869 = 0.76 1 S yy = 14 14 M1 for method for S xy M1 for method for at least oe of S xx or S yy for at least oe of S xy, S xx, S yy correct M1 for structure of r (-0.7 to -0.8 to dp) If x ad y used i rouded form, be geerous with first Structure of r eeds to be fully correct i all parts the first two M1 Sxy marks must have bee eared ad r = applied. S S xx yy OR: cov (x,y) = = 198.56/14 9.9857 10.086 xy x y = 0.454 M1 for method for cov (x,y) rmsd(x) = S xx = (15.657/14) = 1.1184 = 1.0575 M1 for method for at least oe msd rmsd(y) = S yy = (9.869/14) = 0.7049 = 0.896 for at least oe of cov (x,y), msd(x), msd(y) correct If x ad y used i rouded form, be geerous with first cov(x,y) -0.454 r = = rmsd( x) rmsd( y) 1.05750.896 = 0.76 NB: usig oly dp i calculatig x ad y leads to aswer of 0.84 which is still i the acceptable rage M1 for structure of r (-0.7 to -0.8 to dp) 5 Structure of r eeds to be fully correct i all parts the first two M1 cov(x,y) marks must have bee eared ad r = applied. rmsd( x) rmsd( y) 1
(ii) H 0 : ρ = 0 B1 for H 0, H 1 i Codoe hypotheses writte i words ad cotext. H 1 : ρ 0 (two-tailed test) symbols e.g. allow H 0 : There is o correlatio betwee x & y, H 1 : There is correlatio betwee x & y. (i.e. allow x & y as cotext sice these are defied i the questio) where ρ is the populatio correlatio coefficiet B1 for defiig ρ NB If hypotheses give oly i words ad associatio metioed the do ot award first B1 ad last B1 For hypotheses writte i words, cadidates must make it clear that they are testig for evidece of correlatio i the populatio. For = 14, 5% critical value = 0.54 B1 for critical value (+ or -) Oe-tailed test cv = (-) 0.4575 Sice 0.76 > 0.54 the result is ot sigificat. Thus we do ot have sufficiet evidece to reject H 0 There is ot sufficiet evidece at the 5% level to suggest that there is correlatio betwee birth rate ad death rate (iii) The uderlyig populatio must have a bivariate Normal distributio. Sice the scatter diagram has a roughly elliptical shape. M1 for a sesible compariso leadig to a coclusio (provided that -1 < r < 1) for correct result ft their r B1 ft for coclusio i cotext B1 E1 for elliptical shape 6 Compariso should be betwee the cadidate s value of r from part (i) ad a appropriate cv (i.e. the sig of the cv ad the sig of r should be the same). NOTE If result ot stated but fial coclusio is correct award SC1 to replace the fial B1 Not bivariate ad Normal (iv) Because this data poit is a log way from the other data ad it is below ad to the right of the other data. It does brig the validity of the test ito questio sice this extra data poit is so far from the other poits ad so there is less evidece of ellipticity. E1 for a log way E1 for below ad to the right of. E1 for does cast doubt o validity E1 for less elliptical 4 Idicatio that the poit is (possibly) a outlier For idetifyig the positio of this poit (allow i terms of x ad y) Allow o but oly with with suitable explaatio e.g. the sample is still too small to provide evidece either for or agaist the presece of ellipticity. TOTAL 17
