Chapter 8: STATISTICAL INTERVALS FOR A SINGLE SAMPLE. Part 3: Summary of CI for µ Confidence Interval for a Population Proportion p
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1 Chapter 8: STATISTICAL INTERVALS FOR A SINGLE SAMPLE Part 3: Summary of CI for µ Cofidece Iterval for a Populatio Proportio p Sectio 8-4 Summary for creatig a 100(1-α)% CI for µ: Whe σ 2 is kow ad paret populatio is ormal, use a z-value (this works for ay ). x ± z α/2 σ Whe σ 2 is ukow ad is REALLY large, use a z-value ad replace σ 2 with the observed sample variace s 2 (this works for ay paret populatio distributio because is large). x ± z α/2 s 1
2 Whe σ 2 is ukow, is relatively small, ad POPULATION IS N EARLY NORMAL, use a t-value ad the sample variace. x ± t α/2, 1 s Before you make a CI, it is a radom iterval... it depeds o the sample chose, but X will ALWAYS be at the ceter of the 2- sided CI. Example of 16 differet CI s for µ each based o a differet sample: 2
3 Example: Fuel rods i a reactor (problem 8-43 i book) A article i Nuclear Egieerig Iteratioal gives the followig measuremets o the percetage of erichmet of 12 fuel rods i a reactor i Norway: Calculate a 95% CI for the mea percetage of erichmet. Provide a ormal probability plot to verify the assumptio of ormality. ANS: This is a small sample with ukow σ 2, so we will use the procedure with the t- value. x ± t α/2, 1 s 3
4 x= s 2 = s = t α/2, 1 = t 0.025,12 1 = t 0.025,11 = % CI for µ: ± (0.0993/ 12) ± [2.8386, ] 4
5 Checkig ormality with a ormal probability plot: ormal probability plot Theoretical Quatiles Sample Quatiles Looks pretty good. The poits fall radomly aroud the diagoal lie. So, we ca believe that the data were geerated from a ormal distributio (i.e. the paret populatio was at least approximately ormal). 5
6 Populatio Proportio Parameter p Movig o to aother potetial parameter of iterest... A populatio proportio is deoted p. A sample proportio is deoted ˆp. A populatio proportio is based o a yes/o or 0/1 type variable. What proportio of the populatio favor Hillary Clito? yes/o What proportio of a maufactured good is defective? defective/ot defective 6
7 What proportio of the U.S. is republica? republica/ot republica What proportio of studets eterig college successfully complete a degree? succeed/fail To estimate a populatio proportio, we will use a sample proportio ad the bottom lie for proportios is that you just plai eed a large sample. There s sparse iformatio i each observatio (just yes/o), we do t have a ice cotiuum like i a cotiuous radom variable measuremet (like weight). 7
8 Large-Sample Cofidece Iterval for a Populatio Proportio p Sectio 8-4 The costructio of the CI for p relies o the fact that we took a large sample (large ). I will itroduce this cocept usig somethig we are familiar with, the biomial... We will let the category of iterest be called the success category (arbitrary). Let X i = 1 if observatio i falls ito the success category. Let X i = 0 if observatio i falls ito the other or fail category. X i is called a idicator variable. 8
9 Thus, i=1 X i = a cout of all idividuals i the success category. The sample proportio of idividuals fallig ito the success category is ˆP = i=1 X i = # i sample who are i success category ˆP (upper case) is a radom variable ad is the poit estimator for p. ˆp (lower case) is a realized poit estimate from a observed sample. 9
10 Note that p ad are actually the parameters for a biomial distributio (i.e. probability of a success ad umber of trials). There are trials (i.e. idepedet draws of idividuals to form the sample) The probability of gettig a success remais costat as p (assumig we have a large populatio ad is ot too large) Let Y =total umber of successes. So Y Biomial(, p) E(Y ) = p ad V (Y ) = p(1 p) I our otatio, Y = i=1 X i, so Y = i=1 X i Biomial(, p) 10
