Chapter 10. The Chi-Squared Test

Transcription

1 Chaptr 10 Th Chi-Squard Tst Forty for you, sixty for m And qual partnrs w will b. Grald Barzan Th Avonford Star Choos your hospital ward carfully! Statistics rlasd today as part of th nw govrnmnt initiativ Clan Hospitals show that at Avonford Hospital your ward has a major ffct on your chanc of post-oprativ infction. If you ar on th Lnnon ward of Avonford s hospital, you ar mor than four tims as likly to suffr an infction within thr months of bing in hospital than if you ar on th Starr ward. Th Macartny and Harrison wards com in btwn ths two. Ward Lnnon Harrison Macartny Starr Infction No infction Look at th figurs. Thy spak for thmslvs. Th prcntag of patints gtting an infction is a massiv 6.5% for th Lnnon ward compard to a mr 1.6% for th Starr ward. Our advic to our radrs: Don t lt thm put you in Lnnon!? Do you think this is a rsponsibl articl? To giv a full answr to that qustion you will nd to judg whthr th diffrncs btwn th figurs ar larg nough to show likly undrlying diffrncs btwn th wards or whthr thy ar just th sort of random fluctuations you xpct in any statistical data. You can invstigat situations lik this using th χ tst for association. ( χ is th Grk lttr chi; it mans ch and you pronounc it ky as in Kyli.)! This xampl involvd a sampl of 83 of all th Avonford hospital patints. Th us of th χ tst dpnds on th assumption that th data hav com from a random sampl and this is takn to b th cas in th work that follows. Th Chi-Squard Tst 1

2 Th χ tst for association Th tabl in th articl is calld a contingncy tabl. Two variabls ar rcordd for ach itm of data, in this cas th ward in which a prson is placd and whthr or not th prson gts an infction. Th contingncy tabl in this xampl has rows and 4 columns so it is a 4 tabl. It has 8 clls. Th null hypothsis for this tst is that thr is no association btwn th variabls rprsntd in th contingncy tabl. Th altrnativ hypothsis is that thr is an association. SETTING UP THE HYPOTHESIS TEST H 0 : Thr is no association btwn th ward a patint is in and whthr th patint gts an infction. H 1 : Thr is an association btwn th ward a patint is in and whthr th patint gts an infction. 1-sidd tst Th χ tst for association is a 1-sidd tst Significanc Lvl 5% 5% is a suitabl significanc lvl in this situation E! In this book, th χ tst is dscribd as 1-sidd rathr than 1-tail. An xplanation for this is givn on pag *** in th answrs. CALCULATING THE TEST STATISTIC Th first stp in calculating th tst statistic is to work out th totals for ach row and column in th contingncy tabl. Mak an xtra row and column to rcord your answrs. Lnnon Harrison McCartny Starr Total Infction No infction Total This shows you that th ovrall proportion with an infction is without an infction is and th ovrall proportion Sinc th numbr of patints who wr in Lnnon ward was 31, if th null hypothsis is tru you would xpct th numbrs with and without infctions to b as follows. With infctions: Without infctions: = 9.6 (to dcimal placs) = 1.74 (to dcimal placs). 83 Ths ar calld th xpctd frquncis and dnotd by f. By contrast th obsrvd frquncis, f o, wr 14 and 17. Th Chi-Squard Tst

3 ! It may sm strang, but do not round your xpctd frquncis to th narst whol numbr. Th obsrvd frquncis will, howvr, naturally b whol numbrs. Now work out th xpctd frquncis for all th ntris in th contingncy tabl in th sam way. f Lnnon Harrison McCartny Starr Total Infction = = = = No infction = = = = Total Th corrsponding tabl for th obsrvd frquncis was givn in th articl. f o Lnnon Harrison McCartny Starr Total Infction No infction Total Th nxt stag is to work out ( fo f ) f for ach cll in th contingncy tabl. ( f - f ) o f Infction ( ) No infction Lnnon Harrison McCartny Starr ( ) 1.74 Th tst statistic, dnotd by X All groups ( f f ) =.43 = 0.10 ( ) 7.34 ( ) = 0.06 = 0.00 ( 7 6.4) 6.4 ( ) X, is th sum of th ntris in all ths clls. ( ) = 0.05 = 0.00 o = = = 6.37 f = ( ) 39.0 = 0.15 Th Chi-Squard Tst 3

4 INTERPRETING THE TEST STATISTIC As with any othr hypothsis tst, th tst statistic is compard with th critical valu. Bfor you can find th critical valu you nd to find th dgrs of frdom, dnotd by ν, for th situation. ν is th symbol for nu, Grk lttr n; it is pronouncd nw. Dgrs of frdom ar dscribd in mor dtail in th nxt sction of this chaptr. For this typ of tst, ν is givn by ν = m 1 n 1 ( ) ( ) whr m is th numbr of rows in th contingncy tabl and n is th numbr of columns. In this cas m = and 4 n =, so ( ) ( ) ν = = 3. Tabls for th χ tst, show that for 3 dgrs of frdom and a 5% significanc lvl, th critical valu is Extract from th χtabls! Th χ tabls ar givn in columns with hadings of valus of p% of 99, 97.5, 95, 90 and thn 10, 5.0,.5, 1.0, 0.5. You us on of th latr columns (in this cas that hadd 5) to find th critical valu; th arly columns ar usd in som othr applications of th χ distribution, including that dscribd on pag ***. If th tst statistic is lss than th critical valu, th null hypothsis is accptd at th significanc lvl. On th othr hand if th tst statistic is gratr than th critical valu, th null hypothsis is rjctd and th altrnativ hypothsis is accptd at th 5% significanc lvl. In this xampl, sinc 6.37 < 7.815, th null hypothsis is accptd. So th tst suggsts thr is no association btwn th ward a patint is in and whthr th patint gts an infction. So a propr statistical analysis shows that th claims in Avonford Star articl ar not justifid.? Which cll mad th biggst contribution to th valu of X in this xampl? What dos this tll you about th incidnc of infctions at Avonford hospital? Th Chi-Squard Tst 4

