Estimating m when stand deviation is unknown

Estimating m when stand deviation is unknown - STS 2610...

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Unformatted text preview: STS 2610: Quantitative Methods Spring 2014 man“ Id MH’ Estimating [,1 en 6 is Unknown If the variable x is N01, 0'), then J? is N(,u,0‘/‘/;) and is N(0, l) . This is why we use 2 critical values in the formulas for the 2- §_# a/J; If a is Unknown: error margin and confidence interval. pllrli (filma’cr o! 0" , the sample standard deviation. fie. - :y 1-] Must estimate from data using 3 = l n — Problem: if the variable x is N(/.1,0'), then 3 is N(,u,a/~/;) but x _ ,u does not follow a N(0,l) distribution. 2: s/JI—i z cricital values should not be interval. In fact, the correct confidence level will be I used. They will not give use the correct confidence level for our ower than what we are stating. Solution: if the variable X is N(/1,a), then is Now/Jr?) but x - follows a t distribution with n - 1 degrees of freedom. (in) p 5/5 Facts About t Distributions samplrsm - I (”3 tribution with n — 1 degrees of freedom I: If the variable T follows a t dis 0 E(T)=O o The density curve is symmetric. o Varl‘l') > VarlZ) = 1 (Anyt density curve is flatter than the 2 density curve.) As n—boo the t density curves look more and more like the 2 density curve. t an" df= 0-1 NW) owl El Student’ 5 t Distribution i Notortmzasn’mcreases Huntsman: are belt. . antisymmetric. but have law toils that the normal > Estimatingthe Mean it of a Normal Population Distribution (0 Untmwn HOG-am Enos Massin: C 3 t‘lz. 5‘:- l00£l - afitonfldence interval: ‘ + t .5... d‘$® x ‘ é; ‘5 ,e = ‘15 wugmeTableoftCriticalValues ’ l” .65 c"Ia ‘ -0&5 - 0‘ : 8 0‘ t . 63 ‘ I ' f/a : u 0‘ mm‘. WhentoUsetProceduresnPractice {”m‘” “(n namatm . n < :5 Use t procedures only if population distribution has bell shape /""‘"""" 0 I5 5 n < 30 Use t procedures if the data are roughty symmetric with no outliers. o 30 s n < 50 Use t procedures unless data are highly skewed 0: contain outliers. O n > 50 Use t procedures without resetvation. . ...
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