Class 3,4 Jue 7, 9, 05 Pla for Class3,4:. Samplg dstrbuto of sample mea. The Cetral Lmt Theorem (CLT). Cofdece terval for ukow mea.. Samplg Dstrbuto for Sample mea. Methods used are based o CLT ( Cetral Lmt Theorem) ad rules of expectato ad varace. O class 6 we troduced a radom varable havg meag of a w a card game. was defed as follows: If a card draw s er Jack or Quee, (s) = $5 ; f s s er Kg or Ace, (s)=0$. Orwse (s)=-4. Dstrbuto of : -4 5 0 p 36/5 8/5 8/5
I frst row possble values are gve, secod row correspodg probabltes. The Expectato (Expected value, Mea) s defed as follows: E( ) p x ; Expectato also s deoted by E( ). I our problem of drawg a card 8 8 E( ) px 5 0 5 5 The Varace V ( ) s p ( x defed ) as p x 36 4 5. 0.465. The last expresso (rght-had-sde) s so called a shortcut formula for varace. Usg t we fd V ( ) V ( ) p x 6 30.4, ( ) 5.5. 36 5 5 8 5 00 8 5 0.465 30.3.
Example. Assume we are playg ow games depedetly,, deote r ws respectvely. Now cosder sum S Soluto. of S has two games. followg Fd P( S ). dstrbuto : S -8 + 6 0 5 0 p 0.479 0.3 0.3 0.04 0.048 0.04 The resultg table s called Samplg Dstrbuto of Sum. The table gves us P ( S ) 0.479. Example. Fd P ( ). More frequetly dstrbuto of followg close relatve of S s cosdered: ( + ) / whch s called Samplg Dstrbuto of Sample Mea ( + )/ -4 +/ 3 5 7.5 0 p 0.479 0.3 0.3 0.04 0.048 0.04 So, soluto to Example s: P ( ) 0.69.
Example 3. Now we cosder 00 games ad put k { The w k th game}, k,...,00, ad form sample mea :... ( our case equals 00). Fd P( ). Now ma questo s: What s dstrbuto of, or speakg statstcal laguage, what s samplg dstrbuto of sample mea? CLT helps solve out ths problem. A smplfed verso of CLT whch s frequetly used appled statstcs souds as follows: The sum of may depedet detcally dstrbuted radom varables havg a varace, has approxmately a Normal dstrbuto. So CLT tells us that samplg dstrbuto of s approxmately Normal. It remas to fd mea ad varace of. The formulas for ad ; are :. () We ca easly prove above formulas by use of followg rules for expectatos ad varaces:
E( a V ( V ( a )... a... ) a V ( ), ad ) V ( a E( for )... a depedet )... V ( ). E( ); (4) (3)... Remark. We ote that (3) holds wthout codto of depedece, whle (4) depedece s eeded. We also remark (warg!) (4) s ot correct for stadard devatos. Now we come back to our Example 3. Accordg to above facts, 5.5 ~ N ( 0.465,( ) ) 00, ad we fd 4.65 P( ) P( Z ) P( Z.66) 0.0039. 5.5. Homework Assgmet. Exercses 5.6-5., p.83; 5.4
Reducto of arbtrary Normal probablty to that of Stadard Normal. Assume P( a ~ N(, ). The a b b) P( Z ). Example 4. Assume we select at radom a battery from a large batch of alkale batteres. The radom varable of terest s = {The lfetme of a radomly chose battery} Assume t s kow that 0 hours ad hours. Now choose depedetly at radom 00 batteres, cosder 00 depedet radom varables: k { The lfetme of k th battery }, k,...,00, ad form sample mea :... ( our case equals 00). What s P( 9.8)?
Accordg to CLT, has approxmately a Normal dstrbuto. I order to fd desred probablty we have to calculate ad : 0; 0.. () ~ N (0,(0.) ), ad we fd P ( 9.8) 0. 6. Aor sub-problem o your tests mght be: Fd such that ( 9.8) 0.05. P (Aswer: =7). Soluto. P( 9.8) 0.05. Ths mples 9.8 0 /.645. Ths mples P( Z 7. 9.8 0 ) / 0.05. By table, Homework Assgmet o Samplg Dstrbuto of Sample Mea: Exercses 5.6-5., p.83; 5.4.
Cofdece Iterval for ukow mea. Exercse. Cosder followg radom varable ={The mleage per gallo for a radomly chose car of a ew model}, s ot kow. such kd of problem has a very clear practcal meag: t s a populato (global) average of mleage per gallo. The problem s stated as follows: Gve a sample,..., for radomly chose =00 cars of model fd a 95%-cofdece terval for x 3 mles / gallo, 4.6 mles / gallo. The values are gve.
A has For ( ) 00% cofdece cofdece form : z /, where s level Most Come of umber for frequetly level () whch 95% t 0.05, ad back 00, cofdece to sze our 4.6 of sample, ( ) 00% s z / P( z practcal.96. t erval / z cosdered. / ad Z s statstca l example. x 3, z / z for depeds /.96, Ad resultg cofdece terval s 3 0. 90. 0.90 s called boud for error ukow defed ) o. problems. as
Fdg sample sze. Assume we wat to have boud for error 0.50 (stead of 0.90). What s sample sze that esures ths boud for error at same level of cofdece? The problem reduces to followg equato wth respect to : 4.6.96 0.5; 36. Homework Assgmets: Exercses: 6.-6.5, p.307.