Lesso 7: Estimatio 7.3 Estimatio of Populatio Proportio p 1-PropZIterval October 18
Goals Preview Compute a cofidece iterval of the populatio proportio p.
Cofidece Iterval for p We would estimate the populatio proportio p of a attribute by a cofidece iterval. Examples would be 1. proportio p of defective lamps produced i a factory, 2. proportio p of a variety of seeds that will germiate i a particular coditio, 3. proportio p of voters i favor of a cadidate ad others.
The Beroulli(p) variable To estimate such a proportio p, we use the samplig distributio of the sample proportio of success X. The distributio of the sample proportio of success X was discussed i 6.3. As i 6.3, let X represet the correspodig Beroulli(p) radom variable. That meas { 1 if success X = 0 if failure
The Sample Proportio Take a sample X 1,X 2,...,X of size from this Beroulli(p) populatio. So, X i represets the outcome of testig the ith lamp as follows: { 1 if the i X i = th sample is a success 0 if the i th sample is a failure The T = X 1 +X 2 +...+X =the total Number of Success i these trials. So, X = T = the Proportio of Success i these trials.
The Costructio Preview By CLT (see 6.3) Z = X p p(1 p) N(0,1) So, P z α/2 Z = X p p(1 p) z α/2 = 1 α
The Costructio Preview Symplifyig, ( p(1 p) P X z α/2 ) p(1 p) p X +z α/2 = 1 α This looks like the defiitio (equatio) of a (1 α)100 percet cofidece iterval of p. Sice, p is ukow, LEP ad REP is ot computable. Usig X as a estimate, we have, approximately, X(1 X) X(1 X) P X z α/2 p X +z α/2 = 1 α
The Cofidece Iterval: 1-propZIterval So, a approximate, (1 α)100 percet cofidece iterval of p is give by X(1 X) X(1 X) X z α/2 p X +z α/2 We write the same i as a theorem. Theorem A approximate, (1 α)100 percet cofidece iterval of p is give by X(1 X) X e p X +e where e = z α/2 This is called 1-propZIterval.
Two MOE Preview e = z X(1 X) α/2 is called the Margi of Error. As before, defie LEP = X e ad REP = X +e. Also, defie coservative margi of error (Cosevative MOE) as E = z α/2 1 4 It ca be checked that the approximate error e E ad also the actual error z α/2 p(1 p) E
The Required Sample size As i 7.1, we compute sample size eeded for a give (Coservative) MOE. We solve the formula for E to etermie the reqired sample size. Give a specified coservative MOE E, for a (1 α)100 percet cofidece iterval of p, the sample size required is give by = ( zα/2 2E ) 2 rouded to the higher iteger.
Commets: Polls Opiio Polls are cofidece iterval for p, where p represets the proportio of those who is i favor of a propositio. I electios polls, p represets proportio of voters i favor ot a cadidate. I fact, they estimate for both cadidates. A typical poll report i a ellectio cycles looks as follows: Report Cadidate-A is leadig Cadidate-B by 47-45 percetage poits. Margi of error for this survey is ±3.6 percet ad the poll surveyed 754 voters.
Commets: Polls Preview This meas x =.47 is the sample proportio of success for cadidate-a. Here Coservative Margi of Error is E =.036. They did ot give the level of cofidece. Usually, they use 95 percet level. I this case, z α/2 = z.025 = 1.96 we ca compute 1 1 E = z α/2 4 = 1.96 4 754 =.0357 (as give).
A Example of Polls Presidet Clito was impeached i 1998-99 by the House of Represetatives ad acquitted by the Seate. Durig the impeachmet trial, a typical ews item would read as follows: Impeacmet 1998-99 Presidet Clito has 64 percet approval ratig. The poll has a margi of error plus or mius 3.1 percetage poits. The poll surveyed 972 people. This meat the followig: (1) Here p = as the proportio of the populatio who approved Presidet Clito. (2) Sample proportio of people who approve Presidet Clito was x = 0.64.
Cotiued (3) Such polls usually use 95 percet cofidece level. z α/2 = z.025 = 1.96. (4) We compute 1 1 E = z α/2 4 = 1.96 4 972 = 0.031. (as give). (5) Iterestigly, they give both the Coservative MOE ad sample size, while oe ca be computed from the other.
Exercise 7.3.5. Exercise 7.3.5. The proportio p of voters (i a couty) who vote by absetee ballot is to be estimated. You sample 683 voters ad foud that 165 would have voted by absetee ballots. Compute 96 percet cofidece iterval for p. Give the MOE, LEP, REP ad the coservative MOE. Solutio: Here the sample size = 683, the umber of success T = 165 The sample proportio of success X = T = 165 683 =.2416 Level of cofidece = 96 percet. So 1 α =.96,α =.04 ad α/2 =.02.
Cotiued Preview Therefore, z α/2 = z.02 = ivnormal(.98) = 2.0537 So, MOE e = z α/2 X(1 X) = 2.0537.2416(1.2416) 683 For error terms, we retai at least 6 decimal poits. LEP = X e =.2416.033638 =.207962 REP = X +e =.2416+.033638 =.275238 =.033638
Cotiued Preview Coservative MOE 1 1 E = z α/2 4 = 2.0537 4 683 =.039291 For error terms, we retai at least 6 decimal poits.
Exercise 7.3.6. Preview Exercise 7.3.6. It is kow that a vaccie may cause fever as side effect, after oe takes the shot. The proportio p of those who suffer side effect is to be estimated. A sample of 913 idividuals were examied ad it was foud the 153 suffered the side effect. Compute 94 percet cofidece iterval for p. Give the MOE, LEP, REP ad the coservative MOE. Solutio: Here the sample size = 913, the umber of success T = 153 The sample proportio of success X = T = 153 913 =.1676 Level of cofidece = 94 percet. So 1 α =.94,α =.06 ad α/2 =.03.
Cotiued Preview Therefore, z α/2 = z.03 = ivnormal(.97) = 1.8808 So, MOE e = z α/2 X(1 X) = 1.8808.1676(1.1676) 913 For error terms, we retai at least 6 decimal poits. LEP = X e =.1676.023249 =.144351 REP = X +e =.1676+.023249 =.190849 =.023249
Cotiued Preview Coservative MOE 1 1 E = z α/2 4 = 1.8808 4 913 =.031123 For error terms, we retai at least 6 decimal poits.
Exercise 7.3.10. Exercise 7.3.10. The proportio p of those who ear more tha $50 K, i a idustry, is to be estimated withi.07 from the actual value of p, with 90 percet cofidece. What would be the sample size eeded? Solutio: Here the precisio required E =.07 Level of cofidece = 90 percet. So 1 α =.90,α =.10 ad α/2 =.05. Therefore, z α/2 = z.05 = ivnormal(1.05) = 1.6449
Cotiued The sample size = ( zα/2 2E ) ( ) 2 2 1.6449 = = 1067.1111 2.07 We would roud it upward. So, the required sample size = 1068
Exercise 7.3.12. Exercise 7.3.12. The proportio p of those who cosumed more tha 200 CCF i Gas i Jauary, i a couty, would have to be estimated withi.05 from the actual value of p, with 98 percet cofidece. What would be the sample size eeded? Solutio: Here the precisio required E =.05 Level of cofidece = 98 percet. So 1 α =.98,α =.02 ad α/2 =.01. Therefore, z α/2 = z.01 = ivnormal(1.01) = 2.3263
Cotiued The sample size = ( zα/2 2E ) ( ) 2 2 2.3263 = = 541.1672. 2.05 We would roud it upward. So, the required sample size = 542.