Topic 6: Hypothesis Testing (Ch. 10)

Transcription

1 Tpic 6: ypthesis Testig (Ch. 0). Geeral Ccepts Our secd geeral type f statistical iferece is called hypthesis testig. I this secti we will explre testig hyptheses abut a sigle parameter, µ. wever, the ccepts will be geeral ad will apply t subsequet hypthesis tests. Csider a particular prblem. Let x dete the chlesterl level. Suppse we kw, pssibly thrugh extesive experiece, that the mea f x fr U.S. males is. Suppse further that we are iterested i the mea, µ, fr hypertesive males wh smke. I particular, we wuld like t state whether the mea fr smkers is als, r whether it is differet. I rder t make a decisi, suppse we take a radm sample f smkers, perhaps f size = 5. We culd fid their sample mea, fr example, suppse it is x = 0. We kw hw t determie a cfidece iterval. The questi w is hw t test a hypthesis, such as µ =, frm the data. Csider the fllwig framewrk fr a test f hypthesis: ) Determie the ull hypthesis, deted, which is the suppsiti t be tested. e.g. : µ = ) Determie the alt erative hypthesis, deted, which is the suppsiti t be accepted if we reject e.g. : µ 3) Decide the apprpriate test statistic, say TS, which will be used t test e.g. TS = x r ( x µ ) z = σ 4) Chse a sigificace level,α, ad determie the rejecti regi, say RR. The RR are the values f the test statistic which wuld cause e t reject. Fr example, e wuld reject if, give µ =, the prbability f btaiig the bserved x, r a mre extreme value, is sufficietly small. I ther wrds, x values much larger r much smaller tha µ = are ulikely, ad wuld lead us t reject. The exact RR is determied by the sigificace level, α, which is the prbability that we will reject true. Typically, α is set at 0.0, r e.g. Suppse we let α =.05. The apprpriate RR is z > z.05 =.96 r z< z.05 =.96 whe it is

2 5) Make a decisi by first gatherig data, calculatig the TS, ad rejectig if the TS is i the RR. e.g. We culd calculate ( x ) z =, σ 5 ad reject : µ = if z >.96 r z <.96. These are the five geeral steps i hypthesis testig. Oe issue is hw t set α. Csider the fllwig table f pssible actis: NULL YPOTESIS DECISION D Nt Reject (decide µ = ) True ( µ = ) Crrect False ( µ ) Type II errr Reject (decide µ ) Type I errr Crrect Specifically, the defiitis are: ) Type I errr: reject whe is true, i this prblem decidig µ whe i fact µ =. ) Type II errr: failig t reject (fte called acceptig ) whe is false (i this prblem decidig µ = whe i fact µ ). 3) α = Prb (Type I errr), als called rejecti errr. 4) β = Prb (Type II errr), als called acceptace errr. We will study β later. We will see that whe e sets α ad, the size f β is determied. Therefre, the chice f α shuld be iflueced by what size f β we are willig t accept. The p-value is the prbability f btaiig a TS (e.g. x ) as extreme r mre extreme tha the bserved value f the TS give that is true. This is w widely used i scietific research, ad will be illustrated i the fllwig sectis.

3 . Tw -Sided Tests Usig the previus priciples, let s summarize the previus example. Our bjective was t test whether the mea chlesterl level fr males wh smke is. Suppse we kwσ, i this case we assume σ =46, ad ur sample size is =. We wuld the have a z-test as fllws: ) : µ = ) : µ X 3) TS is z = 46/ 4) Suppse α =.05. This is a tw-sided test because we wat t test fr µ > r µ <. Therefre, bth large r small values f z wuld lead us t reject. Our RR is therefre z >.96 ad z < ) Suppse the data shws x = 7. The test statistic is 7 z = = / Because z is t i the rejecti regi, we d t have statistical evidece t reject. The result is t statistically sigificat. T calculate the p-value, we must determie hw prbable is a mea as extreme as x = 7 if µ =. Nte Prb ( x > 7) = Prb (z > 0.45) = Because this is a tw-tailed test, we als are iterested i ptetial extreme values i the lwer tail. Clearly P (z < -0.45)= 0.36 Therefre, P = (0.36)= 0.65 This tells us that a value f x = 7 is t surprisig if µ =. Because p >.05, we d t reject. The hypthesis testig result culd have bee predicted usig the cfidece iterval methdlgy. Befre we fud that a 95% cfidece iterval fr µ based x = 7 was (9, 43) If ur ull value, µ 0, is i the iterval, as µ = is, e wuld t reject. wever, fr a ull hypthesis such as : µ = 60, the hypthesized mea is t i the iterval, ad hece we wuld reject. 3

4 Suppse σ was ukw. I such cases we wuld calculate s ad use a t-test. Fr example, let x dete the alumium level i plasma. Fr rmal childre, the mea is kw t be 4.3. If childre are give a atacid that ctais alumium, wuld the atacid chage the mea? Test at α = 0.05 we assume that x is apprximately rmal. Our framewrk is w as fllws: ) : µ = 4.3 ) : µ 4.3 3) TS : x 4.3 t = s/ 4) RR: We reject fr large r small values f x (r t). ece RR is t > t 0.05 =.6 r t <.6 5) Suppse we test = 0 childre, ad fid x = 37.0 ad s = 7.3 t = = / 0 3. Oe-Sided Tests Clearly e rejects. 6) T fid the p-value, te Pt ( 9 > 4.78) = Clearly t 9= 4.67 is mre extreme, hece p < (.0005) =.00 The iterpretati is that the likelihd f fidig a X value f 37.0 r smethig mre extreme if µ = 4.3 is less tha e i a thusad. Cmputer prgrams usually give precise p-values. Smetimes the ature f the prblem suggests that e is iterested ly i a icrease, r ly i a decrease, i the mea. Fr example let s recsider the last prblem. If childre were give a atacid that ctais alumium, e wuld t expect fr their alumium level i their plasma t decrease. Istead the relevat scietific questi is whether a icrease i mea has ccurred. If we kw a priri that a chage wuld ly be i a sigle directi, a e-sided test is better statistically, as will becme bvius. Csider the previus prblem as a e-sided test. ) : µ 4.3 ) : µ > 4.3 3) x 4.3 TS : t = s/ 4