Questio (i) xf Mea = = 9 50 = 1.84 = 015471610 50 1 Variace = fx x 1 1 58 50 1.84 49 = = 1.81 (to d.p.) (ii) Because the mea is close to the variace B1 B1 for mea M1 for calculatio 1 Use of MSD gets M1 A0 Stadard deviatio gets M0 A0 uless Variace = 1.81 is see. Must compare mea ad their variace as foud i part (i) (iii) (A) P(No sultaas) = e 0 1.84 0! = 0.159 ( s.f.) M1 for probability calc. [1.8 leads to 0.165] (B) P(At least two sultaas) = 1 e 0 1.84 0! e 1 1.84 1! =1 0.159 0.9 = 0.549 (iv) λ = 51.84 = 9. Usig tables: P(X 10) = 1 P(X 9) M1 for P(1) M1 for 1 [P(0) + P(1)] used cao B1 for mea (SOI) M1 for 1 P(X 9) 5 Or attempt to fid P() + P() + P(4) + + P(8) Use of λ = 1.8 loses both accuracy marks [1.8 leads to 1 0.496 = 0.57] Ay λ = 1 0.5611 (= 0.489 NB ANSWER GIVEN)
(v) P( out of 6 cotai at least te sultaas) M1 for p q 4 p + q = 1 6 = M1 dep for coefficiet Coefficiet of 15 as part of a biomial calculatio 0.489 0.5611 4 = 0.864 ft if p rouded from part (iv) (vi) Use Normal approx with μ = p = 60 0.489 = 6.4 σ = pq = 60 0.489 0.5611 = 14.776 0.5 6.4 P(X > 0) = PZ 14.776 B1 for μ B1 for σ B1 for correct cotiuity correctio SOI Allow 6. Allow 14.8 = P(Z > 1.088) = 1 Φ(1.088) = 1 0.8608 = 0.19 M1 for probability usig correct tail. FT their μ & σ cao 5 0 (givig P(Z > 1.091..) = 0.17 sf ) But do ot FT wrog or omitted CC 4
Questio (i) (A) P(X < 5) 5 55 = P Z 5 = P( Z < 0.577) = 1 Φ(0.577) = 1 0.7181 = 0.819 M1 for stadardisig M1 for correct structure CAO NB Whe a cadidate s aswers suggest that (s)he appears to have eglected to use the differece colum of the Normal distributio tables pealise the first occurrece oly Igore spurious cotiuity correctios & allow reversal of umerator i.e. correct tail (icludig below a egative z) Allow aswers which roud to 0.8 (B) P( 00 < X < 400) 00 55 400 55 = P Z 5 5 = P 1.058 Z 0.865 = Φ(0.865) (1 Φ(1.058)) = 0.8065 (1 0.8549) = 0.6614 (0.6615 from GDC) M1 for stadardisig both M1 for correct structure CAO Pealise spurious cotiuity correctios Allow 0.66 if pealised iappropriate table use already Use of stadard deviatio = 5 or 5 ca ear M1 for structure oly i each part max /6 (ii) From tables Φ -1 ( 0.) = 0.8416 B1 for ± 0.8416 see NOT 1 0.8416 k 55 0.8416 5 k = 55 0.8416 5 = 11. M1 for equatio i k CAO Equatio must be equivalet to this. Pealise use of + 0.8416 uless umerator has bee reversed. Codoe other z values but use of probabilities here, e.g. use of 0. or Φ(0.) = 0.579, gets M0 A0 Allow 11 5
(iii) H 0 : μ = 55; H 1 : μ 55. Where μ deotes the populatio mea (reactio time for wome) B1 for use of 55 i hypotheses B1 for both correct B1 for defiitio of μ Use of 55 i hypotheses ad hypotheses give i terms of μ ot p or x, etc. uless letter used is clearly defied as populatio mea Test statistic = 44 55 11 1.058 5 / 5 10.4 M1 must iclude 5 Allow + 1.058 oly if later compared with + 1.96 5% level tailed critical value of z = 1.96 1.058 > 1.96 so ot sigificat. There is ot sufficiet evidece to reject H 0 B1 for 1.96 M1 for a sesible compariso leadig to a coclusio Or -1.96 There is isufficiet evidece to coclude that wome have a differet reactio time from me i this experimet. for correct coclusio i words i cotext 8 Do ot accept me ad wome have same reactio time TOTAL 17 6
