Chapter 8 Part 2. Unpaired t-test With Equal Variances With Unequal Variances

Transcription

1 Chapter 8 Part Upaired t-tet With Equal Variace With Uequal Variace December, 008

2 Goal: To eplai that the choice of the two ample t-tet deped o whether the ample are depedet or idepedet ad for the idepedet ample whether the two variace ca be coidered equal or ot. Skill: Give a quetio ad a dataet ou hould be able to (1) write dow the ull ad alterative hpothee, () pick out the correct Stata commad to aalze our problem ad (3) eplai whether ou accept or reject the ull hpothei ad (4) eplai wh. Cotet: Two ample upaired t-tet with equal variace - Page 1 Pooled variace - Page 4 tet tatitic - Page 5 tadard error - Page 5 cofidece iterval - Page 11 What tep do ou eed kow to correctl aalze the problem - Page 1 How to decide whether to ue uequal or equal variace - Page 14 Micoceptio cocerig overlappig cofidece iterval - Page 16 Two ample upaired t-tet with uequal variace - Page 17 tet tatitic - Page 19 tadard error - Page 19 cofidece iterval - Page 0 degree of freedom - Page 0 Overview table - Page 3 Stata:. ttet chol_bl, b(black) level(99). ttet hit, b(e) uequal. ttet hit, b(e) uequal welch Dataet: allhatchap8_1.dta hitidie.dta

3 Two ample upaired t-tet So far we kow how to deal with oe ample ad with two paired variable. We tarted b covertig the paired variable ito a igle variable b defiig the differece of each pair. Later we aw that Stata would fid the differece ad produce the paired t- tet all i oe commad. See the table o the lat two page of thi hadout for a overview. Whe we look at two ample which are idepedet ad therefore caot be coverted to a igle ample, we ru ito the problem of whether the variace for the two populatio from which the ample were draw are kow or ukow ad the whether the variace are equal or ot. The cae i which the two variace are kow i o rare that we will ot coider it. That leave u with whether the variace are equal or ot. We will approach the impler cae firt, amel the cae i which the variace of the two populatio are coidered equal. Eample: There i ome evidece that Black patiet repod to treatmet for elevated choleterol differetl from o-black patiet. So if we are goig to compare thee two group of patiet with repect to follow-up value of choleterol (alo called total choleterol ad repreeted b TC), we will firt wat to kow if the tarted off with imilar choleterol value. We have two idepedet ample here (baelie TC for Black ad baelie TC for o-black). There i o logical wa to pair thee ample ad the ample are of differet ize. The data et ued i allhatchap8_1.dta.. bort black: um(chol_bl) -> black No-Black Variable Ob Mea Std. Dev. Mi Ma chol_bl > black Black Variable Ob Mea Std. Dev. Mi Ma chol_bl Page -1-

4 The upaired t-tet with equal variace: Thi time we tart with two idepedet radom ample Y Y,,..., Y 1 (ote that ad do t have to be equal). X1, X,..., X ad So ow we have two ample mea ( ad ) ad two ample tadard deviatio ( ad ). Our hpothee are: H0: μ μ or μ μ 0 HA: μ μ or μ μ 0 veru. (ote there i o pairig here, we jut get each of the two mea ad the get the differece). Below we ll tart with the implet form, i.e. whe the variace are the ame. We aume X ~ N μ, σ ad Y ~ N μ, σ where σ i the commo variace for the two populatio. Sice the two ample are idepedet Var( X Y ) Var( X ( Y )) Var( X ) Var( Y ) σ σ 1 1 σ Var( Y ) Var( Y ) (recall that ). Page --

