= p(1 p)/n If the Rule of 15 and Large Population Rule are satisfied.

Transcription

1 Estimatig a Populatio Proportio-Example To estimate the proportio p of college studets who disbelieve i evolutio, take a simple radom sample of 1112 college studets, ad fid the sample proportio, say ˆp = 24%. The sample proportio 24% is the poit estimate for p, but to get a cofidece iterval, eed to ask how differet values of p affect likely values of ˆp i all samples. That is what Lecture 16 told us. If the populatio proportio is p, ad you take may simple radom samples of size, the distributio ˆP of their sample proportios satisfies ormal µ ˆP = p σ ˆP = p(1 p)/ If the Rule of 15 ad Large Populatio Rule are satisfied. From Populatio to Sample So if all is good the distributio of ˆP is ormal with mea p ad s.d p(1 p)/. Remember that 95% of a ormal distributio falls withi 1.96 stadard deviatios of the mea.that meas 95% of samples will give a sample proportio ˆp withi 1.96 p(1 p)/ of the populatio proportio p. That meas i the iterval from p 1.96 p(1 p)/ to p p(1 p)/, p ± 1.96 p(1 p)/ Let s rephrase this fudametal fact to make it look more like a cofidece iterval. From Sample to Populatio I 95% of samples the sample proportio ˆp is withi 1.96 p(1 p)/ of pop. proportio p, i iterval from p 1.96 p(1 p)/ to p p(1 p)/, i p ± 1.96 p(1 p)/ If I am withi 10 feet of you, how close are you to me? Withi 10 feet of course! I 95% of samples the populatio proportio p is withi 1.96 p(1 p)/ of sample proportio ˆp, i iterval from ˆp 1.96 p(1 p)/ to ˆp+1.96 p(1 p)/, i ˆp ± 1.96 p(1 p)/. This is almost a cofidece iterval to estimate p except for oe problem. It depeds o p! 1

2 From Sample to Populatio - Fixig the Glitch I 95% of samples p is withi 1.96 p(1 p)/ of ˆp, i iterval from ˆp 1.96 p(1 p)/ to ˆp p(1 p)/, i ˆp ± 1.96 p(1 p)/. The quatity p(1 p)/ is isesitive to p : modest chages i p produce very small chages i the quatity If the Rule of 15 holds, substitutig ˆp i for p itroduces very little error, so... I 95% of samples p is withi 1.96 / of ˆp, i iterval from ˆp 1.96 / to ˆp /, i ˆp ± 1.96 /. The 95% cofidece iterval for the populatio proportio p is ˆp ± This last formula is all you eed to remember from the precedig argumet, although we will have to go through aalogous steps to derive the other cofidece itervals we will lear for other quatitites. Cofidece Iterval for Proportio To estimate the populatio proportio p for a categorical variable from a sample of size with sample proportio ˆp... The 95% cofidece iterval for the populatio proportio p is ˆp ± (a) SRS - The sample is a simple radom sample. (b) Large Populatio - The populatio is at least 20. (c) Rule of 15 - BOTH ˆp 15 (1 ˆp) 15. AND Notice we adjusted the Rule of 15 : we used ˆp istead of p because that is what we kow. The Rule of 15 is really overkill so as to take the error from this ito accout. Specifically, if ˆp ad (1 ˆp are both more tha 15, it is early certai that p ad (1 p) are both at least 5, which really is good eough for the ormality assumptio. 2

3 Cofidece Iterval for Proportio - Not 95% What about other cofidece levels? We oly used 95% i fact that 95% of data is withi 1.96 s.d.s of the mea. So oly chage that umber. Remember C of data i ormal distributio is withi zc s.ds of mea. The C cofidece iterval for the populatio proportio p is. What is the value of zc? z C = ormiv((1 + C)/2, 0, 1) but you will ot be asked to use that.the oly values for zc I will ask you to kow are C zc 95% % % You will eed to remember those three umbers, but of course you get two pages of otes! Cofidece Iterval for Proportio - Example Give 95% cofidece iterval for proportio p of college studets who disbelieve i evolutio, if a simple radom sample of 1112 colleg studets gives ˆp = 24%. The 95% cofidece iterval for the proportio of all college studets who disbelieve i evolutio is.24(1.24).24± ± =.24 ±.0251 = 24% ± 2.51% Notice that we report this i a setece that gives the cofidece level ad the full parameter icludig parameter, populatio, ad variable, as well as the poit estimate ad margi of error. cofidece iterval calculatios should all be doe i decimals, but I like to report the results i percetages because they are easier to grok. I will ofte ask for you to give me the cofidece iterval i a complete setece icludig the cofidece level, full parameter, mea ad margi of error ad this is what I am ookig for. Geerally you ca just take the wordig of the questio ad spit it back with the aswer iserted ad you will pretty much have everythig i place. 3

4 More Example - Assumptios Give 95% cofidece iterval for proportio p of college studets disbelievig i evolutio, if a simple radom sample of 1112 Rep. voters gives ˆp = 24%. 24% ± 2.51% (a) SRS- It says. Met. (b) Large Pop- Need at least 20 = = 22, 240 college studets. Surely that s Met. (c) Rule of 15- Need BOTH = AND = Met! More Example - Iterpretatio Give 95% cofidece iterval for proportio p of all college studets disbelievig evolutio, if a simple radom sample of 1112 college studets ˆp = 24%. 24% ± 2.51% Remember this meas the followig.each sample we could take of this size would give its ow 95% cofidece iterval. 95% of all samples will give iterval via this procedure that cotai the actual parameter p we are tryig to estimate. So this iterval above has a 95% chace of cotaiig the correct value for the percetage of studets disbelievig i evolutio. The 95% is a percetage of samples, ot a percetage of studets or of beliefs, or of aythig else. Aother Example Give 99% cofidece iterval for proportio p of Coecticut residets with iphoes, if i a simple radom sample of 60 Coecticut residets, 21 have iphoes. The 99% cofidece iterval for the proportio of all Coecticut residets with a iphoe is = ± (1 60 ) = 35% ± 15.9% 60 4

5 (a) SRS- says so. Met. (b) Large Pop- eed more tha = 1200 Coecticut residets. Met. (c) Rule of 15- BOTH ˆp = = AND (1 ˆp) = = so Met. Notice ˆp ad (1 ˆp) are just the umber of yes ad o aswers i the sample, so whe it is expressed this way it could ot be easier. You Fly Solo Give 90% cofidece iterval for proportio p of deer ticks i Fairfield ifected with Lyme disease, if i a sample of 200 deer ticks 135 are ifected. The 90% cofidece iterval for the proportio of all deer ticks i Fairfield ifected with Lyme disease is = ± (1 200 ) = 67.5% ± 5.45% 200 (a) SRS- Does ot say. Not Met. (b) Large Pop- eed more tha = 4000 deer ticks i Fairfield. That seems probable to me. Met. (c) Rule of 15- BOTH ˆp = AND (1 ˆp) = so Met. Lecture 19 Key Poits After watchig this lecture you should be able to Fid the cofidece iterval for a populatio proportio p give the cofidece level, the sample size ad the sample proportio ˆp. Express the result i a complete setece icludig the cofidece level, full parameter, poit estimate ad margi of error. Check the three assumptios. Uderstad what the cofidece iterval meas i this istace. After processig this lecture you should be able to Iterpret the cofidece iterval as a probabilistic statemet. 5