20. CONFIDENCE INTERVALS FOR THE MEAN, UNKNOWN VARIANCE

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1 20. CONFIDENCE INTERVALS FOR THE MEAN, UNKNOWN VARIANCE If the populatio tadard deviatio σ i ukow, a it uually will be i practice, we will have to etimate it by the ample tadard deviatio. Sice σ i ukow, we caot ue the cofidece iterval decribed previouly. The practical verio preeted here ue i place of σ. Eg 1: A radom ample of 8 Quarter Pouder yield a mea weight of x = 02. poud, with a ample tadard deviatio of = 0.07 poud. Cotruct a 95% CI for the ukow populatio mea weight for all Quarter Pouder. (Solutio o ext lide). Remember that i the quare root of the ample variace, 2 2 = 1 xi x x1 x 1 ( ) where are the data value i the ample. i= 1 There are two cae to coider: Large Sample ad Small Sample. If 30 (ad σ i ukow), jut replace σ by ad proceed a before. The CI i X ± z. Thi work becaue X µ i approximately tadard ormal, regardle of the populatio ditributio, whe 30. We ca thik of a the etimated tadard error of X.

2 Suppoe < 30 (ad σ i ukow). To get a valid CI for µ i thi cae, we mut aume that the populatio ditributio i ormal. Thi aumptio i hard to check, ad wa ot required before. The CI i X ± tα 2, where tα 2 i defied below. Do t forget what you leared earlier: If σ i kow, we ue σ X ± z, regardle of the ample ize. So the oly time whe you mut remember to ue t α/2 i whe σ i ukow ad < 30. Otherwie, ue z α/2. SUMMARY OF CONFIDENCE INTERVALS 30 < 30 Sigma Kow X ± z X ± z σ σ Sigma Ukow X ± z X ± t * Mut aume Normal populatio if < 30, σ ukow.

3 What i t α/2, ad why mut we ue it whe <30 ad σ i ukow? Whe < 30, the quatity t = X µ doe ot have a approximately tadard ormal ditributio, eve though we aume here that the populatio i ormal. Itead, t ha a Studet t ditributio with 1 degree of freedom. Thi ditributio wa iveted by W.S. Goet (1908), who wa workig for Guie Brewerie, ad wrote uder the pe ame of Studet. There i a differet t ditributio for each value of the degree of freedom, df. Thee ditributio are ot ormal, although they are ymmetrical aroud zero, ad moud haped.

4 The quatity t α/2 deote the t-value uch that the area to it right uder the Studet t ditributio (with df = 1) i α/2. Note that we ue 1 df, eve though the ample ize i. Value of t α are lited i Table VI, Appedix B. I the table, the degree of freedom are deoted by ν. Eg 1, Solutio: We have = 8, x = 0.2, = 0.07, α = 0.05, df = 1 = 7. From Table VI, t α/2 = t = Therefore, the cofidece iterval i x± t = 0.2± =0.2± 0.059=(0.141,0.259). 8 It iteretig that thi iterval cotai 0.25, o it eem plauible that the burger are really quarter pouder (o average).

5 Relatiohip Betwee t ad z Whe df i mall (the guidelie we're uig i df < 29), the t ditributio ha loger tail (i.e., cotai more outlier) tha the ormal ditributio, ad it i importat to ue the t-value of Table VI, aumig that σ i ukow. Due to the log tail i the t ditributio, t α/2 i larger tha z α/2, o the cofidece iterval baed o t will be wider tha the (icorrect) oe baed o z. So whe df i mall, the t-baed cofidece iterval correctly reflect the added ucertaity due to ot kowig σ, which mut be etimated. A df get large, however, the t ditributio approache the tadard ormal ditributio. Therefore, if i large eough, t α/2 will be extremely cloe to z α/2, ad the two cofidece iterval x ± z ad x ± tα 2 will be virtually idetical, o it will ot matter which oe we ue. But how large i large? The Traditioal Approach (i keepig with the text): Ay value of df 29 i coidered ifiite. We ca the ue the row of Table VI, which give z α (that i, t α with df = ). Miitab Approach: The df i ever truly ifiite, o Miitab alway ue t α/2 i computig cofidece iterval. (Miitab ha it ow iteral table to calculate t α/2 for ay df).

6 To be coitet with the textbook, ad becaue our table doe ot have litig for all value of df, we will ue the traditioal approach wheever we calculate cofidece iterval by had. Miitab approach i actually the correct oe, however, ad i to be preferred wheever t α/2 ad z α/2 differ appreciably. Eg 2: Miitab Decriptive Statitic for Chip Ahoy! data: Variable N Mea Media TrMea StDev SE Mea Chip Let cotruct a 99% cofidece iterval for µ, the mea umber of chip per bag. The Traditioal Approach give x ± z = ± (117.6) / 42 = (1214.9,1308.3) Miitab 1-Sample t output give a wider iterval, ice it ue the exact value, t.005 = for df = 41: T Cofidece Iterval Variable N Mea StDev SE Mea 99.0 % CI Chip ( , ).

7 I the Samplig Lab, each of you cotructed your ow mallample cofidece iterval for the mea. Sice we kow the mea for our populatio, we ca ee what percetage of the iterval actually worked. [Samplig Lab Reult]. Eg 3: The Uited State Che Federatio (USCF) ha a ratig ytem, which they ue to predict the expected outcome of touramet game. Each player ha a ratig betwee 0 ad 3,000 which chage over time. The larger the differece i ratig betwee two player, the greater the probability that the higher rated player will wi. A che game i cored 0 for a lo, 1/2 for a draw, ad 1 for a wi. Figure 3 ue cofidece iterval to how that the predicted wiig expectecy baed o the differece i USCF ratig i quite far from the actual average outcome. I other word, the USCF ratig do t work! Chace, Vol 12, No. 2, 1999.

LP10 INFERENTIAL STATISTICS - Confidence intervals.

LP10 INFERENTIAL STATISTICS - Cofidece iterval. Objective: - how to determie the cofidece iterval for the mea of a ample - Determiig Sample Size for a Specified Width Cofidece Iterval Theoretical coideratio