5 THE p-value An altrnativ way of intrprting this rsult is as a p-valu. As you can s in th diagram blow, th tst statistic of 6.37 lis btwn th 10% critical valu of 6.51 and th 5% critical valu of So th p-valu lis btwn 10% and 5%. Extract from th χtabls Th p-valu is th probability that th rsult, or on mor xtrm, would b obtaind by chanc if th null hypothsis is tru. So th largr th p-valu, th mor willing you ar to accpt th null hypothsis. In this cas th p-valu is gratr than th significanc lvl st for th tst and so th null hypothsis is accptd. Many statistics computr packags giv you th actual p-valu. In this cas it is 9.5 %.! CELLS WITH SMALL EXPECTED FREQUENCIES Th χ tst can b unrliabl in cass whr th xpctd frquncy, f, for on or mor clls is small. A common rul-of-thumb is to say that th valu of f for all clls should b at last 5. In tsts for goodnss of fit, which ar covrd latr in this chaptr, it is common to combin a cll for which f is lss than 5 with a nighbouring cll so that thir combind valu of f is gratr than 5. This, howvr, is not good practic for contingncy tabls; it may b possibl to combin two catgoris, or it may b that a largr sampl is rquird. E YATES CORRECTION Th approximations that ar always prsnt in χ tsts bgin to brak down for contingncy tabls that ar only of siz. To hlp ovrcom this, som popl suggst that, for contingncy tabls th valus of fo f should b mad lss xtrm by 0.5 whn working out th tst statistic. f f and addd to th ngativ valus whn So, 0.5 is subtractd from th positiv valus of ( o ) ( f f ) working out th tst statistic X o =. This is known as Yats corrction; it was first All groups f suggstd by Frank Yats, a grat statistician, in th 1930s. Yats corrction is nvr usd for contingncy tabls that ar biggr than. Th Chi-Squard Tst 5

6 Idas undrlying th χ tst In th χ tst you compar xpctd frquncis, f, with obsrvd frquncis, f o. Th diffrncs ar givn by ( f f ) diffrncs, ( ) f o o f.. For thortical rasons it is appropriat to considr th squars of ths A problm ariss with this masur in that it is not standardisd; th largr th sampl th largr th sum of ths squard diffrncs. This is ovrcom by dividing ach valu of ( ) Summing th rsultant quantitis for all th clls, or groups, givs th tst statistic X All groups ( f f ) o =. f f o f by f. Th tst statistic X has a χ distribution. Th xact shap of th χ distribution dpnds on th dgrs of frdom but, for ν, th gnral shap is that illustratd blow. Th lss wll thory and obsrvations match, th gratr is th valu of X and th furthr to th right on th graph its location. In th diagram th valu of X is gratr than th critical valu, and so to th right of it, and th null hypothsis is rjctd. Th χdistribution Th Chi-Squard Tst 6

7 Exrcis 10A 1. Two studnts invstigat whthr popl ar mor likly to b hlpful in a busy strt or a quit strt. On of thm drops a bag of appls at randomly slctd tims during th day and th othr watchs to s whthr anyon hlps by picking thm up and rturning thm to thir ownr. Th rsults ar as follows. Busy Quit Hlpd 14 0 Not hlpd 10 6 Us a χ tst at th 5% significanc lvl to s whthr thr appars to b any diffrnt btwn th rspons of popl in a busy strt to in a quit strt.. A prviously unknown spcis of bat is discovrd in a rmot cav. A random sampl of 50 of th bats ar collctd for obsrvation and thn rlasd. Som of thm hav strips on thir wings and othrs do not. Th numbrs of ach sx with and without strips ar givn in th tabl blow. f o Strips No strips Mal 16 4 Fmal 9 1 Tst at th 1% significanc lvl whthr thr is any association btwn a bat s sx and its having strips. 3. A poultry brdr is concrnd that many of hr chickn s ggs ar infrtil. A frind suggsts that thir frtility rat will b improvd if th chickns ar givn diffrnt living conditions instad of bing confind to a small run. Sh carris out an xprimnt in which som of hr birds ar kpt as bfor, othrs ar allowd to roam fr and a third group ar put in a fild full of cows. Sh kps random sampls of th ggs from th thr groups sparatly and thn placs thm in incubators for hatching. Th rsults ar as follows. In run Roaming With cows Frtil Infrtil Us a χ tst at th 0.5% significanc lvl to invstigat whthr thr is any association btwn th birds living conditions and th frtility of thir ggs. Th Chi-Squard Tst 7

8 4. Two drugs, A and B, can b usd as tratmnt for a disas. In a trial, 80 patints wr tratd with on or othr of th drugs and thn rviwd on yar latr. Th slction of th patints for th diffrnt tratmnts was mad at random. Th rsults ar shown in th tabl blow. Bttr No chang Wors Drug A Drug B Is thr vidnc, at th 5% significanc lvl, of a diffrnc in th ffctivnss of th two drugs? 5. A group of parnts claim that living clos to lctricity pylons incrass th liklihood of asthma in childrn undr 4 yars of ag. An invstigation is carrid out on a random sampl of childrn, with th following rsults. Distanc of hom from pylons Lss than 1 km 1- km Mor than km Asthma No asthma Dos th study show vidnc of association, at th 10% significanc lvl, btwn childhood asthma and living nar lctricity pylons? 6. Avonford Council ar discussing whthr to introduc spd camras on th roads lading into th town. Councillor Jons says It is a wast of tim. Cars slow down for th camra and spd up again as soon as thy hav passd it. Councillor Smith says Thy may spd up a littl but not vry much so ovrall popl ar safr. Thy agr to an xprimnt. Suitabl locations for spd camras ar chosn on two similar roads into th town; a spd camra is put on on but not th othr. Randomly slctd cars ar thn obsrvd 500 mtrs into town from th two locations. Thy ar classifid as Not spding, Spding mildly or Spding sriously. Th rsults ar as follows. Not spding Spding mildly Spding sriously Spd camra No spd camra 3 46 (i) Us th data to carry out a tst, at th 0.5% significanc lvl, of whthr spd camras do influnc drivrs subsqunt spd. (ii) Which clls mak th largst contribution to th tst statistic? Commnt. Th Chi-Squard Tst 8

9 7. Nwspapr articls with th hadlin Too posh to push hav suggstd that profssional womn ar choosing to hav thir first baby born by Casaran sction rathr than hav a natural birth. In a rcnt survy of attituds of womn prgnant with thir first child, th following rsponss wr givn whn a random sampl of womn wr askd whthr thy would choos a Casaran sction rathr than a natural birth. Casaran Don t mind Natural birth Non-profssional Profssional Is thr vidnc, at th 5% significanc lvl, of association btwn th attituds of womn and thir profssional status? 8. Th following rsults wr obtaind from a survy of attituds to smoking in rstaurants. Th rspondnts wr slctd at random. Against Don t mind For Currnt smokr Ex smokr Nvr smokd (i) Carry out a χ tst at th 5% significanc lvl of whthr thr is association btwn th smoking habits of individuals and thir attitud to smoking in rstaurants. (ii) (iii) Which catgory, currnt smokr, x-smokr or nvr smokd, contributs th last to valu of X? Which 3 clls contribut th most to th valu of X? (iv) What conclusions can you draw from th survy? 9. A survy is carrid out into th lion populations in 5 diffrnt rgions of Africa, dsignatd A, B, C, D and E. Th lions ar classifid as Mal, Fmal or Juvnil. Obsrvations ar carrid out on random sampls from ths lion populations with th following rsults. A B C D E Mal Fmal Juvnil Tst whthr th survy provids vidnc, at th 10% significanc lvl, of association btwn th composition of th lion populations and thir locations? Th Chi-Squard Tst 9

