If σis unknown. Properties of t distribution. 6.3 One and Two Sample Inferences for Means. What is the correct multiplier? t

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1 /8/ Oe a Tw Samle Iferece fr Mea If i kw a 95% Cfiece Iterval i 96 ± ± But i ever kw. If i ukw Etimate by amle taar eviati The etimate taar errr f the mea will be / Uig the etimate taar errr we have a cfiece iterval f ± ( ) The multilier ee t be bigger tha Z (e.g.,.96 fr 95% CI). The cfiece iterval ee t be wier t take it accut the ae ucertaity i uig t etimate. The crrect multilier were figure ut by a Guie Brewery wrker. What i the crrect multilier? t 00( α)% cfiece iterval whe i ukw ± t ( / ) 95% CI 00( 0.05)% cfiece iterval whe i ukw ± t ( / ) Prertie f t itributi The value f t ee hw much ifrmati we have abut. The amut f ifrmati we have abut ee the amle ize. The ifrmati i egree f freem a fr a amle frm e rmal ulati thi will be f. t curve a z curve Quatile f t itributi t table i give i the bk Table B.4 Bth the taar rmal curve N(0,) (the z itributi), a all t(v) itributi are eity curve, ymmetric abut a mea f 0, but t itributi have mre rbability i the tail. A the amle ize icreae, thi ecreae a the t itributi mre clely arimate the z itributi. By 000 they are virtually iitiguihable frm e ather. It ee the egree f freem a well f rbability t

2 /8/009 Cfiece iterval fr the mea whe i ukw t < µ < t Eamle Nie level, Pit etimate fr the average ie level f vacuum cleaer;. 95% Cfiece iterval Sluti, Critical value (0.975 t quatile) with f t % CI ± ± < µ < 77.8 Eamle 8 (age 366) Failure time f 0 rig. The rmal lt lk fairly traight. (If t, try trafrmig r a ifferet itributi, e.g. Weibull) % CI 68.3 ±.833* ± t 87.5 If we were t tet H µ 50 v Ha µ 50, we wul t reject H0, ice 50 i i the cfiece iterval fr µ. T tet H µ # Aalgu t the large amle tet with z tet tatitic # z We wul have T # Determiati f reject / t reject H a well a -value are fu ue T-table with ν f We cul the tet uig t <.74 <.833 Q(0.9) < tet tatitic < Q(.95) 0.05 < Pt ( >.74) < 0. value P( t >.74)* 0.< value < 0. But the cfiece iterval i mre ifrmative.

3 /8/009 O the ther ha Ather meth Rejecti Regi t 0.95,9.833, t 0.05,9.83 If we have a tet tatitic value that i either t mall (<.83) r t big (>.83), the we have trg eviece agait H 0. t.74 which i t t mall r t big (cmare t the cutff value abve/ critical value ), the we cat reject the ull. Alterative Hythee Rejecti Regi µ> µ 0 µ< µ 0 µ µ 0 z>z -α t>t -α z<z α t<t α z>z -α/ r z<z α/ t>t -α/ r t<t α/ Rejecti Regi meth a value meth Paire Data Fr Ha µ<µ 0, if z tet tatitic i le tha.645, the the value i le tha Cmarig the value t 0.05 i the ame a cmarig the z value t.645. Fr t tet we ca al fi me critical value crreig t level f α that we ca cmare t ur tet tatitic. Tet tatitic i the rejecti regi i the ame a value i le tha α. 98 tuy f trace metal i Suth Iia River. 6 ram lcati 3 Tt water zic ccetrati (mg/l) Bbttm water zic (mg/l) T Bttm Paire Mea Differece T cmare T & Bttm Water Zic frm a River Lcati Bttm T That i equivalet t ak i it true that µ ifferece>0? Thi i a ecial cae f the mea f a igle clum f umber. Create a ew clum fr the ifferece betwee variable. T & Bttm Water Zic frm a River Lcati Bttm T Differece

4 /8/009 Check rmality Orere i Z Quatile al Staar Nrma Quatile Zic i River Zic Serie 6 i value f % Cfiece Iterval t ±.57(0.05) ± t 0.56 Nte Eve rmal lt frm ram rmal ata are t erfectly traight By the uual hythei tetig erective, the -value fr H µ 0 v Ha µ 0i le tha 0.05, ice µ 0 i t i a 95% cfiece iterval. Our reult wul be tatitically igificat eviece agait H µ t t >.57 *0.005 < < * < < 0.0 I hythei tetig te that the -value mut be le tha α t claim tatitical igificace. A tet i igificat at the α level if the H value i t i the 00(-α )% cfiece iterval. What abut α 0.0 level? -value > 0.0 D t reject H ± 4.03(0.05) 0.09 ± t ± 0.00 Nt tatitically igificat at α 0.0level Aumti The ulati f ifferece fllw a rmal itributi. A rmal lt f ifferece,, hul be fairly traight. Nte We t ee B r T t be rmal. Rejecti regi eercie Tell whe t reject H 0 μ 0 uig a t tet. Awer wul be f the frm reject H 0 whe t <.746 r maybe reject H 0 whe t >.746 r maybe reject H 0 whe t >.746 (a) H A μ < 0, α 0.05, 0 (b) H A μ > 0, α 0.0, 8 (c) H A μ 0, α 0.0, 9

5 /8/009 Fi value eercie Awer wul be f the frm 0.0 < < 0.05 < 0.00 > 0.8 After fiig the -value i each cae, tell whether t reject r t reject H 0 at the α level. (a) H A µ > 0, 7, t Large Samle Cmari f Tw Mea Glue µ, Glue µ, Bth ulati rmal ieeet value fr glue Nt aire, blcke, ieeet value fr glue (b) H A µ < 0, 7, t -.58 (c) H A µ 0, 7, t -.58 i ur gue at µ µ µ? Hw much might eviate frm Var( ) Var( ) ( ) Var( ) µ Eerimet Mea µ µ µ µ Variace A cfiece iterval fr µ µ i give by z z ± ± ( µ µ ) z A fr hythei tetig But we ever really kw Small Samle Cmari f Tw Mea Cae Bth a are etimatr f. The cfiece iterval ee t be wiee t accut fr aitial ucertaity i a a etimatr f a. Cae Aume equal variace. Cae D t aume variace are ecearily equal. But they may be. Pl a it a le, cmbie etimate f. weighte average f ( ) ( ( ) ( ) ) a, weight by f.

6 /8/009 ± t ± t ± t Fr the table value f t, ue f ( -) ( -) T tet H µ µ #, check if # i cfiece iterval r ue T ( ) # A cmare t T-table with f ( -) ( -). Cae D t aume Lifetime f Srig. Table 6.7 a Figure 6.5 t t ± r ± f T 4 4 ( ) ( ) ( ) # Nte With, ly f chage N rmal Quatile Srig Figure Lifetime 900 Stre 950 Stre Stre 950 Stre Nte Uually lifetime are mre lgrmal tha rmal. T fllw the bk eamle, carry i time cale. Cae Aume f (4.9) 9(33.) Var ( ) ±.0(7.) ± t 8.8 Bae the cfiece iterval, we reject H µ µ 0 v H µ µ 0 at α 0.05 level.

7 /8/009 Alteratively, t tet H µ µ 0 v H µ µ t.7 f < < < < 0.0 T tet H µ µ 0 v H µ µ 0, < < 0.0. > T tet H µ µ 0 v H µ 0 µ <, 0.99 < < Cae Nt aumig f Nte With, ly f chage 46.8 ±.0(7.) Wier CI