5 I geeral, it i ot the cae that the variace of a differece i the um of the variace (it i the idepedece of the two ample that give u thi). Uuall there i a term (called the covariace of the two variable) that take ito coideratio the lack of idepedece of the ample. Sice the um or differece of two ormall ditributed radom variable i alo ormall ditributed we get: 1 1 X Y ~ N( μ μ, σ ) If the ull hpothei i true (i.e. Which implie that μ μ 0. ),the 1 1 X Y ~ N( 0, σ ) X Y N σ 1 1 ~ ( 01, ). Sice we do t kow ad which we do kow. X σ, we will have to etimate it uig the two ample variace Y Sice the ad the are aumed to be from populatio that have the ame variace, ad are coidered to be two etimate of the ame variace. Therefore, it make ee to defie the variace that we eed for the t tatitic a a average of the two ample variace. But ad do t have to be equal. Sice we would epect the larger ample to provide a omewhat better etimate of the commo variace, we take the ample ize ito coideratio ad defie a average weighted b the ample ize o the larger ample ha more weight. The variace obtaied uig the weighted average of the idividual variace i called Page -3-

6 the pooled variace, deoted p ad i defied a below. p ( 1) ( 1) ( 1) ( 1) ( 1) ( 1) Note that 1 ad 1 are ot arbitraril choe but are related to degree of freedom. Recall that the defiitio of i ( i ) i 1 1 which mea that ( ) ( ) 1 i i 1. Thi mea that we have the um of the um of quare for the ad ( ) ( ) i i 1 j 1 j o the top ad the um of the degree of freedom for the ad the o the bottom. The pooled tadard deviatio i the quare root of ad i defied a below. p, o i deoted p p ( 1) ( 1) Page -4-

7 p Note i the pecial cae that (let call thi commo value ), become p ( 1) ( 1) ( 1) ( 1) ( 1)( ) ( 1) So whe the ample are the ame ize, the pooled variace i the imple average of the two variace. Now let go back to the uequal ample ize cae. The t below i the tet tatitic ued to tet the ull v the alterative hpothee H0: μ μ or μ μ 0 HA: μ μ or μ μ 0 veru. whe (1) the ample are idepedet ad () the variace ca be aumed to be equal. t p p p 1 1 with - degree of freedom. Sice t tatitic are of the form, t mea of ome ort tadard error of the mea the SE (tadard error) for the differece i 1 1 p We kow that for two ample (the ame a for oe ample) we ca either do a Page -5-

8 hpothei tet uig the t tatitic ad obtai a p-value or we ca obtai a cofidece iterval. The cofidece iterval for the two idepedet ample, equal populatio variace cae i the give b ( ) t, ( / ) p,( ) t, ( / ) 1 α 1 α p Title Sta [R] ttet -- Mea compario tet Group mea compario tet ttet varame [if] [i], b(groupvar) [optio] optio decriptio Mai * b(groupvar) variable defiig the group uequal upaired data have uequal variace welch ue Welch' approimatio level(#) et cofidece level; default i level(95) * b(groupvar) i required. Note below that the equal variace verio of the tet i the default.. ttet chol_bl,b(black) Two-ample t tet with equal variace Group Ob Mea Std. Err. Std. Dev. [95% Cof. Iterval] No-Blac Black combied diff diff mea(no-blac) - mea(black) t Ho: diff 0 degree of freedom 1071 Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) 0.01 Pr(T > t) Page -6-

9 How do we iterpret thi pritout. Firt let u recall that the ull ad alterative hpothee are H0: μ μ or μ μ 0 HA: μ μ or μ μ 0 Let u aume that the α-level i veru. Becaue we ve aumed a two-tailed tet the reult we wat are i the middle. We have t , p 0.01 ad df 1071 (i.e. NoBK BK ). The p-value (0.01 which we will report a 0.01) i maller tha α (0.05), o ou reject the ull hpothei. Had we choe α to be 0.01, the p (0.01) would be bigger tha α o we would fail to reject the ull hpothei (alo kow a accept the ull hpothei). What I do i the et everal page i jut to how ou how the formula we got above relate to the Stata output. Let u aume the go with the No-Black ad the go with the Black. No-Black i repreeted b o o (7.0853) Black i repreeted b o o ( ) ad Page -7-

10 ( ) ( ) p p SE p Therefore the t tatitic with degree of freedom i t p So we have the t tatitic ad the degree of freedom. All we have to do ow i verif the p-value. If we draw the area for p give a two-tailed tet ad t , we would get the graph o the et page. The fuctio we have available to calculate the p-value i the fuctio ttail. Thi fuctio give ou the area to the right of the cutoff, i thi cae But the area to the right of (ee the graph o the et page) i ot what we wat. We have two choice. We ca ubtitute.5066 for (becaue of the mmetr of the t ditributio) ad obtai the area to the right of The we multiple b two becaue that i ol half of p. Page -8-