10 Dgrs of frdom Jasmin is a psychology studnt. Sh is carrying out an xprimnt in which sh asks a numbr of young mn to choos 10 adults at random and gt thm to fill in a qustionnair. Jasmin is not actually intrstd in th qustionnair. What sh wants to know is whthr hr intrviwrs ar mor likly to slct womn than mn. Hr ar som of hr rsults. (Notic that this is not a contingncy tabl.) Exprimnt on young mn (Each on chooss 10 popl to intrviw) Intrviwr Fmal Mal Andrw 6 4 Charls 3 7 Hanif 5 5 Sunny 7 3 Lroy 10 0 Edward 8 Kwam 7 3 Sanjay 6 4? How many numbrs has Jasmin writtn down?? How many of ths numbrs did sh rally nd to writ down? Th first intrviwr on th list was Andrw and h chos 6 womn. Sinc h intrviwd 10 popl in all, you know that h must hav chosn 4 mn. Similarly for all th othrs; in fact Jasmin did not rally nd hr right hand column at all. In this tabl th numbr of groups, k, is 16. Howvr only 8 of thm ar actually fr bcaus whn you know thm, you know th rst. This is dscribd by saying that thr ar 8 fr variabls in th systm, or that thr ar 8 dgrs of frdom. ACTIVITY Hr is a 3 5 contingncy tabl. Th Total row and th Total column ar also givn. A B C D E Total P 100 Q 150 R 50 Total Start putting numbrs in th mpty clls. How many numbrs ar you fr to choos? How many do you not gt a choic ovr? What would th answrs b for a contingncy tabl with m rows and n columns? Th Chi-Squard Tst 10

11 Th χ tst is usd in a varity of situations, not just as a tst for association in contingncy tabls. You always nd to know th dgrs of frdom. It is oftn hlpful to us th quation Dgrs of frdom (ν ) = Numbr of groups (k) Numbr of rstrictions (r). In th xampl of Jasmin s studnts, ach intrviwr was rquird to carry out xactly 10 intrviws and in ach cas this placd a rstriction on th numbrs in th tabl. So thr wr 8 rstrictions, on for ach intrviwr. So in that cas th quation ν = k r is 8= 16 8.? In th Activity on pag * you workd with a 3 5 contingncy tabl. You should hav found that thr wr 8 dgrs of frdom. How many rstrictions wr thr? How can you xplain that numbr? So far you hav usd th χ tst on contingncy tabls. Th rstrictions ar th row and column totals. In th nxt sction of this chaptr th χ tst is usd to tst how wll particular distributions fit obsrvd data. This is calld tsting for goodnss of fit. In som goodnss of fit tsts you nd to us th data to stimat paramtrs (lik th man) of thos distributions. Each paramtr that you stimat givs anothr rstriction whn you ar working out th appropriat dgrs of frdom. A tabl of dgrs of frdom for diffrnt χ tsts is givn on pag **. Th Chi-Squard Tst 11

12 Exrcis 10B 1. You ar mting a frind on a train and you know sh has dfinitly got on it. Th train has 8 carriags. (i) (ii) How many carriags must you b prpard to sarch to b crtain of finding your frind? How many carriags must you b prpard to sarch bfor you ar crtain which carriag your frind is in?. Sam is doing a probability xprimnt in his mathmatics class. Hr ar th instructions. Tak 3 coins and toss thm togthr. Rcord how many hads you gt. Th answr will b 0, 1, or 3. Do this 50 tims and rcord your rsults. Thn fill in a tabl lik this on. Hads Frquncy How many dgrs of frdom ar thr whn Sam fills in th tabl? Explain your answr. 3. Ruth, Srna and Tamara ar told to roll a di 60 tims ach as part of a probability xprimnt. Thy hav to fill in this tabl of rsults Total Ruth 60 Srna 60 Tamara 60 Total Thy dcid to talk instad and to invnt som numbrs to complt th tabl. (i) How many mpty clls ar thr in th tabl as shown? (ii) How many numbrs ar thy fr to invnt whn thy fill th clls? (iii) How many numbrs ar thy not fr to invnt whn thy fill th clls? Th Chi-Squard Tst 1

13 4. In a qustionnair, studnts ar askd to rat a lcturr on a 1 to 4 scal. 1 Poor Fair 3 Good 4 Excllnt Th rsults ar rcordd in th frquncy row in a tabl lik this on. Scor Frquncy In ach of situations (i) and (ii): (A) stat how many of th clls can b filld in frly; (B) stat how many rstrictions thr ar; (C) dscrib th rstrictions. (i) 100 studnts fill in th qustionnair. (ii) 100 studnts fill in th qustionnair and th avrag is.4. Invstigation Look at this Sudoku puzzl. Whn it is compltd, ach row, ach column and ach of th nin squar blocks contains ach of th digits 1,, 3, 4, 5, 6, 7, 8 and 9 xactly onc. You can stat this as thr ruls. 1 Th sam numbr may not appar twic in any row. Th sam numbr may not appar twic in any column. 3 Th sam numbr may not appar twic in any of th squar blocks Us only ths 3 ruls to answr th following qustions. (Exprts at solving Sudokus us othr tchniqus as wll as ths ruls.) (i) How many numbrs can you plac in th following squar at th start of th puzzl? (A) Th top lft squar (B) Th middl squar (C) Th bottom right squar (ii) (iii) How many dgrs of frdom do you hav for filling in ach of th thr squars in part (i)? Find a squar for which thr ar 7 dgrs of frdom at th start of th puzzl. Th Chi-Squard Tst 13

14 (iv) A good Sudoku playr only fills in a squar for which thr is 1 dgr of frdom. How many such squars ar thr at th start of this puzzl? Copyright not: this Sudoku appard in th Indpndnt on Sunday on 19/3/06. Th Chi-Squard Tst 14

15 Goodnss of fit tsts Th Avonford Star Snaks galor Ystrday aftrnoon Jimmy Rawlings had to b rushd to Avonford Hospital aftr h inadvrtntly trod on a snak. A hospital spoksprson said this is th fourth cas this yar; Thr sms to b a spat of thm and all from diffrnt parts of th rgion. Is this vidnc, th Star asks, of global warming in th Avonford ara?? Do you think it is a bit far ftchd using snak bits as vidnc of global warming? You would xpct som random variation in th numbr of cass tratd pr yar by a hospital, and this yar may just b on of thos yars that has mor than usual. Bfor you can dcid whthr this is th cas, you nd a modl for th probability distribution of diffrnt numbrs of cass, and suitabl data. A snak bit is a rar vnt in Britain and so you might wll xpct th numbr of cass to b modlld by Poisson distribution. Following th articl in th Avonford Star, a doctor who has bn carrying out rsarch into th frquncy with which diffrnt typs of accidnts ar tratd, writs in with th following data, covring th last 100 yars. Snak bits Obsrvd frquncy, f o Th qustion thn ariss Is th Poisson distribution a rasonabl modl for ths data? You can answr this by carrying out a χ goodnss of fit tst, as follows. SETTING UP THE HYPOTHESIS TEST H 0 : Th Poisson distribution is a good modl. H 1 : Th Poisson distribution is not a good modl. 1-sidd tst Th χ Goodnss of Fit tst is a 1-sidd tst Significanc Lvl 5% 5% is a suitabl significanc lvl in this situation CALCULATING THE TEST STATISTIC Sinc th paramtr that dfins th Poisson distribution is qual to its man, th first stp is to find th man of th sampl data. (31 0) + (34 1) + (0 ) + (9 3) + (4 4) + ( 5) Man = 100 = 1.7 Th Chi-Squard Tst 15

