(7 One- and Two-Sample Estimation Problem )

Transcription

1 34 Stat Lecture Notes (7 Oe- ad Two-Sample Estimatio Problem ) ( Book*: Chapter 8,pg65) Probability& Statistics for Egieers & Scietists By Walpole, Myers, Myers, Ye

2 Estimatio 1 ) ( ˆ S P i i Poit estimate: Is a sigle umerical value to estimate parameter Example : 1 ) ( ˆ S P i i Iterval estimate Is two umerical values to estimate parameter It meas to be i a iterval s.a Lower boud upper boud (L) (U) ( U ) L

3 Iterval estimatio A iterval estimate of a populatio parameter is a iterval of the form ˆ L ˆ U where ˆL ad ˆU deped o the value of the statistic ˆ for a particular sample ad also o the samplig distributio of ˆ.Sice differet samples will geerally yield differet values of ˆ ad therefore differet values of ad ˆU. From ˆL the samplig distributio of ˆ we shall be able to determie ˆL ad ˆU such that the P( ˆ ˆ L U ) is equal to ay positive fractioal value we care to specify. If for istace we fid ad such that: P( ˆ ˆ ) 1 for 0 1 L U the we have a probability of (1 ) of selectig a radom sample that will produce a iterval cotaiig. ˆL ˆU

4 The iterval ˆ called a L U (1 )100% ˆ computed from the selected sample, is the cofidece iterval. The fractio cofidece. (1 ) is called cofidece coefficiet or the degree of The ed poits ˆL ad ˆU are called the lower ad upper cofidece limits. For Example: whe 0.05 we have a 95% cofidece iterval ad so o, that we are 95% cofidet that is betwee ˆL, ˆU

5 9.4 Sigle Sample: Estimatig the Mea: 1-Cofidece Iterval o µ ( σ² Kow): If is the mea of a radom sample of size from a populatio with kow variace σ², a ( 1)100% cofidece iterval for µ is give by: Z 1 / Z 1 / (1) where is the -value leavig a area of to the right Z 1 / ˆ L ˆ Z 1 /, U Z 1 / ()

6 E (1): The mea of the quality poit averages of a radom sample of 36 college seiors is calculated to be.6. Fid the 95% cofidece itervals for the mea of the etire seior class. Assume that the populatio stadard deviatio is 0.3.

7 Solutio: 36,.6, % cofidece iterval for the mea µ : Z 1 See Ex 9. pg 71 Z ( ) Thus, we have 95% cofidet that µ lies betwee.50 ad.698

8 Theorem 9.1: If is used as a estimate of µ, we ca be ( 1 cofidet that the error will ot be exceed. Z 1 )100% For example (1): e= (1.96) (0.3/6)=0.098 or Theorem (): If is used as a estimate of µ, we ca be ( 1)100% cofidet that the error will ot exceed a specified amout, the sample size is: The fractio of is rouded up to ext whole umber. Z 1 (3) e e whe

9 E (): How large a sample is required i Ex. (1) if we wat to be that our estimate of µ is off by less tha 0.05? Solutio 95% cofidet Z 1.96, 1 0.3, e 0.05 (1.96)(0.3) is rouded up to whole umber.

10 - Cofidece Iterval of whe σ² is Ukow <30 : If ad s are the mea ad stadard deviatio of a radom sample from a ormal populatio with ukow variace cofidece iterval for µ is give by: σ², a ( 1)100 % where area of t 1 S t t 1, 1 1, 1 is the t-value with -1 degrees of freedom leavig a to the right. S (4)

11 Ex 9.5 pg 75 The cotets of 7 similar cotaiers of sulfuric acid are 9.8, 10., 10.4, 9.8, 10, 10., 9.6 liters. Fid a 95% cofidece iterval for the mea of all such cotaiers assumig a approximate ormal distributio.