11 So the followig give u the p-value.. di *ttail(1071,.5066) The other choice i to ue ttail(1071, ) ad recogize that the complemet to thi area i half of the value of p (ee graph o et page).. di *(1 - ttail(1071,-.5066)) t ditributio with df 1071 Black/No-Black compario.3 ivttail(1071, ) light area to right of dark area (1/)p t How do we obtai the cutoff poit for the area α 0.05 (i.e o either ed) for a two-tailed tet? We wat thi cutoff becaue it i the t-value i the cofidece iterval (i.e. t 1, ( α / ) ). I thi cae we kow the ize of the area (0.05) ad the degree of freedom ad we wat to kow the cutoff. The ivttail allow u to plug i the area ad degree of freedom ad get the umber that cut off that area. We eed half of the area i the upper tail ad half i the lower tail o I put 0.05 (half of 0.05) i the fuctio. Page -9-

12 . di ivttail(1071,0.05) Thi awer hould t be a urprie ice with 1071 degree of freedom thi t ditributio hould be ver cloe to a N(0,1). Note that the cutoff for the other tail i therefore So the olid area i the graph below i α 0.05 with 0.05 i each tail. The two tailed p-value that goe with the cutoff i (i.e i twice the area to the left of ) a we ve dicued earlier. Notice that the area for the p-value fit withi the area for α 0.05 (i.e. the p-value i le tha 0.05 o we reject the ull hpothei). Or aother wa of aig thi i that the t -.5 fit i the rejectio regio..4.4 t ditributio with df 1071 t ditributio with df 1071 Black/No-Black compario Black/No-Black compario ttail(1071, ) ivttail(1071,0.05) ivttail(1071,0.05) Sice df 1071, Sice df 1071, the cutoff for a i the cutoff for a i Blue area (1/)p t t The 100*(1 - α)% cofidece iterval for the differece i give b the equatio below ( ) t, ( / ) p,( ) t, ( / ) 1 α 1 α p Page -10-

13 For α 0.05 we foud above that t 1071, SE p So the 95% CI for (which i the mea for the No-Black miu the mea for the Black) i ( ( ), ( )) (-.44, -0.30). di (1.96*0.545) di (1.96*0.545) Zero i ot i the CI o we reject the ull hpothei at α 0.05 (we are lookig for zero i the iterval becaue μ μ 0 implie μ μ ) i favor of the alterative μ μ. Whe we have the p-value, we ca chage the α - level ad till kow whether to reject or ot without makig a more computatio. Of coure, ou hould t be chagig the α - level after ou ve aalzed the data, but a reader of our paper who did t agree with our choice of α - level could eail compare the p-value to hi/her preferred α - level. Such a reader would have to recompute the cofidece iterval to be able to ue a differet α - level (paper rarel provide eough iformatio for a reader to be able to do thi). O the other had, oce we ve calculated the cofidece iterval we ca ee how far apart the mea ca be ad till be coidered equal. Page -11-

14 Now the quetio i - what do I epect ou to do? Certail ot all the tuff above. 1) What i the quetio? We wih to kow if the Black ad No-Black have imilar baelie value for total choleterol. ) Do we ue a two-tailed tet or a oe-tailed tet? Remember we go with a two-tailed tet ule we have prior evidece of a particular directio (i.e % of time we ll ue two-tailed tet). 3) So what i our formal tatemet of the ull ad alterative hpothee. H 0 : μ (No-Black) μ (Black) or μ - μ 0 veru H A : μ μ or μ - μ 0 4) What are we goig to elect for the α - level? We make the choice here (i.e. before we look at the data). A commo choice i I m goig to pick α 0.01 o we ca ee how a differet choice of α-level ca impact the reult. I the commad below otice that I have added level (99) to the commad I ra earlier. Thi i becaue the default i a 95% cofidece iterval (i.e. α 0.05) but we wat a 99% cofidece iterval to go with α ) I thi a paired or upaired problem? Upaired. 6) How do we kow a t-tet i the appropriate choice? Firt, baelie choleterol i a cotiuou variable ad the ample ize i large o the ditributio of mea (amplig ditributio) i likel to be ormall ditributed. Secodl, we are comparig two group (Black ad No-Black) with repect to mea. 7) How do we kow we hould ue the equal variace verio of the t-tet? We do t et but we will. Page -1-

