Topic 6 Estimation. Correction Pages

Transcription

1 Topic 6 Estimatio Correctio Pages

2 Page 0 Corrected Recall that the basic structure of the required cofidece iterval is poit estimate + safety et: Lower limit = ( poit estimate ) - ( multiple ) (SE of poit estimate) Upper limit = ( poit estimate ) + ( multiple ) (SE of poit estimate) Poit Estimate of µ is the Sample Mea X =30 X =30 Xi = i= 1 = 051. =30 The Stadard Error of X is σ SE(X σ 05. ) = variace(x ) = = = 30 =30 = The Cofidece Coefficiet For a 95% cofidece iterval, this umber will be the 97.5th percetile of the Normal (0,1) distributio. See the table o page 19 ad locate that value is Desired Cofidece Level Value of Cofidece Coefficiet Here 1.96 = (1-.05/)100th = 97.5th percetile of the Normal(0,1) distributio Puttig this all together Lower limit = ( poit estimate ) - ( multiple ) (SE of poit estimate) = ( 1.96 ) ( ) = 0.33 Upper limit = ( poit estimate ) + ( multiple ) (SE of poit estimate) = ( 1.96 ) ( ) = 0.69

3 Page 8 corrected 7. Normal: Cofidece Iterval for µ whe σ is Ukow Whe σ is ot kow, the computatio of a cofidece iterval for the mea µ is ot altered much. We simply replace the cofidece coefficiet from the N(0,1) with oe from the appropriate Studet s t-distributio (the oe with df = -1) We replace the (ow ukow) stadard error with its estimate. The latter looks early idetical except that it utilizes s i place of σ Recall Thus, s= ( Xi X) i=1 ( 1) Cofidece Iterval for µ i two settigs of a sample from a Normal Distributio σ is KNOWN σ is NOT Kow X ± (z 1 α/ )(σ/ ) X ± (t 1;1 α/ )(s/ )

4 Page 38 corrected 1. Obtai the poit estimate of σ. It is the sample variace S To get the sample variace S, we will eed to compute the sample mea first. X i i=1 X= 35.5 = ad ( ) Xi X i= 1 S = = Determie the correct chi square distributio to use. It has df = (4-1) = Obtai the correct multipliers. Because the desired cofidece level is 0.95, we set 0.95 = (1-α). Thus α =.05 For a 95% cofidece level, the percetiles we wat are (i) (σ/)100 th =.5 th percetile (ii) (1 α/)100 th = 97.5 th percetile From Roser Table 6 o page 758, use the row for degrees of freedom=3 (i) (ii) χ = = χ = = df 3, df 3, Put it all together, obtai (i) Lower limit = (ii) Upper limit = (-1)S χ 1- α/ χ α / (-1)S = (3)(3.67) = = (3)(3.67) =

5 Page 43 corrected Solutio for a 99% Cofidece Iterval for µ d Step 1 Poit Estimate of µ d is the Sample Mea d =30 d i i= 1 d=30 = = 0.51 =30 Step The Estimated Stadard Error of d is S d ˆ ˆ S d SE(d =30) = variace(d =30) = = = Step 3 The Cofidece Coefficiet For a 99% cofidece iterval, this umber will be the 99.5th percetile of the Studet s t-distributio that has degrees of freedom = (-1) = 9. This value is.756. Step 4 Substitute ito the formula for a cofidece iterval Lower limit = ( poit estimate ) - ( cof coeff. ) (SE of poit estimate) = (.756 ) ( ) = Upper limit = ( poit estimate ) + ( cof coeff. ) (SE of poit estimate) = (.756 ) ( ) = 0.780

6 Page 56 corrected How to Use the F Distributio Table i Roser (5 th Editio) Percetiles of selected F Distributios are provided i Table 9, begiig page 76. Each row defies a differet deomiator df. Each colum defies a differet umerator df. The body of the table gives values of selected pecetiles of the F- distributio. Oly the upper-tail percetiles (.90,,.999) are provided to you. Example - What is the 95th percetile value of a F distributio radom variable with umerator degrees of freedom equal to 4 ad deomiator degrees of freedom equal to 3? Locate o page 76 the colum for umerator df = 4 ad row for deomiator df=3. Withi this block, fid the 95th percetile value = 8.64 Example - What is the 5th percetile value of a F distributio radom variable with umerator degrees of freedom equal to 8 ad deomiator degrees of freedom equal to 6? To obtai this percetile value requires usig the woderful result just described. Locate o page 76 the colum for umerator df = 6 ad row for deomiator df=8. Withi this block, fid the 95th percetile value. This is F 6,8;.95 = 3.58 Thus, F 8,6;.05 =1/ F 6,8;.95 =(1/3.58)=0.79

7 Page 66 corrected 3. The Cofidece Coefficiet is agai a Percetile from the Normal(0,1) z.975 = Puttig it all together. Lower = (estimate ) - ( multiple ) (SE) = 0.38 (1.96)(0.093) = Upper = (estimate ) + ( multiple ) (SE) = (1.96)(0.093) = 0.56 Cofidece Iterval for a differece betwee two idepedet proportios [π 1 - π ] Two Idepedet Biomial Distributios where the required calculatios are ( ) ˆ ( ) [ ˆ π ˆ π ] ± z SE ˆ π ˆ π 1 1- α/ 1 (1) X Y X= ad Y= N 1 N () ˆ π 1 = X ad ˆ π = Y (3) ˆ SE= ( ) Y1-Y ( ) X1-X N + N 1 (4) For small umber of trials (N < 30 or so) i either group, use ˆ SE= ( ) ( ) N N 1