Questio 4 (i) H 0 : o associatio betwee pebble size ad site H 1 : some associatio betwee pebble size ad site; B1 Must be i cotext NB if H 0 H 1 reversed, or correlatio metioed, do ot award first B1 or fial E1 EXPECTED Site A Site B Site C Large 1.70 9.44 1.86 Medium..96.70 Small 4.96 9.60 4.44 M1 A for expected values (to dp) (allow for at least oe row or colum correct) 1d.p.ca get M1A0 M1A ca be implied by correct cotributios/fial aswer CONTRIB N Site A Site B Site C Large 0.16 0.6940 1.071 Medium 0.85 1.5484.7861 Small 0.79 0.91 1.744 M1 for valid attempt at (O-E) /E NB These (M1) marks caot be implied by a correct fial value of X. for at least 1 row/colum correct X = 10.1 M1 for summatio for X Depedet o previous M1 Refer to X 4 B1 for 4 deg of freedom Critical value at 5% level = 9.488 B1 CAO for cv Award oly if o icorrect workig see Result is sigificat B1 ft their sesible X ad critical value Allow reject H 0. B0 if critical value of 0.711 (lower tail) or.776 (t distributio) used. There is evidece to suggest that there is some associatio betwee pebble size ad site E1 must be cosistet with their X 1 Depedet o previous B1 SC1 (to replace B1E1 if first B1B1 eared where sigificat ot stated but fial statemet is correct) 7
(ii) Site A Cotributes least to X showig that frequecies are as expected if there were o associatio. OR Cotributio of 0.85 implies that there are (slightly) fewer medium pebbles tha expected. Site B Cotributio of 1.5484 implies that there are fewer medium pebbles tha expected. E,1,0 E,1,0 NOTE For each site, some referece to cotributios eeded (explicitly or implicitly). Award E oly if o icorrect additioal commet made. Allow large/small as expected or more tha expected ad medium as expected or less tha expected for E1 (if cotributio ot metioed) Award E oly if o icorrect additioal commet made. Allow large/small as expected or more tha expected ad medium less tha expected for E1 (if cotributio ot metioed) Site C Cotributio of.7861 implies that there are a lot more medium tha expected. E,1,0 Need a lot more for E Award E oly if o icorrect additioal commet made. Allow large/small fewer tha expected ad medium more tha expected for E1 (if cotributio ot metioed) NB MAX /6 for aswers ot referrig to cotributios (explicitly or implicitly). TOTAL 18 8
Additioal otes re Q1(ii) For those carryig out a oe-tailed test, B0 B1 B1 M1 B1 is available provided that workig is cosistet with a oe-tailed test beig used. For the fial B1 to be eared, the coclusio should refer to alterative hypothesis used. e.g. There is ot sufficiet evidece at the 5% level to suggest that there is a egative correlatio betwee birth rate ad death rate. If the cv is take from the Spearma s Test table (i.e. -0.585 ad -0.467) the the third B1 will be lost. If other sesible cvs are used the oly B1 B1 B0 M1 A0 B0 available. Use of t distributio leads to B1 B1 B0 M0 A0 B0 max. Additioal otes re Q(iii) Critical Value Method 55 1.96 5 5 gets M1B1 = 4.6 gets 4.6 < 44 gets M1for sesible compariso still available for correct coclusio i words & cotext Cofidece Iterval Method CI cetred o 44 + or - 1.96 5 5 gets M1 B1 = (.6, 64.84) cotais 55 gets M1 still available for correct coclusio i words & cotext Probability Method Fidig P(sample mea < 44) = 0.1451 gets M1 B1 0.1451 > 0.05 gets M1 for a sesible compariso if a coclusio is. 0.1451 > 0.05 gets M1 A0 uless usig oe tailed test still available for correct coclusio i words & cotext. Codoe P(sample mea > 44) = 0.8549 for M1 but oly allow if later compared with 0.975 at which poit the fial M1ad are still available Oe-tailed test Max B1 B0 B1 M1 B1 (for cv = -1.645) M1 (provided that the coclusio relates to H 1 : μ < 55, e.g. there is isufficiet evidece to suggest that wome have a lower reactio time tha me i this experimet). Cosistet use of σ = 5 Do ot pealise i parts (ii) ad (iii). 9