15 . ttet chol_bl,b(black) level(99) Two-ample t tet with equal variace Group Ob Mea Std. Err. Std. Dev. [99% Cof. Iterval] No-Blac Black combied diff diff mea(no-blac) - mea(black) t Ho: diff 0 degree of freedom 1071 Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) 0.01 Pr(T > t) Thi pritout i the ame oe ou got before ecept for the chage i the cofidece iterval. A chage i α - level doe t chage the value of the t tatitic or p-value. It ma chage our reult (reject v fail to reject). Becaue of the wa we have tated the problem we will coider ol the two-tailed reult. The two-tailed tet i the ceter oe with t ad p The value of p (0.01) i larger tha the value of α (0.01) o we fail to reject the ull hpothei (for α 0.05 we would reject the ull hpothei for the two-tailed tet). I would report the reult a: The aali ued wa a two-tailed t-tet ad the reult were t -.51, df 1071 ad p We therefore cocluded that TC wa the ame for the Black ad No-Black give α OR The 99% cofidece iterval i alo a two-tailed tet at the α 0.01 level. The 99% cofidece iterval for the differece i (-.771, 0.038). If zero i i the iterval, we fail to reject the ull hpothei. If zero i ot i the iterval, we reject the ull hpothei. Zero i i the iterval o we fail to reject the ull hpothei. Thi i the ame awer we got with the p-value (if ou get a differet awer, ou d better look cloel at what ou ve doe becaue ou hould get the ame awer). Wh are we cocered about whether or ot zero i i the iterval? Becaue μ μ 0 i equivalet to μ μ. The lat thig ou ll eed to do with thi problem i to make ure ou have appropriatel ued a tet aumig equal variace. Sta tued to thi tatio for further detail. Page -13-

16 Aide for oe-tailed tet: Notice that had we aumed a oe-tailed lower tailed tet H 0 : μ (No-Black) μ (Black) or μ - μ 0 veru H A : μ < μ or μ - μ < 0 we would have rejected the ull hpothei becaue α 0.01 i larger tha p ad cocluded that the TC of the No-Black wa le tha the TC of the Black. Had we aumed a oe-tailed upper tailed tet H 0 : μ (No-Black) μ (Black) or μ - μ 0 veru H A : μ > μ or μ - μ > 0 we would have failed to reject the ull hpothei becaue α 0.01 i maller tha ad cocluded that the TC for Black ad No-Black wa the ame. p If ou ue a oe-tailed tet ou have to be ure ou kow which tail to pick prior to ou aali. Now back to how to decide o equal variace veru uequal variace. Old Fahioed Method The Rule of Thumb i a follow: If the two ample each have the ame ample ize, uig the equal variace verio of the t-tet i appropriate regardle of what the two variace look like. If the ample ize are ot equal ad 1) the ample with the maller ample ize ha the larger variace ad ) that variace i more tha twice the ize of the variace of the larger ample, the ue the uequal tet. Otherwie, ue the equal variace verio of the t-tet. Thi mea the equal variace verio i what i ued mot of the time. For our particular problem the maller ample (Black with ample ize 388) ha the maller variace ( ). So the equal variace verio i the appropriate oe to ue. More Moder Method - Eve eaier tha the rule of thumb. Ru both the equal variace ad the uequal variace verio of the t-tet. If ou reject the ull hpothei with both verio or ou accept the ull hpothei with both verio, report the equal variace verio otherwie, report the uequal variace Page -14-

