The steps of the hypothesis test

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1 ttisticl Methods I (EXT 7005) Pge 78 Mosquito species Time of dy A B C Mid morning Mid Afternoon Dusk The Chi squre test sttistic is the sum of the chi squre vlues in ech cell. Observed Expected with d. f. = (rows )*(col ) =* = 4 The steps of the hypothesis test Expected ) Mosquito species occurrence is independent of time of dy ) Mosquito occurrence is NOT independent 3) Assume IID r.v. 4) 0.05, the criticl vlue with (r )(c ) = 4 d.f. for 0.05 is ) From the smple we get clculted = ) The clculted test sttistic exceeds the tbulr vlue with (r )(c ) = 4 d.f.. The criticl tbulr vlues re; 0.05, = sttisticlly significnt 0.0, = highly significnt 0.00, = Wow! Using this terminology we see we hve highly significnt deprture in this cse. o wht is the P vlue, the chnce of finding vlue of or greter by rndom chnce? This is P(> )= , bout in million. 7) o we conclude tht the occurrence of mosquito species is not independent of the time of dy. It ppers to be very dependent. A exmple 3 of the Chi squre Test of Independence in A Two vrieties of prticulr moth species occur in two colors (brown nd white). A biologist in North Crolin wnts to know if the distribution of the two vrieties vries with the re of the stte. He collects individuls from ech region of the stte nd note the number of ech vriety. ee computer output Jmes P. Geghn Copyright 0

2 ttisticl Methods I (EXT 7005) Pge 79 Chi qure Goodness of Fit This test is similr to the Chi qure test of independence, but insted of deriving the expected vlues from the row nd column sums, the expected vlues re derived from some theoreticl distribution. In flowering plnt with incomplete dominnce cross between red flowered prent (RR) nd white flowered prent (rr) is expected to yield offspring with pink flowers (Rr). If the offspring re then crossed with ech other, rtio of 9:6: or red to pink to white flowered offspring should result. In one prticulr experiment the offspring produced the observed results of 53 red, 7 pink nd 7 white plnts. Does this result conform to the expected results? ) H 0 : Results follow the expected proportions ) H : Results do not follow the expected proportions 3) Assume IID r.v. 4) et (t sy 0.05 or 0.0 s before). Clculte the criticl vlue where the chi squre results for this test hve c = degrees of freedom (where c is the number of column or the number of ctegories, 3 in this cse). The criticl vlue for = 0.05 is 5.99 nd for = 0.0 it is ) Conduct n experiment to obtin results. We got 4 flowers from our experiment. Red Pink White Totl Expected rtio Expected proportions Observed numbers Expected numbers Chi squre ) Compre the clculted chi squre vlue to the test sttistic. The clculted vlue (6.) does exceeds the test sttistic vlue of = 0.05, but not for = 0.0. Tbulr Chi squre vlues lph Chi squre vlue n = d.f. = Actul P(> ) ) In this cse we reject the null hypothesis nd might term our result s sttisticlly significnt deprture from the expected result under the null hypothesis. As published result we my wish to stte tht the P vlue ws 0.045, clerly indicting tht our result ws significnt t the 0.05 level of, but not t higher level. A exmple 3b of the Chi squre Goodness of Fit test in A ee computer output A exmple 3c testing simple rtio with Chi squre ee computer output Jmes P. Geghn Copyright 0

3 ttisticl Methods I (EXT 7005) Pge 80 A finl note on the Chi squre tests of hypothesis. Although tests of hypothesis bout vrinces my be either directionl or non-directionl, the chi squre tests of independence re directionl. mll vlues of the chi squre sttistics for these tests indicte tht the null hypothesis is met very well since the observed vlues re very close to the expected vlues. It is only with lrger deprtures of the expected vlues from the observed vlues tht the chi squre sttistic should be rejected. Therefore, it is only excessively lrge chi squre vlues tht cuse the null hypothesis to be rejected nd this would be one tiled test. ummry The Chi squre distribution The Chi squre distribution cn be derived s the squre of the Z distribution. mple vrinces re Chi squre distributed The Chi squre distribution cn be used to test hypotheses bout vrinces. The distribution hs only one prmeter,, (nd is different for every ) the distribution is non negtive nd symmetricl. The vrince of the distribution is. Hypothesis testing employs the form of the distribution For testing vrince we ssume the vrible i is Normlly nd Independently distributed rndom vrible (NID r.v.) In the Chi squre tbles Degrees of freedom re on the left nd different distribution is given in ech row. elected probbilities in the upper TAIL of the distribution is given in the row t the top of the tble. The distribution is NOT symmetric, so the probbilities t the top must be used for both upper nd lower limits. In ddition to tests of vrinces, the Chi squre cn be used to do Test of Independence Test of Goodness of Fit For these tests we ssume i is n Identiclly nd Independently distributed rndom vrible (IID r.v.) In A the Chi squre test of independence cn be done with PROC FREQ. 0 Jmes P. Geghn Copyright 0

