Exam Advice. You will find helpful advice about common errors in the Examiners Reports. Some specific examples are dealt with here.

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Francine Robinson
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1 Exam Advice You will fid helpful advice about commo errors i the Examiers Reports. Some specific examples are dealt with here. All modules The OCR Report o the Uits tae i Jue 2006 cotais a statemet of the rules examiers use whe they see replaced or crossed out wor. Our advice is that you should show this statemet to your studets. The fial paragraph reads as follows. If there are two or more attempts at a questio which have ot bee crossed out, examiers should mar what appears to be the last (complete) attempt ad igore the others. A few cadidates adopt a policy that, whe i doubt betwee two alterative aswers to a questio, the best course of actio is to give both, o the assumptio that the examier will select the better oe. The wordig above maes it clear that this is ot the case; such a cadidate will lose mars uless the better aswer happes to be the last oe. This is a quite differet situatio from a cadidate beig geuiely uhappy with a aswer ad replacig it, whether or ot the earlier wor is crossed out. These evetualities are covered i two earlier paragraphs of the OCR statemet. If a cadidate attempts a questio more tha oce, ad idicates which attempt he/she wishes to be mared, the examiers should do as the cadidate requests. If two or more attempts are made at a questio, ad just oe is ot crossed out, examiers should igore the crossed out wor ad mar the wor that is ot crossed out. It is ormal practice to mar wor that is crossed out ad ot replaced, assumig that the crossig out has ot bee so thorough as to mae it illegible. The use of Tippex i examiatios ca be a expesive way of losig mars. Use of calculator May cadidates are losig mars because they do ot uderstad the meaig of the word exact. A example of this occurred i the Questio 3 of the Jue 2006 C2 paper. θ is a acute agle ad siθ = 4. Fid the exact value of taθ. Cadidates were expected to draw a right-agled triagle, with hypoteuse 4 uits ad the side opposite the agle θ uit, ad the to use Pythagoras theorem 4 to show that the legth of the third (adjacet) side is 5. So the exact value of taθ is 5. θ May cadidates, however, used their calculators to fid arcsi(0.25) ad the used them agai to fid ta of the aswer. Cadidates who did this commoly gave their aswers as 0.25 or , either of which is the exact aswer which was ased for i the questio. The Examier s Report says It must be made clear to cadidates, as part of the preparatio for this paper, that a request for the exact value of aythig usually implies that calculators must ot be used. The same advice is give o page 32 of the Studets Hadboo. Updated 0/2/0 page

2 Use of Calculator 2 The September 2006 Newsletter icluded a remider that questios which as for a exact value should be doe without a calculator. Calculators are allowed i all A Level uits other tha C, but cadidates do lose mars i all uits if they fail to show sufficiet details of their worig. For example, a questio i FP2 might as cadidates: 5 Show that cosh 2 x dx= Havig doe the first step correctly cosh 2 x dx = sih 2x 2 ad got full credit for it, cadidates might tur to their calculators ad complete the questio thus: 5 sih 2x = sih ( 2l2) sih ( 2l2) 2 =. 2 2 Typig sih ( 2l2) sih ( 2l2) ito may scietific calculators does give the 2 2 exact aswer 5 but o credit ca be give for this part of the questio because the aswer was prited ad cadidates could just have copied it dow. Cadidates are expected to show worig, for example: 2l2 2l2 l4 l4 e e e e 5 sih(2) 5 = = = 4 = sih( 2) = So sih 2 x = =. The examiers reports for the hyperbolic fuctios questios o FP2 show studets ofte drop mars o this id of questio because they do ot show sufficiet worig. Pure Modules Asymptotes i MEI Structured Mathematics The word asymptote is ot icluded i the atioal Subject Core for AS ad A Level Mathematics ad testig of asymptotes is excluded from C. However, studets are liely to have see asymptotes already, for example, through setchig the curve y = i GCSE. x I C2 studets meet asymptotes whe they draw graph of y = taθ. This curve, together with trasformatios of it, is explicitly i the syllabus. So it is liely that, by ow, studets will have met the word asymptote, at least i the cotext of vertical asymptotes, sice teachers will have used the word whe describig these curves. Whe doig C3 coursewor, studets are expected to ow that failure of a chage of sig method ca occur for solvig f ( x ) = 0 whe y= f ( x) has a vertical asymptote, as stated i Note C3e2 o page 6 of the specificatio. Updated 0/2/0 page 2