16 For th Poisson distribution with paramtr 1.7, th probabilitis and xpctd frquncis for 100 yars ar as follows. Snak bits Probability = = ! = ! = = ! 1 = Expctd frquncy, f ? Th probabilitis in this tabl hav bn workd out on a calculator and thn roundd to 3 dcimal placs. What is th accuracy of th xpctd frquncis?! CELLS WITH SMALL EXPECTED FREQUENCIES Notic that th xpctd frquncis for 4 snak bits and for 5 + snak bits ar both lss than 5 and indd com to lss than 5 in total. Whn you ar carrying out a χ goodnss of fit tst, it is usual to combin clls to nsur that in ach cas th xpctd frquncy is gratr than 5. So 3, 4 and 5 + ar combind into on group; this is dscribd as 3 + in th tabl blow. Th xpctd frquncy of this group is = 13.6 and th obsrvd frquncy is = 15. Snak bits Obsrvd frquncy, f o Expctd frquncy, f You ar now in a position to work out th tst statistic X All groups ( f f ) o =, as shown blow. f Snak bits ( f f ) o f (31 8.1) 8.1 = 0.99 ( ) 35.7 = (0.6).6 = 0.99 ( ) 13.6 = Th Chi-Squard Tst 16

17 So, X All groups ( f f ) o = = f = INTERPRETING THE TEST STATISTIC Th nxt stp is to compar th tst statistic, X = 0.83, with th critical χ valu. To find th critical valu, you first nd to know th dgrs of frdom. You know that Dgrs of frdom (ν ) = Numbr of groups (k) Numbr of rstrictions (r) In this cas, thr ar rstrictions: th total of th frquncis (100) th Poisson paramtr (1.7). Thr ar 4 groups, so ν = 4 =. Thr ar dgrs of frdom. Th critical valu for dgrs of frdom at th 5% significanc lvl is found from th χ tabls. It is Sinc th tst statistic, 0.83, is lss than th critical valu, 5.991, th null hypothsis is accptd. Th Poisson distribution is indd a good modl for th incidnc of snak bits in th Avonford ara. THE AVONFORD STAR ARTICLE This xampl bgan with th Avonford Star articl. In on yar 4 popl had suffrd snak bits. Th hypothsis tst has shown that is rasonabl to assum a Poisson distribution so you ar now in a position to stimat th probability of th obsrvd rsult, or a mor xtrm cas, happning by chanc. It is 1 Probability of 0, 1, or 3 snak bits and so 1- ( ) = This is a probability of about 1 5, so it is unusual but by no mans impossibl.? Whr did th figurs, 0.81, 0.357, 0.6 and 0.096, in th calculation abov com from?? How could this analysis b st up as a hypothsis tst? What would you conclud from such a tst at (i) th 5% significanc lvl (ii) th 1 % significanc lvl? In th prvious xampl th χ tst was usd to invstigat th goodnss of fit of a Poisson distribution to a givn data st. Similar χ tsts can b usd for othr distributions. In th nxt Th Chi-Squard Tst 17

18 xampl th tst is carrid out to invstigat th goodnss of fit of a modl in which th outcoms ar in a givn proportion; in th following xampl a binomial distribution is usd. Exampl (frquncis in a givn proportion) A botanist is invstigating a particular typ of plant which can hav rd, pink or whit flowrs. Sh collcts data from a random sampl of 160 plants. Flowr colour Rd Pink Whit Total Frquncy Sh has a thory that th diffrnt colour flowrs should occur in th ratio, Rd: Pink: Whit = 1: 6: 9. Tst, at th 5% significanc lvl, whthr hr thory is consistnt with th data. Solution SETTING UP THE HYPOTHESIS TEST H 0 : Th diffrnt colour flowrs occur in th ratio, Rd: Pink: Whit = 1: 6: 9. H 1 : Th diffrnt colour flowrs do not occur in this ratio. 1-sidd tst Significanc Lvl: 5% CALCULATING THE TEST STATISTIC Sinc = 16, th fractions of th diffrnt colour flowrs should, according to th null hypothsis, b, and. So th xpctd frquncis ar calculatd as follows Flowr colour Rd Pink Whit 1 Expctd frquncy, f = = = Th obsrvd frquncis, f o, ar thos givn in th qustion. Flowr colour Rd Pink Whit Obsrvd frquncy, f o And so th tst statistic is givn by X All groups ( f f ) ( 0 10) ( 69 60) ( 71 90) o = = + + = = f Th Chi-Squard Tst 18

19 INTERPRETING THE TEST STATISTIC In this situation thr ar 3 groups and 1 rstriction (th total frquncy of 160) so th dgrs of frdom ar givn by ν = 3 1=. Notic that in this cas you did not us th data to stimat any population paramtr, as you did in th prvious xampl, so thr ar no furthr rstrictions. Th critical valu for th χ tst at th 5% significanc lvl for ν = is Sinc th tst statistic, , is gratr than th critical valu of 5.991, th null hypothsis is rjctd. Th vidnc dos not support th botanist s thory. Exampl (binomial distribution) A mdical rsarchr wants to dcid whthr a common mdical condition runs in familis (and so might hav a gntic caus), or whthr it occurs indpndntly and with a fixd probability, p. H slcts 00 sts of 6 closly rlatd popl at random and dtrmins how many popl in ach st hav th condition. Th rsults ar as follows. Numbr in st with th condition Total Frquncy Th rsarchr says that if th condition occurs indpndntly and with a fixd probability, this distribution should b wll modlld by a binomial distribution, and so h wishs to chck whthr this is indd th cas. Solution SETTING UP THE HYPOTHESIS TEST H 0 : Th undrlying distribution is binomial (th probability, p, has yt to b dtrmind) H 1 : Th undrlying distribution is not binomial 1-sidd tst Significanc lvl: 5% CALCULATING THE TEST STATISTIC In ordr to work out th xpctd frquncis, you nd to us th xprimntal data to calculat th probability, p. Th total numbr of popl with th condition in th sampl is (15 0) + (40 1) + (68 ) + (47 3) + (4 4) + (5 5) + (1 6) = 444. Th total numbr of popl in th sampl is 00 6 = 100. So th probability, p, is givn by 444 p = = Th Chi-Squard Tst 19