11 E( i=1 X i ) = p ad V ( i=1 X i ) = p(1 p) Thus, if is large, we have ( i=1 X i ) Z = ( i=1 X i ) p = p p(1 p) p(1 p) = ˆP p p(1 p) where Z is approximately stadard ormal. (NOTE: This is a ormal approximatio to the biomial. Ad for the approximatio to be reasoable, we should have p 5 ad (1 p) 5.) 11
12 Normal approximatio for a sample proportio ˆP If is large, the distributio of Z = ˆP p p(1 p) is approximately stadard ormal. Or similarly, ˆP N(p, p(1 p) ) This meas the samplig distributio for ˆP is ormal... See applet at: 12
13 Thus, we ca ca use a z-value to form a 100(1-α)% CI for p: P ( z α/2 ˆP p p(1 p) z α/2 ) = 1 α rearragig... P ( ˆP z α/2 p(1 p) ) p ˆP p(1 p) + z α/2 = 1 α ad we have the lower ad upper bouds... Lower boud: ˆP zα/2 p(1 p) Upper boud: ˆP + zα/2 p(1 p) BUT WE DON T KNOW p, SO WE CAN T GET ACTUAL VALUES FOR THE BOUNDS! Solutio replace p with ˆP i the formulas. 13
14 Approximate 100(1-α)% CI for a populatio proportio p If ˆp is the proportio of observatios i a radom sample of size that belogs to a class of iterest, a approximate 100(1-α)% CI o the proportio of p of the populatio that belogs to this class is ˆp ± z α/2 ˆp(1 ˆp) where z α/2 is the upper α/2 percetage poit of the stadard ormal distributio. Thigs you eed for the appropriate behavior of ˆP : Populatio is large, ad you do t take too may idividuals for your sample. Maybe o more tha 10% of the total populatio. 14
15 The sample is a simple radom sample. p 5 ad (1 p) 5. This statemet meas that if you have a really rare evet, you re goig to eed a very large sample... just logical though. Example: Iterpolatio methods are used to estimate heights above sea level for locatios where direct measuremet are uavailable. After verifyig the estimates, it was foud that the iterpolatio method made large errors at 26 of 74 radom sample test locatios. Fid a 90% CI for the overall proportio of locatios at which this method will make large errors. 15
16 ANS: 16
17 Choice of sample size for estimatig p For a specified error E = p ˆP i your estimate, the previously stated behavior of ˆP suggests you should choose a sample size as: = ( zα/2 E ) 2 p(1 p) But sice we do t kow p (that s what we re tryig to estimate!), we ca t compute a sample size from this formula... uless we estimate p first. Here, we choose to err o the coservative side. It turs out the largest variace for ˆP [where V ( ˆP ) = p(1 p) ] occurs whe p = 0.5 o matter what the was: 17
18 Plot of variace of ˆP vs. p p(1 p)/ p So, if we do t kow p, we ll plug-i p = 0.5 to make sure we do t uder-estimate the variace of our estimate (usig worst case sceario idea). 18
19 Thus, the workig sample size formula for estimatig p with a 100(1 α)% CI ad a error of E is: ( ) zα/2 2 = 0.25 E We use this method because, before we collect data, we do t have ay iformatio o p, so there s o observed ˆp to plug-i. I cotrast, whe doig a CI for p, we DO have a observed estimate ˆp for p, so i that case, so use our plug-i estimate. 19
20 Example: I the iterpolatio example, how large of a sample would you eed if you wated to be at least 95% cofidet that the error i your estimate (i.e. p ˆp ) is less tha 0.08? ANS: 20
21 Example: Gallup Poll Betwee February 9-11, 2007, adults were radomly sampled (by phoe) ad asked: Would you favor or oppose Cogress takig actio to set a time-table for withdrawig all U.S. troops from Iraq by the ed of ext year. Let p = the proportio of the populatio watig a withdrawal time-table. How large is their error if they sample 1000 people ad form a a 95% CI? 21
22 ANS: See USA Today, February 12, What ca go wrog i the estimatio? Bias i selectig the sample for oe thig. Gallup idetifies flaws i 2012 electio polls 22
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