Samplig Distributios ad Estimatio T O P I C #
Populatio Proportios, π π the proportio of the populatio havig some characteristic Sample proportio ( p ) provides a estimate of π : x p umber of successes i the sample sample size If two outcomes, p has a biomial distributio
Samplig Distributio of p Approximated by a ormal distributio if: where π 5 ( π) 5 P(p).3.. 0 Samplig Distributio 0..4.6 8 p µ p π ad σ p π( π) (where π populatio proportio)
z-value for Proportios Stadardize p to a z value with the formula: z p π σ p p π π( π) If samplig is without replacemet ad is greater σ p tha 5% of the populatio size, the must use the fiite populatio correctio factor: σ p π( π) N N Chap 7-4
Example If the true proportio of voters who support Propositio A is π.4, what is the probability that a sample of size 00 yields a sample proportio betwee.40 ad.45? i.e.: if π.4 ad 00, what is P(.40 p.45)?
Example if π.4 ad 00, what is P(.40 p.45)? (cotiued) Fid : σ p π( π).4(.4) σ p.03464 00 Covert to stadard ormal: P(.40 p.45).40.40 P.03464 z.45.40.03464 P(0 z.44)
Example if π.4 ad 00, what is P(.40 p.45)? (cotiued) Use stadard ormal table: P(0 z.44).45 Samplig Distributio Stadardized Normal Distributio.45 Stadardize.40.45 0.44 p z
Cofidece Itervals for the Populatio Proportio, π A iterval estimate for the populatio proportio ( π ) ca be calculated by addig a allowace for ucertaity to the sample proportio ( p )
Cofidece Itervals for the Populatio Proportio, π (cotiued) Recall that the distributio of the sample proportio is approximately ormal if the sample size is large, with stadard deviatio σ π π( π) We will estimate this with sample data: s p p( p)
Cofidece iterval Upper ad lower cofidece limits for the populatio proportio are calculated with the formula p ± z p( p) where z is the stadard ormal value for the level of cofidece desired p is the sample proportio is the sample size
Example A radom sample of 00 people shows that 5 are left-haded. Form a 95% cofidece iterval for the true proportio of left-haders
Example A radom sample of 00 people shows that 5 are left-haded. Form a 95% cofidece iterval for the true proportio of lefthaders. (cotiued). p 5/00.5. 3. Sp p( p)/.5(.75)/00.5 ±.96 (.0433).0433 0.65... 0.3349
Fidig the Required Sample Size for proportio problems Defie the margi of error: e z π( π) z π( π) Solve for : e π ca be estimated with a pilot sample, if ecessary (or coservatively use π.50)
What sample size...? How large a sample would be ecessary to estimate the true proportio defective i a large populatio withi 3%, with 95% cofidece? (Assume a pilot sample yields p.)
What sample size...? Solutio: For 95% cofidece, use Z.96 E.03 p., so use this to estimate π z π( e π) (.96) (.)(.) (.03) (cotiued) 450.74 So use 45
Estimatio for Two Populatios Estimatig two populatio values Examples: Populatio meas, idepedet samples Group vs. idepedet Group Populatio proportios Proportio vs. Proportio
Differece Betwee Two Meas Goal: Form a cofidece iterval for the differece betwee two populatio meas, µ µ The poit estimate for the differece is x x
σ ad σ kow (cotiued) The cofidece iterval for µ µ is: * σ x ± z α/ + ( ) x σ
σ ad σ ukow, large samples (cotiued) The cofidece iterval for µ µ is: x ± tα/ sp + ( ) x Where t α/ has ( + ) d.f., ad s p ( ) s + ( ) + s
Two Populatio Proportios Populatio proportios Goal: Form a cofidece iterval about the differece betwee two populatio proportios, π π Assumptios: π 5, (-π ) 5 π 5, (-π ) 5 The poit estimate for the differece is p p
Cofidece Iterval for Two Populatio Proportios Populatio proportios The cofidece iterval for π π is: ( p p ) ± z p ( p ) + p ( p )
Note: Samplig Distributio of x Z-value for the samplig distributio of : (x µ) z σ x where: x µ σ sample mea populatio mea populatio stadard deviatio sample size
Fiite Populatio Correctio Apply the Fiite Populatio Correctio if: the sample is large relative to the populatio ( is greater tha 5% of N) ad Samplig is without replacemet Note that if σ ukow replace it by s sample stadard deviatio z σ (x N N µ) The