Statitical Iferece Procedure Cofidece Iterval Hypothei Tet Statitical iferece produce awer to pecific quetio about the populatio of iteret baed o the iformatio i a ample. Iferece procedure mut iclude a tatemet of how cofidet we ca be that the awer i correct. Thi preetatio coider oly cofidece iterval for µ. The bai for thi importat topic i the amplig ditributio of. X Cofidece Iterval A cofidece iterval i a formula that tell u how to ue ample data to etimate a populatio parameter. The cofidece level idicate the percetage of iterval cotructed i thi maer which cotai the value of the parameter. Cofidece Iterval for µ Sice X i a ubiaed etimator of the populatio mea, µ, we ue X to etimate µ. The amplig ditributio of X, provide the tool we eed to build a cofidece iterval for µ. A 100(1 α)% Cofidece Iterval for µ (1 α) i the cofidece coefficiet. It i the probability that the iterval etimator ecloe the populatio parameter. If the cofidece coefficiet i (1 α), the 100(1 α)% i the cofidece level. The cofidece level i the percetage of the iterval cotructed by the formula will cotai the true value of µ. 1
A 100(1 α)% Cofidece Iterval for µ Aume the populatio tadard deviatio i kow. Aume that X i ormally ditributed with mea, µ, ad tadard deviatio,. For ample of ize, i ormally ditributed with mea, µ, ad tadard deviatio,. X Let Ue What We Kow P µ z < x < µ + z = P < z ( z < z ) α/ What i z α/? z α/ i the value of z that awer the quetio what i the value of z uch that 100(1 α)% of the value of z lie betwee z α/ ad z α/. If 100(1-α) = 95, the P(- z α/ < z < z α/ ) =.95. Therefore, z α/ = 1.96. What i the formula for a 100(1-α)% cofidece iterval for µ? Uig algebra which you are ot repoible for doig, I ca take the expreio below ad chage it from a probability tatemet about X to oe about µ. P µ z P X z < X < µ + z < µ < X + z = I thi expreio we are talkig about the variable X, ot a oberved value of the variable. x z The formula for a C% cofidece iterval for µ i < µ < x + z α Whe we ubtitute the value of the ample mea for a particular ample, we ca o loger talk about the probability. Itead we ay that we are 100(1 α)% ure that the true mea µ lie betwee the two value we obtaied by uig the ample mea. A Example We wat to etimate the mea time required to perform a tak with 95% cofidece. We elect a radom ample of ubject ad fid that x = 43 ecod. Aume that = ecod. We mut alo aume that the time required to perform thi tak i ormally ditributed. Give a 95% Cofidece Iterval for µ x zα/ < µ < x + zα 43 1.96 < µ < 43 + 1.96 4.0 < µ < 43.98 We are 95% cofidet that the mea time required to perform thi tak i betwee 4.0 ad 43.98 ecod.
Thi Formula i uually writte x ± z Thi formula i i the form we ue for may cofidece iterval, etimate margi of error. Give a 95% CI for µ x ± zα 43± 1.96 43±.98 4.0 < µ < 43.98 Thi i the way I wat you to how your work. Iterpret the Cofidece Iterval We are 95% cofidet that the mea time to perform thi tak i betwee 4.0 ad 43.98 ecod Propertie of the Margi of Error All other thig beig equal, the margi of error of a cofidece iterval decreae a the cofidece level 100(1-α) decreae. the ample ize icreae. the populatio tadard deviatio decreae. What happe whe we do ot kow the populatio tadard deviatio? We ue the ample tadard deviatio a a ubtitute for. I thi cae the formula to be ued deped o the ample ize,. If < 30, we eed to kow the propertie of a ew ample tatitic called t. x µ t = Compare the z-core for X the t-core for X. x µ z = t = x µ to 3
The t Ditributio The t ditributio are a family of ditributio oe for each umber of degree of freedom. We ue df to deote the umber of degree of freedom. Each t ditributio i ymmetric about 0. A table givig value of t α for variou probabilitie ad umber of degree of freedom i give i the frot cover of your book. The table etrie are t α uch that P(t > t α ) = α. Example: If α =.0 ad df = 14, t α =.145. The formula for a 100(1 α)% cofidece iterval for µ whe i ot kow ad < 30 i x t < µ df < x + t = 1 α / Thi Formula i uually writte x ± tα, df = 1 Thi formula i i the form we ue for may cofidece iterval, etimate margi of error. A Example uig t We wat to etimate the mea time required to perform a tak with 95% cofidece. We elect a SRS of ubject ad fid that X = 43 ecod ad = ecod. We mut aume that the time required to perform thi tak i ormally ditributed. Give a 95% CI for µ t x tα / < µ < x + tα 43.064 < µ < 43 +.064 41.968 < µ < 44.03 We are 95% cofidet that the mea time required to perform thi tak i betwee 41.968 ad 44.03 ecod. Give a 95% CI for µ x ± tα, df = 1= 4 43 ±.064 43± 1.03 41.968 < µ < 44.03 Thi i the way I wat you to how your work. 4
What happe whe X i ot ormally ditributed? The Cetral Limit Theorem provide a guide. If i large eough the amplig ditributio of X i approximately ormally ditributed. Remember, we claim that i large eough if i at leat 30. If we kew the we could ue the x ± z formula. What if we do ot kow ad i at leat 30? We imply ue the formula x ± zα/ ad ubtitute for. Summary There are formula ued to create a cofidece iterval for µ. How do we chooe which formula to ue? We ak the followig quetio: I X ormally ditributed? I kow? What i? If X i ormally ditributed ad i kow, the for every ue x ± z If X i ormally ditributed ad i ot kow, the for < 30 x ± t, df = 1 a 5
If X i ormally ditributed ad i ot kow, the for greater tha or equal to 30 ue x ± z If X i ot ormally ditributed ad i ot kow, the for greater tha or equal to 30 ue x ± z If X i ot ormally ditributed ad i ot kow, the for le tha 30 either of thee formula ca be ued. 6