ࡱ> @ ~bjbjFF ,,-~4999h:<T;$$<DhA(AAAf6Hf$lf  $ӟR%09h0^e"hh0AAiEEEhp lAA Eh EE,A< ÿ9rL4>0gxhfZf@Eg4Ng]fff00,l4D}Rl4ask LEVEL "WHAT LEVEL?" ask SUBJECT "WHAT SUBJECT?" Comparison of key skills specifications 2000/2002 with 2004 standardsask CODE "WHAT CODE?" X015461ask DATE "WHAT DATE?" July 2004ask ISSUE "WHAT ISSUE?" Issue 1     GCE Mathematics (6691/01) June 2007 6691 Statistics S3 Mark Scheme  Question Scheme Marks number  1. (a) M1A1 ABCDEFGH P Rank26431785 Q Rank28163574 EMBED Equation.DSMT4 04994411  EMBED Equation.DSMT4  M1A1  EMBED Equation.DSMT4  M1  EMBED Equation.DSMT4  A1 (6)  (b)  EMBED Equation.DSMT4  ( EMBED Equation.DSMT4  is OK) both B1  EMBED Equation.DSMT4  1 tail 5% critical value is 0.6429 (Independent of their  EMBED Equation.DSMT4 ) B1 (+ is OK) 0.619 < 0.6429 or not significant M1 So insufficient evidence of a positive correlation between judges Or competitor is justified A1f.t. (4) 10  (a) 1st M1 for attempting to rank both P and Q. 1st A1 for both correct (could be reversed) 2nd M1 for attempting  EMBED Equation.DSMT4  2nd A1 for  EMBED Equation.DSMT4 . 3rd M1 for correct use of formula for  EMBED Equation.DSMT4  (b) M1 for a correct comparison or statement about significance (o.e.) Follow through their  EMBED Equation.DSMT4  provided 0<  EMBED Equation.DSMT4 <1 A1f.t. for a conclusion in context. Must mention judges or marks or competitor. If they use correlation they must say it is positive. Follow through their positive  EMBED Equation.DSMT4  with their positive c.v. and ignore hypotheses. So  EMBED Equation.DSMT4  =0.667 they could say competitors claim is not justified etc. S.C. No ranking Typical answer (-3.82) can get mark for use of  EMBED Equation.DSMT4 formula and hypotheses in (b) only (a) M0A0M0A0M1A0 (b) B1B1M0A0   Question Scheme Marks number  2. (a)  EMBED Equation.DSMT4 Maths grades are independent of English grades or No association  EMBED Equation.DSMT4  Maths and English grades are dependent or There is an association B1 2027.512.52027.512.5 Expected Frequencies e.g.  EMBED Equation.DSMT4  M1 A1  EMBED Equation.DSMT4 , = 3.9545 AWRT 3.95 or 3.955 M1, A1  EMBED Equation.DSMT4 4.605 B1; B1 3.95< 4.605 or not significant or do not reject  EMBED Equation.DSMT4  (allow reject  EMBED Equation.DSMT4  ) M1 Insufficient evidence of an association between English and maths grades or there is support for the Directors belief A1 (9) or Students grades in maths and English are independent (b) May have some expected frequencies <5 (and hence need to pool rows/cols) B1 (1) 10  (a) 1st B1 for both hypotheses in terms of independence or association and in context. Must mention Maths and English in at least one of the hypotheses. relationship or correlation or connection or link is B0 1st M1 for some correct calculation seen 1st A1 for all expected frequencies correct. Accept answers without formula seen. 2nd M1 for some evidence seen of attempt to calculate test statistic. At least one correct term seen. Follow through their expected frequencies. 2nd A1 for AWRT 3.95. Answers only please escalate! 3rd M1 for correct comparison or statement may be implied by correct conclusion. 3rd A1 for conclusion in context using association or independence in connection with grades. Dont insist on seeing English or maths mentioned here. Use ISW for comments if a false statement and correct statement are seen. (b) B1 If they just say expected frequencies are small they must go onto mention need to pool.  