20 Now you can work out th probabilitis of th various outcoms, as follows. Numbr with th condition Probability, p C = C = 0.0 C = C = 0.53 C = 0.11 C = 0.06 C = To find th xpctd frquncis, f, you multiply th probabilitis by th total numbr of sts (ach of six popl), in this cas 00. Numbr with th condition & 6 Expctd frquncy, f ? Why hav th columns for 5 and 6 bn combind togthr? Th corrsponding tabl of obsrvd frquncis is Numbr with th condition & 6 Obsrvd frquncy, f o and so th tst statistic X X can now b calculatd. ( f f ) ( ) ( ) ( ) ( ) ( 4.3) ( 6 5.8) o = = All groups f = INTERPRETING THE TEST STATISTIC In this situation thr ar 6 groups and rstrictions: th total frquncy of 00 and th stimatd probability. So th dgrs of frdom ar givn by ν = 6 = 4. Notic that in this cas you usd th data to stimat on population paramtr (th probability) and this providd th scond rstriction. Th critical valu for ν = 4 at th 5% significanc lvl is Sinc < 9.488, th null hypothsis, that th undrlying distribution is binomial is accptd. In th binomial distribution th valu of p is constant and so, in this cas, dos not vary from on family to anothr. Th vidnc dos not support th hypothsis that th disas runs in familis. Consquntly a gntic link would not b xpctd.? This xampl was about a mdical rsarchr wanting to stablish whthr a particular condition ran in familis and so had a possibl gntic link. Dscrib th shap of distribution you would xpct from this xprimnt if this wr th cas. Th Chi-Squard Tst 0

21 Goodnss of fit tsts for othr distributions Th sam gnral mthod is usd for goodnss of fit tsts for othr distributions. Th main diffrnc from on distribution to anothr is in trms of what population paramtrs you nd to stimat and th ffct on th dgrs of frdom. Thus if you ar tsting for a Normal distribution you nd to stimat both th man and standard dviation from th data, making rstrictions. A third rstriction coms from th total numbr, so in that cas ν = k 3. Th tabl blow summariss th dgrs of frdom appropriat to commonly usd χ tsts. Distribution Groups Totals Paramtrs ndd Total numbr of rstrictions Dgrs of frdom, ν Spcifid k k 1 proportion Binomial k 1 1 p k Poisson k 1 1 λ k Normal k 1 1 μ, σ 3 k 3 Contingncy m n m+ n m n 1 tabl, m n m 1 n 1 + ( ) ( ) E Figurs that match too wll Th Avonford Star Is Jmmy a fraud? Ovr rcnt yars, local man Jmmy Dicman Turpin has oftn claimd to b abl to throw dic with wondrful accuracy. Only last wk, in th privacy of his own hous, h thrw a di 60 tims. Jmmy says Th prfct rsult would hav bn for ach numbr to com up 10 tims. I was off form; 1,, 5 and 6 all cam up 10 tims but 3 only cam up 8 tims and 4 cam 1 tims. Now Avonford Collg studnt Angla Ghosh has challngd Jmmy to rpat his prformanc in public. Statistics don t work lik that. sh says You xpct mor random variation. Th Star joins Angla in asking Is Jmmy a fraud?. Suppos you apply a χ tst to Jmmy s rsults. Th xpctd and obsrvd rsults ar as follows. Scor Expctd frquncy, f Obsrvd frquncy, f o Th Chi-Squard Tst 1

22 Th valu of X All groups X is givn by ( f f ) ( 10 10) ( 10 10) ( 8 10) ( 1 10) ( 10 10) ( 10 10) o = = f = 0.8 Thr ar 6 groups and only 1 rstriction so thr ar 5 dgrs of frdom. (In this cas th frquncis ar in a givn proportion.) Look at th lin in th tabls for ν = 5. You can s that th valu of 0.8 for X lis in th lft hand part of th tabl, btwn 99% and 97.5%. Extract from th χtabls Whil it dos not prov that th data hav bn collctd dishonstly, a valu of th tabl should arous suspicion. Thr ar 4 possibl xplanations. Th data hav bn invntd Data which do not fit wll hav bn rjctd X in this part of Th data hav bn usd to construct th thory and ar thn bing usd to confirm it Th data hav bn honstly obtaind and just happn to fit vry wll. Evn if th data hav bn collctd honstly, popl will b disinclind to bliv thm if thy fit so wll. Rpating th xprimnt should rmov such doubt. So Angla Ghosh, th Avonford studnt, is quit right to call for Jmmy to rpat his prformanc, and in public.! Whn a small valu X ariss, it dos not man that you could, or should, suddnly chang from a goodnss of fit tst into on of whthr th data wr honstly collctd. Thr is no such tst. All it tlls you is that it might b wis to chck that th data collction was carrid out satisfactorily.! Rmmbr that th χ tst is an uppr tail on-sidd tst. Thus, if a tst at th 5% significanc lvl, th critical rgion is on th right and rprsnts th full 5%. Quit a common mistak is to mak th 5% into two tails, ach of ½%. Th Chi-Squard Tst

23 Exrcis 10C 1. A plant has two colours of flowrs, blu and whit. A botanist has a thory that th two colours should occur in th ratio 1:3. Sh collcts data from a random sampl of 80 flowrs. 66 of thm ar whit and th rst blu. Sh plans to carry out a χ tst to dtrmin whthr th data support hr thory. (i) (ii) Explain why thr is 1 dgr of frdom in this cas. Carry out th tst at th 5% significanc lvl, stating th null and altrnativ hypothss and th conclusion.. A gnticist is studying a situation whr thr ar thr typs of offspring, T 1, T and T 3. In a random sampl of siz 70, thr ar 8 of typ T 1, 6 of typ T and 36 of typ T 3. A simpl modl for this situation is that T 1 occurs with probability 1, T 4 with probability 1 and T 3 with probability 1 4. Writ down th appropriat xpctd frquncis. Tst at th 5% lvl of significanc th hypothsis that this modl applis. [MEI S3 Jan 99 part] 3. Ptr and Jan ar playing th wll known gam Papr, Scissors and Ston. Each playr chooss on of thm at random and thy both rval thir choic at th sam tim. Th ruls ar: o Papr bats Ston; Ston bats Scissors; Scissors bat Papr o If both playrs mak th sam choic, that round is a draw. (i) Find th probabilitis of th possibl outcoms (Ptr wins, Jan wins, thr is a draw) in any round, if Ptr and Jan both choos at random. In a gam of 30 rounds Ptr wins 5 tims, Jan wins 16 tims and thr ar 9 draws. Jan claims that sh can rad Ptr s mind and that is why sh is so succssful. (ii) Stat null and altrnativ hypothss rlating to Jan s claim, and carry out a suitabl χ tst at th 5% significanc lvl. 4. Th numbrs of missions from a radio-activ sourc in 1000 intrvals, ach of 10 sconds, ar rcordd using a Gigr countr. Numbr of missions Total Obsrvd frquncy, f o (i) (ii) Find th man and varianc of ths data. Commnt on your answrs. Tst at th 5% significanc lvl whthr th Poisson distribution, with th man you found in part (i) as paramtr, provids a suitabl modl for th data. Th Chi-Squard Tst 3