4 ttisticl Methods I (EXT 7005) Pge 8 The F test This test cn be used to either, test the equlity of popultion vrinces our present topic test the equlity of popultion mens (this will be discussed lter under ANOVA) The F test is the rtio of two vrinces (the rtio of two chi squre distributions) Given two popultions Men Vrince Popultion Popultion Drw smple from ech popultion Popultion Popultion mple size n n d.f. mple men mple vrince To test the Hypothesis H: 0 H: (directionl, or one sided, hypotheses re lso possible) The test sttistic is F which hs n expected vlue of under the null hypothesis. In prctice there will be some vribility, so we need to define some resonble limits nd this will require nother sttisticl distribution. The F distribution ) The F distribution is nother fmily of distributions, ech specified by PAIR of degrees of freedom, nd. is the d. f. for the numertor is the d. f. for the denomintor Note: the two smples do not hve to be of the sme size nd usully re not of the sme size. ) The F distribution is n symmetricl distribution with vlues rnging from 0 to, so [0 F ]. 3) There is different F distribution for every possible pir of degrees of freedom. 4) In generl, n F vlue with nd d.f. is not the sme s n F vlue with nd d.f., so order is importnt. i.e. F, F, usully Jmes P. Geghn Copyright 0

5 ttisticl Methods I (EXT 7005) Pge 8 5) The expected vlue of ny F distribution is if the null hypothesis is true. The F distribution F distribution with, 5 d.f. F distribution with 00, 00 d.f. F distribution with 5, 0 d.f. The F tbles The numertor d.f. ( ) re given long the top of the pge, nd the denomintor d.f. ( ) re given long the left side of the pge. ome tbles give only one F vlue t ech intersection of nd. The whole pge would be for single vlue nd usully severl pges would be given. Our tbles will give four vlues t the intersection of ech nd, ech for different vlue. These vlues re given in the second column from the left. Our tbles will hve two pges. Only very few probbilities will be vilble, usully 0.05, 0.05, 0.0 nd 0.005, nd sometimes Only the upper til of the distribution is given. Prtil F tble d.f. P>F Jmes P. Geghn Copyright 0

6 ttisticl Methods I (EXT 7005) Pge 83 Working with F tbles The F tbles re used in fshion similr to other sttisticl tbles. elect probbility vlue t the intersection of the two degrees of freedom. Be sure to keep trck of which one is the numertor degrees of freedom (top of tble) nd which is the denomintor degrees of freedom (left side of tble). At the intersection of the degrees of freedom there re four F vlues corresponding to vlues of 0.05, 0.5, 0.0 nd Find the corresponding F vlue for the desired. Exmple of F tble use, one tiled exmple: Find F with (5,0) d.f. for 0.05 find F 0.05 such tht P[F F 0.05, 5, 0 d.f. ] = where = 5 nd = 0 For F 5, 0 d.f. The tbulr vlues re listed s 3.33, 4.4, 5.64, 6.87 These represent P[F 5, 0 d.f..5] = 0.00 (Not in your tble) P[F 5, 0 d.f. 3.33] = P[F 5, 0 d.f. 4.4] = 0.05 P[F 5, 0 d.f. 5.64] = 0.00 P[F 5, 0 d.f. 6.87] = so the vlue we re looking for is For this vlue P[F 5, 0 d.f. 3.33] = Note tht since this is tiled vlue, then P[F 5, 0 d.f. 3.33] = 0.950, so the two sides sum to Also note tht if we reverse the d.f., the tble shows tht P[F 0, 5 d.f. 4.74] = 0.050, so the F vlues generlly differ when d.f. re reversed More working with F tbles Only the upper til of the distribution is given. There re three resons for this. Most F tests, including those for Anlysis of Vrince (ANOVA), re one tiled tests, where the lower til is not needed. The need to clculte the lower til cn be eliminted in some two-tiled cses. The vlue of F for the lower til cn be found by trnsformtion of vlues from the upper til. Clculting lower til vlues for the F distribution To obtin the lower til for vlue F, for prticulr vlue of First obtin the vlue in the upper til for F, for the sme vlue of (note the chnge in order of the d.f.) Then clculte /F, to get the lower til. Jmes P. Geghn Copyright 0