12 Solutio: cofidece iterval for the mea : , 10, S 0.83, at : t t0.05,6 =.447 t0.975, , 1 1, 1 S 0.83 t ( ) 10 (.447)( ) P ( ) 0.95

13 9:4 Two Samples: Estimatig the Differece betwee Two Meas: ad 1 1- Cofidece Iterval for 1 whe Kow: 1 ad If are the meas of idepedet radom samples of size 1 ad from populatios with kow variaces 1 ad respectively, a ( 1)100% cofidece iterval for 1 is give by: 1 ( 1 ) Z (5) 1 1 where Z is the -value leavig a area of to the right. 1

14 E (4): A stadardized chemistry test was give to 50 girls ad 75 boys. The girls made a average grade of 76, while the boys made a average grade of 8. Fid a cofidece iterval for the differece 1 % where is the mea score of all boys ad 1 is the mea score of all girls who might take this test. Assume that the populatio stadard deviatios are 6 ad 8 for girls ad boys respectively. 96

15 Solutio: girls Boys cofidece iterval for the mea : 96% Z.05 1 ( 1 ) Z See Ex 9.10 pg (8 76) (.05) P ( )

16 -Cofidece Iterval for 1 whe 1 Ukow but equal variaces: If ad 1 are the meas of idepedet radom samples of size 1 ad respectively from approximate ormal populatios with ukow but equal variaces, a ( 1)100% cofidece iterval for 1 is give by:, where 1 1 ( 1 ) t S P (6) 1, v 1 S P ( 1) S ( 1) S (7) is the pooled estimate of the populatio stadard deviatio ad is the t t value with vdegrees of 1, v 1 freedom leavig a area of to the right.

17 E (9.11- pg 88): The idepedet samplig statios were chose for this study, oe located dow stream from the acid mie discharge poit ad the other located upstream. For 1 mothly samples collected at the dow stream statio the species diversity idex had a mea value ad a stadard deviatio while 10 mothly samples had a mea idex value ad a stadard deviatio S Fid a 90% cofidece iterval for the differece betwee the populatio meas for the two locatios, assumig that the populatios are approximately ormally distributed with equal variaces..04 S

18 Solutio: Statio 1 Statio S S % cofidece iterval for the mea 1 : S P 11(0.771) 9(0.448) at t t 0.95, , , ( ) (1.75)(0.646) P ( ) 0.90

19 9:10 Sigle Sample Estimatig a Proportio: Large Sample Cofidece Iterval for P: pˆ If is the proportio of successes i a radom sample of size ad a approximate ( 1)100% cofidece iterval for the biomial parameter is give by: pq ˆˆ 1 pˆ Z (8) qˆ 1 pˆ Z Where 1 is the Z- value leavig a area of to the right.

20 E(6): A ew rocket lauchig system is beig cosidered for deploymet of small, short rag rockets. The existig system has p=0.8 as the probability of a successful lauch. A sample of 40 experimetal lauches is made with the ew system ad 34 are successful. Costruct a 95% cofidece iterval for p

21 Solutio: a 95% cofidece iterval for p , pˆ 0.85, qˆ at Z pq ˆˆ (0.85)(0.15) pˆ Z 0.85 (1.96) 0.85 (0.111) p P (0.739 p 0.961) 0.95 See Ex 9.14 pg 97

22 Theorem 3: If pˆ is used as a estimate of p we ca be ( 1)100% cofidet that the error will ot exceed pq e Z. 1 ˆˆ

23 E 7: I Ex. 7, fid the error of p. Solutio: The error will ot exceed the followig value: e pq ˆˆ (0.85)(0.15) Z (1.96)

24 Theorem : pˆ If is used as a estimate of p we ca be ( 1 cofidet that the error will be less tha a specified amout e whe the sample size is approximately: )100% Z 1 e pq ˆˆ (9) The the fractio of is rouded up.

25 E(8): How large a sample is required i Ex. 7 if we wat to be 95% cofidet that our estimate of p is withi 0.0? Solutio: e 0.0, Z 1.96, pˆ 0.85, qˆ (1.96) (0.85)(0.15) (0.0)

26 Two Samples: Estimatig the differece betwee two proportios ( 1)100% Z 1

27 Ex 9.17 pg 301 A certai chage the process for maufacturig compoet parts is beig cosidered. Samples are take uder both the existig ad the ew process results i a improvemet. If 75 of 1500 items from the existig process are foud to be defective, ad 80 of 000 items from the ew process foud to be defective, fid a 90% cofidece iterval for the rue differece i the proportios of defectives for the existig ad ew process respectively.

28 Z 1