17 verio. Equal variace verio of the t-tet for two idepedet ample:. ttet chol_bl,b(black) level(99) Two-ample t tet with equal variace Group Ob Mea Std. Err. Std. Dev. [99% Cof. Iterval] No-Blac Black combied diff diff mea(no-blac) - mea(black) t Ho: diff 0 degree of freedom 1071 Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) 0.01 Pr(T > t) Uequal variace verio of the t-tet for two idepedet ample:. ttet chol_bl,b(black) level(99) uequal Two-ample t tet with uequal variace Group Ob Mea Std. Err. Std. Dev. [99% Cof. Iterval] No-Blac Black combied diff diff mea(no-blac) - mea(black) t Ho: diff 0 Satterthwaite' degree of freedom Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) Pr(T > t) Uig the two-tailed tet ou would fail to reject the ull hpothei (aumig α 0.01) ad reject the ull hpothei (aumig α 0.05) with both the equal ad uequal variace verio o report the equal variace reult. Notice that whe we are coiderig a -level of 0.01, we requet the 99% cofidece iterval. Whe coiderig a α -level of 0.05, we requet the 95% cofidece iterval. Thi would be the ed of the problem (FINALLY). α Page -15-

18 A commo micoceptio regardig hpothei tetig ad cofidece iterval i the two idepedet group tet: You will otice that i the output for the t-tet of Black veru o-black with repect to total choleterol, ou get ot ol the cofidece iterval for the differece of the two group mea but alo a cofidece iterval for the mea choleterol i the Black ad for the mea choleterol i the o-black. You will frequetl ee i the literature ivetigator claimig that ice the cofidece iterval for the mea of the two idepedet group overlap, the two group mea are ot tatiticall differet. Thi i ot correct. There ca be ubtatial overlap of the two cofidece iterval ad et the tatitic (mea i thi cae) ca till be tatiticall differet. Let u uppoe that we have two idepedet ample ad the mea of oe i 10 (Group 1) ad the mea of the other i (Group ). Let u further uppoe that the tadard deviatio are the ame for both group ad the ample ize are alo the ame for both group. Thi mea the tadard error are the ame for both group SE SD ice. Let u aume that the value for the commo SE i 4. Our hpothee are the: H 0 : 1 veru H A : 1 μ μ μ μ Let u further aume that the ample ize are large eough that we are comfortable uig the ormal ditributio to obtai our cofidece iterval. If we are uig 95% cofidece iterval, thi mea the critical poit (or cutoff poit) i So our cofidece iterval are: ( (4), (4)) (., 17.8) ad ( (4), 1.96(4)) (14., 9.8). So we have coiderable overlap betwee the two cofidece iterval. However, uig the hpothei tetig approach, we get Page -16-

19 z Sice.1 > 1.96, the reult i clearl igificat at the α 0.05 level. Or uig the cofidece iterval for the differece we get ( - 10) 1.96 (0.9, 3.1). ± 4 4 Sice 0 i ot i the iterval for the differece, we reject the ull hpothei i favor of the alterative. Although there i coiderable overlap, the mea of the two group are tatiticall differet. So for the two idepedet group problem ou eed to coider the cofidece iterval for the differece or the hpothei tet but ot the relatiohip of the cofidece iterval of the idividual mea. The upaired t-tet with uequal variace: Small Stud: The followig data repreet total hitidie ecretio i mg from 4-hour urie ample for 5 me ad 10 wome.. lit id e hitidie,oob id e hitid~e Me 9 Me 36 3 Me Me 17 5 Me Wome Wome 4 8 Wome Wome Wome Wome Wome Wome Wome Wome 138 Page -17-

20 The quetio i - Do the total hitidie level of me differ from thoe of wome? Sice the ample of me i idepedet of the ample of wome, we ll ue the upaired t-tet. Sice we kow othig about the data at thi poit, we ll ue the t-tet with equal variace.. ttet hit,b(e) Two-ample t tet with equal variace Group Ob Mea Std. Err. Std. Dev. [95% Cof. Iterval] Me Wome combied diff diff mea(me) - mea(wome) t Ho: diff 0 degree of freedom 13 Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) Pr(T > t) Note that the maller ample (i.e. the me) ha a variace ( ) that i more tha twice that of the larger ample ( ). So let u look at the t-tet for uequal variace.. ttet hit,b(e) uequal Two-ample t tet with uequal variace Group Ob Mea Std. Err. Std. Dev. [95% Cof. Iterval] Me Wome combied diff diff mea(me) - mea(wome) t.5667 Ho: diff 0 Satterthwaite' degree of freedom Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) Pr(T > t) Note that we would ot reject the ull hpothei at a α-level of 0.05 uig the uequal variace t-tet but would reject for the equal variace t-tet. So here we would report the uequal variace tet becaue there i a mimatch betwee to two aale. Page -18-