Question Scheme Marks number  3.  EMBED Equation.DSMT4  B1, B1  EMBED Equation.DSMT4  AWRT 1.94 M1, A1 5% one tail c.v. is z = (-) 1.6449 or probability (AWRT 0.026) (+) 1.6449 B1 - 1.94 < -1.6449 or significant or reject  EMBED Equation.DSMT4  or in critical region M1 There is evidence that the (mean) time to complete the puzzles has reduced Or Robert is getting faster (at doing the puzzles) A1f.t. 7 1st & 2nd B1 must see m and 18 1st M1 for attempting test statistic, allow + . Or attempt at critical value for  EMBED Equation.DSMT4 :  EMBED Equation.DSMT4  1st A1 for AWRT  1.94. Allow use of |z| = +1.94 to score M1A1. Or critical value = AWRT 16.7. 3rd B1 for AWRT 0.026 (i.e. correct probability only) or + 1.6449. (May be seen in cv formula) 2nd M1 for correct comparison or statement relating their test statistic and 1.6449 or their probability and 0.05. Ignore their hypotheses if any or assume they were correct. 2nd A1f.t. for conclusion in context which refers to speed or time. Depends only on previous M   Question Scheme Marks number  4. (a)  EMBED Equation.DSMT4 ,  EMBED Equation.DSMT4  (Accept  EMBED Equation.3  or 2 per 20) M1, A1 (2) (b) e.g.  EMBED Equation.DSMT4  M1 r = 28.5, s = AWRT 9 A1, A1 (3) x0123> 4 EMBED Equation.DSMT4 1731191419 EMBED Equation.DSMT4 12.227.028.519.013.3 EMBED Equation.DSMT4 1.890.593.171.322.44 (c) Pooling M1  EMBED Equation.DSMT4  AWRT 9.4 M1A1c.a.o.  EMBED Equation.DSMT4  B1ft, B1ft  EMBED Equation.DSMT4 Binomial distribution is a good/suitable model/fit [Condone: B(20, 0.1) is]  EMBED Equation.DSMT4  Binomial distribution is not a suitable model both B1 (Significant result) Binomial distribution is not a suitable model A1cao (7) (d) defective items do not occur independently or not with constant probability B1ft (1)  13 (a) M1 for attempt to find mean or  EMBED Equation.DSMT4 (as printed or better). The 0.1 must be seen in part (a). (b) M1 for correct expression for r or s using the binomial distribution. Follow through their  EMBED Equation.DSMT4 . (c) 1st M1 for some pooling (accept x > 5, obs.freq. 14, 9, 10 and exp.freq. 19.0, s, 4.3) 2nd M1 for calculation of test statistic (N.B. x > 5 gives 14.5). One correct term seen. 1st B1ft for number of classes 2 (N.B. x > 5 will have 6 2 = 4) 2nd B1ft for the appropriate tables value, ft their degrees of freedom. (NB  EMBED Equation.DSMT4 ) 3rd B1 (for hypotheses) allow just X ~ B(20, 0.1) for null etc. 2nd A1 for correctly rejecting Binomial model. No ft and depends on 2nd M1. (d) B1ft for independence or constant probability must mention defective items or defectives Follow through their conclusion in (c). So if they do not reject they may say defectives occur with probability 0.1. Stating the value implies constant probability.    Question Scheme Marks number  5. (a)  EMBED Equation.DSMT4  M1, A1  EMBED Equation.DSMT4  M1A1ft AWRT 1.51 A1 (5) (b)  EMBED Equation.DSMT4  B1 B1 Denominator M1  EMBED Equation.DSMT4  z dM1 = (+) 1.8689 AWRT (+) 1.87 A1 One tail c.v. is z = 1.6449 (AWRT 1.645 or probability AWRT 0.0307 or 0.0308) B1 (significant) there is evidence that diet A is better than diet B or evidence that (mean) weight lost in first week using diet A is more than with B A1ft (7) (c) CLT enables you to assume that  EMBED Equation.DSMT4  are normally distributed B1 (1) (d) Assumed  EMBED Equation.DSMT4  (either) B1 (1) 14  (a) 2nd M1 for a correct attempt at s or  EMBED Equation.DSMT4 , A1ft for correct expression for  EMBED Equation.DSMT4 , ft their mean. N.B.  