7 ttisticl Methods I (EXT 7005) Pge 84 F tble : two-tiled exmple Find both upper nd lower limits for F with (8, 0) d.f. for 0.05 Find F, 8, 0 lower nd F, 8, 0 such tht P[F 0.975, 8, 0 F F 0.05, 8, 0 ]=0.950 where = 8 nd = 0 For the upper til, the vlue we re looking for cn be red directly from the tble. It is P[F 8, ] = 0.05 note tht we use only s the probbility for one of the two tils To find the lower til, we reverse the d.f., we find tht P[F 0, ] = 0.05 nd then clculte F 8, 0 d.f. lower limit = /F 0,8 = /4.30 = 0.36 Note the reversl of the order of the degrees of freedom Numericl exmple of n F-test of hypothesis The concentrtion of blue green lge ws obtined for 7 phytoplnkton-density smples tken from ech of two lke hbitts. Determine if there is difference in the vribility of phytoplnkton density between the two hbitts. ) ) H: 0 H: 3) Assume: Independence (rndomly selected smples) nd tht BOTH popultions re normlly distributed. 4) 0.05 nd the criticl limit is Clculte P[F lower F F upper ] = P[F lower F] P[F F upper ] = = 0.95 P[F F upper ] = 0.05 we cn get directly from the tble, F =0.05, 6, 6 = 5.8 P[F lower F] = 0.05 we clculte s /P[F F upper ] = /5.8 = 0.78 P[0.78 F 5.8] = 0.95 This cse is uncommon becuse d.f. upper = d.f. lower, so the F vlues re the sme. 5) Drw smple of 7 from ech hbitt; clculte the vrinces nd the F rtio. Jmes P. Geghn Copyright 0

8 ttisticl Methods I (EXT 7005) Pge 85 mple dt ummry sttistics Hbitt Hbitt Observtion Observtion Observtion Observtion Observtion Observtion Observtion ttistic Hbitt Hbitt i i Men ( ) Then clculte the F vlue s F = = 6.95 = ) Compre the clculted vlue (0.73) to the criticl region. Given = 0.05 nd TWO TAILED lterntive, nd knowing tht the degrees of freedom re =6 nd =6, (note tht both re equl), the criticl limits re P[0.78 F 5.8] = ince our clculted F vlue is between these limit vlues we would fil to reject the null hypothesis, concluding tht the dt is consistent with the null hypothesis. But it ws close. Mybe there is difference nd we did not hve enough power. ome notes on F tests NOTE tht in this exmple the smller vlue fell in the numertor. As result, we were compring the F vlue to the lower limit. However, for two tiled tests, it mkes no difference which flls in the numertor, nd which in the denomintor. As result, we cn ARBITRARIL decide to plce the lrger vlue in the numertor, nd compre the F vlue to the upper limit. lrger The need to clculte the lower limit cn be eliminted if we clculte F. smller However, don't forget tht this rbitrry plcing of the lrger vrince estimte in the numertor is done for TWO TAILED TET ONL, nd therefore we wnt to test ginst F. There re three common cses in F testing (ctully two common nd one not so common). ) Frequently, prticulrly in ANOVA (to be covered lter), we will test H: 0 ginst the lterntive, H:. In this cse we ALWA form the F vlue s F =. Jmes P. Geghn Copyright 0

9 ttisticl Methods I (EXT 7005) Pge 86 We put the vrince tht is expected to be lrger in the numertor for one tiled test! Don t forget tht this is one tiled nd ll of is plced in the upper til. In the event tht F < we don't even need to look up vlue in the tble, it cnnot be significnt. ) Norml tiled tests (used in smple t-tests to be covered lter) will test the lterntive, H:. Here we cn form the F vlue s F H: ginst 0 lrger. Don t forget tht this is -tiled test nd it is tested ginst the upper til with only hlf of (i.e. ) in the upper til. When the lrger vlue is plced in the numertor there is no wy tht we cn get clculted F <. 3) If both the upper nd lower bounds re required (not common, found mostly on EXAM in bsic sttistics) then we will be testing H: 0 ginst the lterntive H:. We cn form the F vlue ny wy we wnt, with either the lrger or smller vrince in the numertor. This is tiled test with in ech til, nd F cn ssume ny positive vlue (0 to ) ummry smller The F distribution is rtio of two vrinces (i.e. two Chi squre distributions) nd is used test used to test two vrinces for equlity. The null hypothesis is H:. 0 The distribution is n symmetricl distribution with vlues rnging from 0 to, nd n expected vlue of under the null hypothesis. The F tbles require two d.f. (numertor nd denomintor) nd give only very few criticl vlues. Mny, perhps most, F tests will be directionl. For the tests the vrince tht is expected to be lrger nd hypothesized to be lrger goes in the numertor whether it is ctully lrger or not. This vlue is tested ginst the upper til with probbility equl to. For the non-directionl lterntive we my rbitrrily plce the lrger vrince in the numertor nd test ginst the upper til, but don't forget to test ginst. Probbility Distribution interreltionships The probbility tbles tht we hve been exmining re interrelted. One of these interreltionships is ctully pretty importnt! If you exmine the F tble it turns out tht in ddition to F vlues the first column is equl to vlues of t, the lst vlue in the first column corresponds to Z nd the lst row is Chi squre vlue divided by degrees of freedom. Jmes P. Geghn Copyright 0