21 So what i the etup for the upaired t-tet with uequal variace: X, X,..., X 1) Thi time we tart with two idepedet radom ample ad YY,,..., Y 1 1 (ote that ad do t have to be equal). We have μ ad where σ Y ~ N( μ, σ ) X ~ N(, ) σ σ ) So ow we have two ample mea ( ad ) ad two ample tadard deviatio ( ad ). X Y 3) Sice the ad the are aumed to have differet variace, it o loger make ee to ue a pooled etimate. X 4) So i aumed to be ormall ditributed with mea ad variace σ σ We get the above for the variace becaue the ample are idepedet o Y μ μ Var( X Y ) Var( X ) Var( Y ) μ μ μ μ 0 5) The ull hpothei i or. 6) Now we get t The problem here i the degree of freedom that will ue for the t ditributio i ot o clear. Page -19-

22 7) Satterthwaite give u a approimatio for the degree of freedom that give u a appropriate tpe I error. Ufortuatel the proof that thi make ee i beod the cope of thi book o I ll jut preet the reult. We ll let d' be the degree of freedom. The d' ( 1) 1 ( ) d' Notice that will ot alwa be a iteger. Roer ugget that ou roud dow to the earet iteger, but ewer computig techique allow ou to ue a o-iteger form of d' (i.e. Stata ivttail ca be computed for d' 7.6 for eample). 8) The cofidece iterval i the give b ( ) td ', 1 ( α/ ),( ) td ', 1 ( α/ ) Notice that I have ued d ' rather tha Roer d''. Page -0-

23 9) The Welch degree of freedom are give b The Welch formula for the degree of freedom i le commol ued tha the Satterthwaite. ( ) df 1 1 ( ) Page -1-

24 . ttet hit, b(e) uequal welch Two-ample t tet with uequal variace Group Ob Mea Std. Err. Std. Dev. [95% Cof. Iterval] Me Wome combied diff diff mea(me) - mea(wome) t.5667 Ho: diff 0 Welch' degree of freedom Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) Pr(T > t) ttet hit, b(e) uequal Two-ample t tet with uequal variace Group Ob Mea Std. Err. Std. Dev. [95% Cof. Iterval] Me Wome combied diff diff mea(me) - mea(wome) t.5667 Ho: diff 0 Satterthwaite' degree of freedom Ha: diff < 0 Ha: diff! 0 Ha: diff > 0 Pr(T < t) Pr( T > t ) Pr(T > t) Notice that the differece i the degree of freedom mea that the cofidece iterval for the differece i differet uig the Welch tha uig the Satterthwaite. The p-value are alo differet. Otherwie the two pritout are the ame. Page --

25 Oe ample Two ample Depedet d i i i Two ample Idepedet Equal variace Two ample Idepedet Uequal variace H H H : μ μ 0 0 H A : μ μ0 H H H H : μ 0 0 d A : μ 0 d : μ μ 0 0 : μ μ 0 A : μ μ 0 0 : μ μ 0 A t μ0 t t t d d 0 ( ) p ( ) 0 Oe ample t μ0 df 1 Two ample Depedet d i i i Two ample Idepedet Equal variace Two ample Idepedet Uequal variace t t t d d 0 ( ) p ( ) 0 df df 1 Satterthwaite Welch Page -3-

26 H 0 :μ SBP 145 Oe ample ttet bp_bl 145 Two ample Depedet di i i Two ample Idepedet Equal variace Two ample Idepedet Uequal variace H 0 : μ SBP BL μ _ SBP _ 1r H 0 :μ Bk μ NoBk H 0 :μ Bk μ NoBk ttet bp_bl bp_1 Paired i the default ttet bp_bl,b(black) Equal variace i the default ttet bp_bl, b(black) uequal Satterthwaite i default Page -4-