EMBED Equation.DSMT4 is M1A1ft (b) 1st B1 can be given for  EMBED Equation.DSMT4 , but 2nd B1 must specify which is A or B. 1st M1 for the denominator, follow through their 1.51. Must have square root can condone  EMBED Equation.DSMT4  but  EMBED Equation.DSMT4  is M0. Allow  EMBED Equation.DSMT4  leading to AWRT 1.85 to score M1M1A0 in (b) and can score in (d). 2nd dM1 for attempting the correct test statistic, dependent on denominator mark 1st A1 for AWRT + 1.87, may be implied by a correct probability. 2nd A1ft ft their test statistic vs their cv only if  EMBED Equation.DSMT4  is correct and both Ms are scored (c) B1 for stating either  EMBED Equation.DSMT4  (but not A or B) are normally distributed (d) B1 for either, can be stated in words in terms of variances or standard deviations.   Question Scheme Marks number  6.  EMBED Equation.DSMT4  B1 2.5758 B1  EMBED Equation.DSMT4  M1 AWRT 1.96 B1  EMBED Equation.DSMT4  M1 So 95% C.I. = 139.1 + 11.87= (127.22, 150.97) AWRT (127, 151) A1  6 1st B1 for mean = 139.1 only 1st M1 for UL mean or mean LL set equal to z value times standard error or some equivalent expression for standard error. Follow through their 2.5758 provided a z value. May be implied by  EMBED Equation.DSMT4  = 6.056 [N.B.  EMBED Equation.DSMT4 ] Condone poor notation for standard error if it is being used correctly to find CI. 2nd M1 for full method for semi-width (or width) of 95% interval Follow through their z values for both M marks N.B. Use of 2.60 instead of 2.5758 should just lose 2nd B1 since it leads to AWRT (127, 151)    Question Scheme Marks number  7. (a) Let X = L 4S then E(X) = 19.7 4 EMBED Equation.DSMT4 ,= 0.1 M1, A1 Var(X) =  EMBED Equation.DSMT4  M1, M1 = 0.89 A1 P(X > 0) = [P(Z > -0.10599)] M1 = AWRT (0.542 0.544) A1 (7) (b)  EMBED Equation.DSMT4  (May be implied by 0.16) M1  EMBED Equation.DSMT4  E(T) = 19.6 B1 Var(T) = 0.16 or  EMBED Equation.DSMT4  A1 (3) (c) Let Y = L T E(Y ) = E(L) E(T) = [ 0.1] M1 Var(Y) = Var(L) + Var(T) = [ 0.41] M1 Require P( - 0.1 < Y < 0.1) M1 = P(Z <0) P(Z < - 0.31..) or 0.5 P(Z < - 0.31..) or P(Z < 0.31..) P(Z < 0) M1 = 0.1217 (tables) or 0.1226.. (calc) AWRT (0.122 0.123) A1 (5) 15  (a) 1st M1 for defining X and attempting E(X) 1st A1 for 0.1. Answer only will score both marks. 2nd M1 for Var(L) +.. 3rd M1 for . EMBED Equation.DSMT4 Var(S). For those who dont attempt L 4S this will be their only mark in (a). 2nd A1 for 0.89 4th M1 for attempting a correct probability, correct expression and attempt to find, which should involve some standardisation: ft their  EMBED Equation.DSMT4  and their 0.1. If 0.1 is used for E(X) answer should be > 0.5, otherwise M0. (c) 1st M1 for a correct method for E(Y), ft their E(T). 2nd M1 for a correct method for Var(Y), ft their Var(T). Must have +. 3rd M1 for dealing with the modulus and a correct probability statement. Must be modulus free. May be implied by e.g. P(Z <  EMBED Equation.DSMT4 ) 0.5, or seeing both 0.378 (or 0.622) and 0.5 4th M1 for correct expression for the correct probability, as printed or better. E.g. 0.5 + 0.378.. is M0 A1 for AWRT in range.       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