3 I examiatio questios i C3 ad C4, for example Jue 2007 C3 questio 7, the correct techical term asymptote is used to avoid a legthy explaatio. Teachers are advised to loo carefully at the style used i previous questios o these papers ad to treat them as establishig precedet. Cosiderable care was tae whe settig such questios to mae sure that the meaig was clear from the accompayig text ad/or diagrams ad this will always be the case whe the word asymptote is used o the C3 ad C4 papers. I FP vertical ad horizotal asymptotes are explicitly i the syllabus (Note FPC2 o page 75 of the specificatio) ad are routiely tested i examiatio questios. Oblique asymptotes are first met i FP2 (Note FP2C4 o page 9). FP Studets ofte drop mars uecessarily i proof by iductio questios by givig ustructured proofs ad/or missig out vital steps. I FP these questios will always relate to provig formulae for the sums of series or for the th term of a sequece. Below is a example questio (adapted from a questio i the FP textboo), together with a mar scheme. This is iteded to highlight how the mars are allocated. Note that failure to state explicitly that the cojecture is assumed true for =, which is a fudametal logical step i a proof by iductio, will cost a miimum of 3 mars, eve if everythig else is correct. Example questio A sequece of itegers 5 u, u2, u3,..., u u = ad u = 3u 2 + for. u is defied by Prove by iductio that for all positive itegers. Mar scheme u = 5 ad whe =, u = 5, so cojecture true for = Mar B Notes Clear demostratio for the case = Assume true for =, so that u u + = 32 ( + 3) 2 + = = 2(3 ) But this is the give result with + replacig. Therefore if it is true for, it is true for +. Sice it is true for =, it is true for =, 2, 3, ad so it is true for all positive itegers. E M A A E E Assume true for = Use of iductive defiitio to fid u + Correct substitutio Correct simplificatio (correct aswer oly) Depedet o previous E ad immediately previous A Depedet o B ad both previous E mars At the simplificatio stage, some teachers ecourage the method of settig up a target to aim for. This is perfectly acceptable ad is mared equivaletly. Updated 0/2/0 page 3

4 S2 ad Z3 Null hypothesis for correlatio questios For tests usig Spearma s ra correlatio coefficiet, the ull hypothesis should have the form: H 0 : There is o associatio betwee the variables i the uderlyig populatio There is o uiversally agreed way of writig this algebraically so the ull hypothesis must be give i words. There will be suitable correctios i the ext reprit of the MEI S2 text boo. For tests usig Pearso s product momet correlatio coefficiet, the ull hypothesis ca be writte as: H 0 : ρ = 0, where ρ is the populatio correlatio coefficiet Cadidates should say what ρ stads for, emphasisig that it refers to the uderlyig populatio. They should say what ay letter they itroduce ito a questio stads for but a mar is ofte give specifically for this i the case of hypothesis tests. t tests i S3 I the specificatio for S3, the otes about t tests outlie whe it is appropriate to use such a test for a mea: i situatios where the sample is small ad the populatio variace is uow, but the populatio may be assumed to have a Normal distributio. Aswers to some of the questios that studets may as about this are give below. How small does the sample eed to be to use a t test? Here is o exact aswer, it depeds o the situatio but a reasoable rule of thumb is that samples of fewer tha about 30 are small whereas samples of over 30 are large. What if the sample is large? If the populatio is Normal, the populatio variace is uow ad the sample size is large the a Normal test should be used. The populatio variace is estimated from the sample. Studets leart this i S2. What if the populatio variace is ow? Whe the populatio is Normal ad the variace is ow, a Normal test should be used, whether the sample is large or small. Studets leart this i S2. What if the populatio does ot have a Normal distributio? If the sample size is large the the Cetral Limit Theorem implies that the sample mea will be approximately Normally distributed. The Normal test ca be used, either with ow or estimated variace. Updated 0/2/0 page 4

5 How do we ow whether the populatio has a Normal distributio? There are techiques for testig whether the uderlyig populatio is Normal. For example, usig Normal probability paper or usig a Kolmogorov-Smirov test or a χ² goodess of fit test. Oly the χ² goodess of fit test is i the S3 Syllabus. However, cadidates may ot have the data eeded to use this test i a examiatio questio. So, whe cadidates eed to decide what test to use, the examiers will idicate whether the variable ca be modelled by a Normal distributio. Is it OK to use a t test if the sample is large, the variace is uow ad the populatio ca be modelled by a Normal distributio? If the populatio is ot exactly Normal (ad, i real life, it rarely is), the it is better to use a Normal test for a large sample tha a t test. For that reaso, cadidates should always use Normal tests for large samples, eve though the tables available i examiatios give percetage poits of the t distributio for = 50 ad = 00 As ca be see from the specificatio (S3I7 page 63), the same rules apply to cofidece itervals. Updated 0/2/0 page 5

M1 for method for S xy. M1 for method for at least one of S xx or S yy. A1 for at least one of S xy, S xx, S yy correct. M1 for structure of r

M1 for method for S xy. M1 for method for at least one of S xx or S yy. A1 for at least one of S xy, S xx, S yy correct. M1 for structure of r 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