10 ttisticl Methods I (EXT 7005) Pge 87 d.f t (two tiled) (4.96) Z (3.84) F vlues =0.83 more F vlues The distributions nd reltionships we hve discussed re: i 0 ) Zi for observtions nd Z for testing hypothesis bout mens. n ) = Z with d.f. = Z with n d.f. = / with n d.f. 3) t with n d.f. n 4) F = Interreltionships with n, n d.f. ) / with d.f. = F with, d.f. which follows n F distribution with, d.f. ) F with, d.f. if H 0 is true given from prt bove Jmes P. Geghn Copyright 0

11 ttisticl Methods I (EXT 7005) Pge 88 then nd therefore, nd with d.f. F with n, n =, d.f. if H 0 is true, then, then F with, or n, n d.f. 3) t with = follows Z distribution, since s increses the smple vrince ( ) pproches the popultion vrince ( ). Tht is, s the smple size pproches infinity the t distribution, i t pproches the Z distribution i Z. 4) Z = F with, d.f. we sw tht Z = with d.f. we sw tht / = F with, d.f. then F = / = Z / = Z / = Z. 5) t with d.f. = F with, d.f. This cn be shown in severl wys. First, we just sw tht Z = F with, d.f. This suggests tht t = F with, d.f. Another type of proof is given below. F with, d.f. i with n d.f., nd where ˆi is the men of subgroup of the dt prtitioned into two (or more) groups (the bsis of Anlysis of Vrince). recll, n n ˆ ˆ ˆ ˆ i i i i i i i d.f re n = (n ) + Let n i i with d.f. n i Let i with n d.f. n both of which re unbised estimtes, n then F with, n d.f. nd n F t n n with n d.f. Jmes P. Geghn Copyright 0

12 ttisticl Methods I (EXT 7005) Pge 89 ummry ) / with d.f. = F with, d.f. ) F with, d.f. if Ho is true 3) t with = follows Z distribution 4) Z = F with, d.f. 5) t with d.f. = F with, d.f. ome Exmples in the F tbles (ll 0.05, two tils for Z nd t vlues since sign is lost in squring) F with, 0 d.f. = 4.96 = t with 0 d.f. = (.8) = 4.96 F with, d.f. = 3.84 = Z = (.96) = 3.84 F with 0, d.f. =.83 = / = 8.3 / 0 =.83 with 0 d.f. Confidence intervls nd mrgin of error The confidence intervl is n expression of wht we believe to be rnge of vlues tht is likely to contin the true vlue of some prmeter is clled confidence intervl. The width of this intervl bove nd below the prmeter estimte is clled the mrgin of error. We cn clculte confidence intervls for mens () nd vrinces ( ). Confidence intervls for t nd Z distributions t nd Z distribution confidence intervls strt with t or Z probbility sttement. P( t t t ) cn lso be written P( t t ) which is modified to express n intervl bout insted of t (or Z). P( t t ) P( t t ) The finl form is given below. P ( t t ) The expression for Z hs n identicl derivtion. P ( Z Z ) Jmes P. Geghn Copyright 0

13 ttisticl Methods I (EXT 7005) Pge 90 A common short nottion for the intervl in the probbility sttement is given s t the probbility sttement is preferble s finl result. The vlue t mens nd t for intervls on for intervls on individul observtions is hlf of the intervl width from the lower limit to the upper limit nd is clled the mrgin of error. Confidence intervls for vrince Vrinces follow Chi squre distribution. The confidence intervl for vrince is bsed on the Chi qure distribution. P or ( lower upper ) P( lower ) upper which is solved to isolte. P( ) P lower upper ( ) upper lower giving the expression, ( ) P upper lower Notice tht the upper tbulr Chi squre vlue comes out in the lower bound nd the lower Chi squre in the upper bound. ( ) P upper lower Notes on confidence intervls One sided intervls re possible, but uncommon. Confidence intervls re one of the most common expressions in sttistics, frequently occurring in publictions. Mrgins of error nd confidence intervls re not lwys clculted in sttisticl softwre progrms, but they cn esily be done by hnd. From the previous A Exmple c We receive shipment of pples tht re supposed to be premium pples, with dimeter of t lest.5 inches. We will tke smple of pples, nd plce confidence intervl on the men. The smple vlues for the pples re;.9,.,.4,.8, 3.,.8,.7, 3.0,.4, 3.,.3, 3.4, but Jmes P. Geghn Copyright 0