24 5. At th last gnral lction, th Grn Party candidat for th constituncy of Avonford rcivd 37% of th vot. An opinion poll is takn to dtrmin whthr thr has bn any chang in th lvl of support for th Grn Party in Avonford sinc th lction; 400 popl ar slctd at random for th poll and th rsults ar as follows. Numbr xprssing support for th Grn Party 166 Numbr xprssing support for othr partis 34 (i) Calculat th xpctd frquncis corrsponding to th abov obsrvd frquncis on th assumption that that support for th Grn Party has not changd. (ii) Carry out an appropriat χ tst at th 5% significanc lvl, stating clarly th null and altrnativ hypothss undr tst, and th conclusion rachd. 6. A fruit farmr s appls ar gradd on a scal from A to D bfor sal. Lngthy past xprinc shows that th prcntags of appls in th four grads ar as follows. Grad A B C D Prcntag 9% 38% 7% 6% Th farmr introducs a nw tratmnt and applis it to a small numbr of trs to s if it affcts th distribution of grads. Th appls producd by ths trs ar gradd as follows. Grad A B C D Numbr of appls (i) (ii) Writ down suitabl null and altrnativ hypothss for a χ tst. Calculat th xpctd numbr of appls in ach grad undr th null hypothsis. (iii) Carry out th χ tst at th 5% lvl of significanc. Stat th conclusions of th tst clarly. [MEI S3 Jan 94] 7. A train oprating company collcts data on th numbr of its trains that arrivd at thir final dstinations mor than on hour lat. Th monthly figurs for ach month on calndar yar ar as follows. J F M A M J J A S O N D (i) Carry out a χ tst at th 5% significanc lvl to dtrmin whthr or not lat running is mor likly to occur in som months than othrs. (You may tak all months to b of th sam lngth.) (ii) Which on of th following is likly to hav bn a major contributory caus to your answr to part (i): ovrhating, frozn points, lavs on th rails? Th Chi-Squard Tst 4

25 8. Th NRS social grad dfinitions ar usd to dscrib social class. A random sampl of 400 rsidnts of Avonford is takn and ach prson in th sampl is assignd to on of th classs. Th tabl blow givs th numbrs in th various classs togthr with th national prcntags of popl in th various classs. Band Numbr of popl in sampl, f o National prcntag A 3 3.4% B % C % C % D % E 6 8.7% Total % (i) Find th xpctd frquncy within ach band in Avonford basd on th national proportions. (ii) Carry out a χ tst, at th 0.5% significanc lvl, to invstigat whthr th class distribution in Avonford is diffrnt from that in th nation as a whol. (iii) Commnt on th Avonford distribution in th light of your conclusions. 9. It is said that th Poisson distribution provids a good modl for th numbr of goals scord in football matchs. Th data blow rfr to th total numbrs of goals in on sason s matchs of a football lagu Numbr >7 Frquncy (i) (ii) Find th man and varianc of ths data. Tst whthr th Poisson distribution provids a suitabl modl for th data at th 5% significanc lvl. (Us th man that you found in part (i) as paramtr.) A football nthusiast claims that it is wll known that in th long trm th man numbr of goals pr lagu gam is 1.7. (iii) Us th Poisson distribution with paramtr 1.7 to draw up a frquncy tabl for th xpctd numbr of goals. (iv) Explain why th dgrs of frdom ar givn by ν = k 1 and not ν = k in this cas. (v) Carry out th tst and commnt on your conclusion. Th Chi-Squard Tst 5

26 10. A local council has rcords of th numbr of childrn and th numbr of housholds in its ara. It is thrfor known that th avrag numbr of childrn pr houshold is It is suggstd that th avrag numbr of childrn pr houshold can b modlld by th Poisson distribution with paramtr In ordr to tst this, a random sampl of 1000 housholds is takn, giving th following data. Numbr of childrn Numbr of housholds (i) (ii) Find th corrsponding xpctd frquncis obtaind from th Poisson distribution with paramtr Carry out a χ tst, at th 5% lvl of significanc, to dtrmin whthr or not th proposd modl should b accptd. Stat clarly th null and altrnativ hypothss bing tstd and th conclusion which is rachd. [MEI S3 Jun 9] 11. A psychology rsarchr has th thory that mn prfr to work in groups with othr mn and womn prfr to work with othr womn. H is con ducts an xprimnt at a GCSE rvision cours attndd by 180 studnts agd 15 or 16; most of thm did not prviously know anyon ls attnding. At on stag th studnts ar askd to form into 36 groups ach of 5 popl to work togthr on a projct. Th psychologist counts th numbrs of boys and girls in ach group. Group 5G 0B 4G 1B 3G B G 3B 1G 4B 0G 5B Frquncy (i) (ii) Find th numbrs of girls and boys on th cours, and stat th probability, p, that a randomly slctd studnt is a girl. Us th binomial distribution to find th frquncis you would xpct in ths groups if boys and girls joind groups at random. Th rsarchr is going to carry out a χ tst to s how wll th binomial distribution modls ths data. H uss dgrs of frdom. (iii) Explain how h coms up with this numbr. (iv) Carry out th χ tst at th 1% significanc lvl, stating th null and altrnativ hypothss and th conclusion. (v) Th rsarchr claims that this shows that mn and womn rally do not lik working togthr. Explain why his claim may not b justifid. Th Chi-Squard Tst 6

27 1. Chiptch produc computr chips for us in microprocssors. Th quality control dpartmnt rgularly taks random sampls of siz 10 and nots th valu of X, th numbr of faulty chips. Th rsults for 100 such sampls wr as follows. X f (i) (ii) (iii) (iv) Explain why X could rasonably b modlld by a binomial distribution. Us th data to stimat p, th probability that a chip chosn at random will b faulty. Hnc calculat th xpctd frquncis for th binomial modl. Carry out a suitabl tst at th 5% significanc lvl, to dtrmin whthr or not th modl fits th data wll. Commnt on th principal points of diffrnc btwn th data and th modl. [MEI S3 Jan 98] Th Chi-Squard Tst 7