5 4) Let α = We wuld reject ly fr large values f x, r equivaletly f t. ece the RR is t > t 9,.05 =.833 5) With the previus data, we had t 9 = ece e wuld reject, ad cclude that there is statistical evidece that µ has icreased (frm 4.3). 6) We are lkig w ly fr extremes i the right tail. Because Pt ( 9 > 4.78) =.0005, we kw p < The bk has a example f testig i the lwer tail. 4. Types f Errrs ad Pwer Let s csider ather prblem, ad illustrate ther ccepts. certai brad f cigarettes carries the statemet.5 mg ictie average per cigarette. Let x dete the ictie ctet f a cigarette, ad µ its mea. The claim is that µ =.5. The FD clearly has a iterest i such claims. Let s call this a lw-ictie cigarette. Fr simplicity, let s assume that the variability i ictie ctet amg cigarettes is σ = 0.0. Suppse that we decide t use α =0.0 (why s lw?). It is prpsed that we use a radm sample f =36 cigarettes, if that is suitable statistically. I terms f ur framewrk, e has: ) : µ.5 ) : µ >.5 x.5 x.5 3) TS: z = = RR: z > z =.33 4).0 This framewrk is usually sufficiet. wever, fr calculatig β, te that we culd state the RR smewhat differetly, x.5 i.e. we wuld reject if >.33 r x >.5+.33(0.0333) = Let s fid the errr pssibilities. ) We set α = 0.0. ece the prbability f rejectig whe it is true, i.e. f decidig the cigarettes are high ictie whe they are actually lw ictie, is 0.0. The legal implicatis f such a errr are sizeable, s we set α very lw t be very cfidet f ur pssible claim. ) But what is β, i.e. prbability f a Type II errr, which i this case wuld be decidig that they are lw ictie whe i fact they are t. Suppse e pssible alterative 5

6 ictie ctet is µ =.6. β is the the prbability f decidig that µ =.5, i.e. failig t reject, whe i fact µ =.6, i.e. the is false. T fid this prbability, we must determie the prbability that x <.578 (why is this s?) if µ =.6 ad σ = 0.0. Nte that Px< (.578) = P z < = Pz ( < 0.66) = I ther wrds, we have a 5% chace f decidig erreusly that high ictie cigarettes (with µ =.6) are lw ictie cigarettes. If this errr is t large, we eed t either icrease α (which may be uacceptable) r icrease. Why wuld these wrk as claimed? Pwer is defied as the prbability f rejectig β. I this example, pwer = = whe it is false. ece, pwer = - Ofte, e wats t determie the pwer fr a rage f alterative meas. Let µ dete sme specific mea i the, e.g. we fud pwer befre fr µ =.6. Suppse e wats t determie pwer als fr µ =.55 ad µ =.65. Nte fr µ =.55: β = P x <.578if µ =.55 ( ) = P z < = Pz ( < 0.84) = Pz ( > 0.84) = 0.8 ece pwer= 0.8 = 0.. Fr µ =.65 β = Px ( <.578 µ =.65) = Pz ( < ( )/0.0333) = Pz ( <.6) = 0.05 plt f pwer versus varius pssible µ values is called a pwer curve. It is widely used i egieerig. The text gives a differet example. Suppse we kw that the mea chlesterl level fr 0-4 year ld males is 80 mg. We wat t test whether this same mea hlds fr all US male adults. Our ull hypthesis is 0 : µ = 80. 6

7 Suppse we wat the pwer f the test if the true mea is µ =. Suppse als that α =.05 ad we kw σ = 46. Let s calculate the RR. Nte we wuld reject ly i the upper tail, fr z >.645. This implies we reject if X 80 > / 5 r X > (9.) = 95.. This rejecti regi is illustrated i Fig. 0.. Pwer is the prbability f rejectig the ull hypthesis whe it is false, i this case whe µ =. The calculati is Pwer = PX ( 95. if µ = ) = PZ (.73) = PZ ( <.73) = 0.04 = Calculatig this fr may values f µ gives us the pwer curve illustrated i Fig Sample Size Csider a e-sided test, 0: µ = µ 0 with specified α. Suppse e als wats t specify β fr the same particular alterative mea, µ. The sample size required t achieve give α ad β is ( zα + zβ) σ = µ µ 0 Fr example, i ur cigarette prblem, µ 0 =.5 ad α =.0. Suppse fr µ =.6 we desire β =.05. With σ = 0.0, e has ( )0.0 = 0. = 6.6 ece, e wuld have t icrease t 63 t meet this specificati. Fr a tw-sided test, the frmula is 7

8 ( zα/ + zβ ) σ = µ µ 0. Fr the abve prblem with α =.0 ad β =.05, e wuld have ( )0.0 = 0. = 7.4 8