28 Th Chi-Squard Tst 8

29 Ky Points Th χ tst can b usd as a tst for association and a tst for goodnss of fit. Th data ar groupd. If th xpctd frquncy for a group is small (typically lss than 5), it is usually combind with anothr group. In th tst for association th groups ar th clls of a contingncy tabl. In th tst for association, th null hypothsis is that th variabls ar not associatd; th altrnativ hypothsis is that thy ar associatd. In goodnss of fit tsts, th null hypothsis is that th distribution in qustion is th undrlying distribution for th data (or provids a good modl); th altrnativ hypothsis is that this is not th cas. Th χ tst is a 1-tail tst. Th xpctd frquncy, f, and th obsrvd frquncy, f o, for ach group ar compard. ( ) fo f Th tst statistic is X =. All groups f Critical valus ar obtaind from tabls, for th appropriat significanc lvl and dgrs of frdom. If th tst statistic is lss than th critical valu, th null hypothsis is accptd. Th dgrs of frdom for diffrnt situations ar givn in th tabl blow Distribution Groups Totals Paramtrs ndd Total numbr of rstrictions Dgrs of frdom, ν Spcifid k k 1 proportion Binomial k 1 1 p k Poisson k 1 1 λ k Normal k 1 1 μ, σ 3 k 3 Contingncy m n m+ n m n 1 tabl, m n m 1 n 1 + ( ) ( ) Th Chi-Squard Tst 9

30 Answrs Pag 1 Discussion point Th hospital would b unlikly to b abl to oprat if vryon rfusd to go into on of its 4 wards so it would sm to b irrsponsibl to suggst that popl rfus to us it. On th othr hand, if it is gnuinly th cas that on ward is dangrous thn popl should b warnd about it. Pag Caution In most tsts th tst statistic can tak valus that ar gratr or lss than a cntral valu and so thr ar possibl tails. This is not, howvr, th cas with th χ tst. Th xtrm cas is that th data fit prfctly and th tst statistic, X, is a masur of how far th data ar away from this idal in whatvr dirction. It dos not distinguish btwn cass whr th frquncy in a particular group is highr than xpctd, or lowr. So, 1-sidd is a mor appropriat dscription of th tst than 1- tail. Pag 4 Discussion point Th largst ntry coms from th Starr ward for infctions. Looking at th data shows that th rason for this is that this ward has fwr infctions than would b xpctd. Exrcis 10A In all th qustions in this xrcis, th null and altrnativ hypothss hav th sam form. H 0 : Thr is no association btwn th variabls. H 1 : Thr is an association btwn th variabls. 1. X =.38, ν = 1. Sinc.38 <3.841, H 0 is accptd at th 5% significanc lvl. Thr is no diffrnc in rsponss for busy and quit strts.. X = 1, ν = 1. Sinc 1 > 6.635, H 0 is rjctd at th 1% significanc lvl. Th vidnc supports th hypothsis that thr is an association btwn th sx of a bat and th probability that it has strips. 3. X = 19.51, ν =. Sinc > 10.60, H 0 is rjctd at th 0.5% significanc lvl. Th vidnc supports th hypothsis that thr is an association btwn th living conditions of hns and th frtility of thir ggs. 4. X = 14.71, ν =. Sinc > 5.991, H 0 is rjctd at th 5% significanc lvl. Th vidnc supports th hypothsis that thr is a diffrnc btwn th outcoms from th two tratmnts. Th Chi-Squard Tst 30

31 5. X = 1.485, ν =. Sinc < 4.605, H 0 is accptd at th 10% significanc lvl. Th vidnc dos not support th hypothsis that thr is association btwn asthma and how clos you liv to lctricity pylons. 6. (i) X = 37.4, ν =. Sinc 37.4 > 10.60, H 0 is rjctd at th 0.5% significanc lvl. Th vidnc supports th thory that spd camras hav a lasting ffct. (ii) Th clls rlating to srious spding hav th largst ffct. Th data support Councillor Smith s viw. 7. X = 8.10, ν =. Sinc 8.10 > 5.991, H 0 is rjctd at th 5% significanc lvl. Th vidnc supports th hypothsis that thr is association btwn attituds and profssional status. 8. (i) X = 18.95, ν = 4. Sinc > 9.488, H 0 is rjctd at th 5% significanc lvl. Th vidnc supports th hypothsis thr is an association btwn smoking habits and attituds to smoking in rstaurants. (ii) Thos who hav nvr smokd. (iii) Thos who hav givn up smoking and ar against smoking in rstaurants. Thos who hav givn up smoking and ar in favour of smoking in rstaurants. Currnt smokrs who ar in favour of smoking in rstaurants. (iv) Th strongst ractions, for or against, com from thos who hav xprincd smoking. Among thos who hav givn up, mor than would b xpctd ar against smoking in rstaurants and fwr than xpctd ar in favour. Th opposit is tru among currnt smokrs; mor than would b xpctd ar in favour of smoking in rstaurants and fwr than xpctd ar against. 9. X = 1.7, ν = 8. Sinc 1.7 < 13.36, H 0 is accptd at th 10% significanc lvl. Th vidnc dos not support th hypothsis that thr ar diffrncs btwn th composition of lion groups at diffrnt locations. Pag 10 Pag 10 Pag 10 Discussion point 1 16 numbrs, for ach of th ight intrviwrs. Discussion point 8 numbrs Activity You can fill in 8 numbrs frly but you hav no choic for th rmaining 7 if you ar to mak obtain th corrct row and column totals. So thr ar 8 dgrs of frdom. For an r c tabl thr ar r 1 ( c 1) dgrs of frdom. Th numbr of clls ovr which you hav no choic is givn by r+ c 1. ( ) Pag 11 Discussion point Thr ar 7 rstrictions. Th Chi-Squard Tst 31

32 At first sight it might sm that thr should b 8, on for ach of th 3 row totals and on for ach of th 5 column totals. Howvr on rstriction is lost sinc th total of th row totals and th total of th column totals ar th sam. So th numbr of rstrictions is 8 1= 7. Exrcis 10B 1. (i) 8 (ii) (i) 5 (ii) 15 (iii) (i) (A) 3 (B) 1 (C) Th total numbr of rplis (ii) (A) (B) (C) Th total numbr of rplis (100) and th total of th scors (100.4 = 40 ) Invstigation (i) (A) 1 (B) (C) 5 (ii) (A) 1 (B) (C) 5 (iii) row 3, column 8 (iv) 5 in all: row 1, column 1; row 6, column 3; row 7, column 9; row 8 column 4; row 9, column 6. Pag 15 Discussion point Ys, you hav to b vry carful about linking caus to ffct. You can us statistics to judg how unusual it is to hav a givn numbr of snak bits in a yar (in this cas 4) but you hav to b much mor carful in claiming what causd this. Thr ar many possibl xplanations, only on of which is global warming, and it could just b random variation. Pag 16 Discussion point Th xpctd frquncis ar accurat to on dcimal plac. Pag 17 Discussion point 1 Thy ar th xpctd probabilitis in th tabl on pag 14. Pag 17 Discussion point Th tst would b st up as H 0 μ = 1.8 th situation is as it always was with th man numbr of snak bits pr yar unchangd H 1 μ > 1.8 th man numbr of snak bits pr yar has incrasd 1-tail tst Th probability of 1 = 0.04 is lss than 5% so at th 5% significanc lvl th null hypothsis is 5 rjctd (answr (i)). Howvr it is gratr than 1% so at th 1% significanc lvl th null hypothsis is accptd. Although many tsts ar carrid out at th 5% significanc lvl, you should b awar that whn th null hypothsis is tru, somtims (1 tim in 0) a tst at this lvl will lad you to rjct it incorrctly. Pag 0 Discussion Point 1 Th Chi-Squard Tst 3

33 Th groups hav bn combind bcaus othrwis thr would b on group, that for 6 popl with th condition, with vry small xpctd frquncy (much lss than 5). Pag 0 Discussion Point You would xpct a bimodal distribution. Th sts of popl blonging to familis without th gntic link would contain fw, if any, popl with th condition; by contrast th sts of popl blonging to familis with th gntic link would contain svral popl with th condition. So you would xpct on mod corrsponding to fw (if any) popl with th condition and anothr corrsponding to svral with it. This would not b a binomial distribution. Th Chi-Squard Tst 33

34 Exrcis 10C 1. (i) Thr ar clls and on rstriction, th total, so ν = 1= 1. (ii) H 0 : Th colours ar in th statd proportion of 1:3 H 1 : Th colours ar not in th statd proportion of 1:3 X =.4 Sinc.4<3.84, H 0 is accptd. Th botanist s thory is accptd at th 5% significanc lvl.. Expctd frquncis: T , T 35, T H 0 : Th modl applis H 1 : Th modl dos not apply X = 7.03, ν = Sinc 7.03 > 5.991, H 0 is rjctd at th 5% significanc lvl. Th vidnc dos not support th modl. 3. (i) All thr possibl outcoms hav th sam probability of 1 3. (ii) H 0 : Th probabilitis ar 1 3 whn Ptr and Jan play. H 1 : Th probabilitis ar not 1 3 whn Ptr and Jan play. X = 6., ν = Sinc 6. > 5.991, H 0 is rjctd. Th vidnc supports Jan s claim at th 5% significanc lvl. 4. (i) Man.898, varianc.88 Ths ar vry clos togthr supporting a Poisson modl. (ii) H 0 : Th undrlying distribution is Poisson. H 1 : Th undrlying distribution is not Poisson. X = 7.83, ν = 9 = 7 (Th last two groups ar combind) Sinc 7,83 < 14.07, H 0 is accptd. Th data support th Poisson modl at th 5% significanc lvl. 5. (i) 148, 5 (ii) H 0 : Thr is no chang in th lvl of support for th Grn Party H 1 : Thr is chang in th lvl of support for th Grn Party X = 3.47, ν = 1 Sinc 3.47 < 3.84, H 0 is accptd. At th 5% significanc lvl th vidnc dos not support th viw that thr has bn a chang of support for th Grn Party. 6. (i) H 0 : Th proportions in th diffrnt grads ar in th sam as thos spcifid. H 1 : Th proportions in th diffrnt grads ar diffrnt from thos spcifid. (ii) Grad A B C D Numbr of appls (iii) X =.997, ν = 3 Sinc.997 < 7.815, H 0 is accptd. At th 5% significanc lvl, th vidnc suggsts thr has bn no chang in th proportions in th diffrnt grads. Th Chi-Squard Tst 34

35 7. (i) H 0 : Th proportions of lat-running trains in th diffrnt months ar qual. H 1 : Th proportions of lat running trains in th diffrnt months ar not all qual. X = 9.63, ν = 1 1 = 11 Sinc 9.63 > 19.68, H 0 is rjctd. At th 5% significanc lvl, th vidnc suggsts that thr ar mor dlays in som months than othrs. (ii) Th largst contribution to th valu of X coms from Octobr suggsting that lavs on th rails may b a major caus. 8. (i) Band Expctd numbr of popl in sampl, f A 13.6 B 86.4 C C 84 D 64.8 E 34.8 Total 400 (ii) H 0 : Th proportions in th various bands ar th sam in Avonford as nationally. H 1 : Th proportions in th various bands ar not th sam in Avonford as nationally. X = 39.4, ν = 6 1 = 5 Sinc 39.4 > 11.7, H 0 is rjctd. At th 0.5% significanc lvl, th vidnc suggsts that th proportions oof popl in th diffrnt bands in Avonford ar diffrnt from th national proportions. (iii) Avonford has mor popl in th xtrm bands, for xampl A and E, and fwr in th middl bands. 9. (i) Man 1.5, varianc (ii) H 0 : Th distribution is Poisson. H 1 : Th distribution is not Poisson. X = 3.33, ν = 6 = 4 (5, 6, 7 and >7 goals ar put togthr as on group) Sinc 3.33 < 9.488, H 0 is accptd. At th 5% significanc lvl, th vidnc suggsts that th distribution is indd Poisson. (iii) Numbr Frquncy (iv) In this cas th valu of th paramtr, 1.7, is givn and not calculatd from th data. (v) H 0 : Th distribution is Poisson with paramtr 1.7. H 1 : Th distribution is not Poisson with paramtr 1.7. X = 1.87, ν = 6 1 = 5 Sinc 1.87 > 11.07, H 0 is rjctd. Th vidnc suggsts that th distribution is not Poisson(1.7). It sms that th football nthusiast is wrong whn h claims that th man numbr of goals is 1.7. Th Chi-Squard Tst 35

36 10. (i) Numbr of childrn Numbr of housholds (ii) H 0 : Th distribution is Poisson with paramtr H 1 : Th distribution is not Poisson with paramtr X = 3.5, ν = 6 1 = 5 Sinc 3.5 > 11.07, H 0 is rjctd at th 5% significanc lvl. Th vidnc suggsts that th distribution is not Poisson(1.40). 11. (i) 94 girls and 86 boys, 96 p = 180 (ii) Group 5G 0B 4G 1B 3G B G 3B 1G 4B 0G 5B Frquncy, f (iii) H combins th groups to avoid thos with small valus of f to gt Group 5G 0B or 4G 1B 3G B G 3B 1G 4B or 0G 5B Frquncy, f This rducs th numbr of groups, k, to 4. Thr ar rstrictions (th total and th obsrvd probability that a randomly slctd studnt is a girl) so ν = k = 4 =. (iv) H 0 : Th distribution is binomial. H 1 : Th distribution is not binomial. X = 7.44, ν = 4 = Sinc 7.44 > 5.991, H 0 is rjctd. Th tst supports th rsarchr s thory at th 5% significanc lvl. (v) Th xprimnt assumd that mn and womn in gnral will bhav in th sam way as 15- and 16- yar old boys and girls who do not know ach othr. This may vry wll not b th cas. 1. (i) It is rasonabl to assum that faults occur at random and with fixd probability. (ii) 1 p = 4 (iii) H 0 : Th distribution is binomial. H 1 : Th distribution is not binomial. X = 10., ν = 6 = 4 (5, 6 and 7 ar combind into on group) Sinc 10. > 9.488, H 0 is rjctd. At th 5% significanc lvl, th data do not support th thory that th distribution is binomial. (iv) Th frquncy is gratr in th xtrm groups, 0 and 5+, than would b xpctd for a binomial distribution. Th Chi-Squard Tst 36