Objectives. Sampling Distributions. Overview. Learning Objectives. Statistical Inference. Distribution of Sample Mean. Central Limit Theorem

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1 Objectives Samplig Distributios Cetral Limit Theorem Ivestigate the variability i sample statistics from sample to sample Fid measures of cetral tedecy for distributio of sample statistics Fid measures of dispersio for distributio of sample statistics. Fid the patter of variability for sample statistics Overview Distributio of Sample Mea A ew sample mea ca be calculated each time a ew sample is take I this way, the sample mea ca be aalyzed as a radom variable Beig able to calculate (approimately) the distributio of the sample mea is a critical tool for iferece Learig Objectives Uderstad the cocept of a samplig distributio Describe the distributio of the sample mea for samples obtaied from ormal populatios Describe the distributio of the sample mea for samples obtaied from a populatio that is ot ormal Statistical Iferece Ofte the populatio is too large to perform a cesus so we take a sample How do the results of the sample apply to the populatio? What s the relatioship betwee the sample mea ad the populatio mea? What s the relatioship betwee the sample stadard deviatio ad the populatio stadard deviatio? This is statistical iferece 1

2 Estimatio We wat to use the sample mea X to estimate the populatio mea µ If we wat to estimate the heights of eight year old girls, we ca proceed as follows Radomly select 100 eight year old girls Compute the sample mea of the 100 heights Use that as our estimate This is usig the sample mea to estimate the populatio mea Sample Mea is a Variable Usually, we just take oe sigle sample to estimate populatio parameter. However, if we take a series of differet radom samples from a target populatio Sample 1 we compute sample mea 1 Sample we compute sample mea Sample 3 we compute sample mea 3 Etc. Each time we sample, we may get a differet result The sample mea X is a radom variable! Distributio of Sample Mea Because the sample mea is a radom variable The sample mea has a probability distributio We ca obtai the ceter ad spread of the probability distributio of the sample mea This is called the distributio of the sample mea Because the sample mea is a sample statistic, a distributio of a sample statistic is ofte called a samplig distributio. So, the distributio of the sample mea is also called samplig distributio of the mea. Eample 1 Cosider a populatio of a uiformly distributed variable X with all of the values i the set {1,, 3, 4} occurrig equally likely: 1) Calculate the mea ad stadard deviatio of the populatio. ) Make a list of all samples of size that ca be draw from this set (Sample with replacemet) 3) Costruct the samplig distributio for the sample mea for samples of size 4) Calculate the ceter ad spread of the samplig distributio for the sample mea 5) Compare 1) ad 4) Eample 1 (cotiued) Eample 1 (cotiued) Mea of the populatio distributio of the variable X, deoted by µ or simply µ: µ = p( ) = = Variace of the populatio distributio of the variable X, deoted by or simply : = P( ) µ = = Stadard deviatio of the populatio distributio of the variable X, deoted by or simply : = = This table lists all possible samples of size, the mea for each sample, ad the probability of each sample occurrig (all equally likely) Sample {1,1} {1,} {1,3} {1,4} {,1} {,} {,3} {,4} {3,1} {3,} {3,3} {3,4} {4,1} {4,} {4,3} {4,4} Sample Mea Probability

3 Samplig Distributio of the Sample Mea P ( ) /16.0 3/16.5 4/ / / Eample 1 (cotiued) Summarize the iformatio i the previous table to obtai the samplig distributio of the sample mea : P ( ) Histogram: Samplig Distributio of the Sample Mea Notice that the samplig distributio of the sample mea is ormal. Eample 1 (cotiued) Mea of the samplig distributio of the sample mea X, deoted by µ is: µ = = Variace of the samplig distributio of the sample mea X, deoted by is : = = Stadard deviatio of the samplig distributio of the sample mea X, deoted by is: = = Eample 1 (cotiued) From the above eample, we coclude that Samplig distributio of the sample mea teds to be bell-shaped. The mea of the samplig distributio of the sample mea is the same as the uderlyig populatio mea. That is µ = µ The stadard deviatio of the samplig distributio of the sample mea is less tha the stadard deviatio of the populatio stadard deviatio. I fact, = We have the data Eample 1, 7, 11, 1, 17, 17, 17, 1, 1, 1,, ad we wat to take samples of size = 3 First, a histogram of the etire data set Check: = Eample (cotiued) A histogram of the etire data set Eample (cotiued) Takig some samples of size 3 Defiitely skewed left ot bell shaped The first sample, 17, 1, 1, has a mea of 16.7 The secod sample, 17, 7, 17 has a mea of 13.7 The third sample,, 11, 1 has a mea of 18 3

4 Eample (cotiued) More sample meas from more samples We calculate the mea for each sample as show below: Eample (cotiued) Fially, a histogram of 0 sample meas Eample (cotiued) Eample (cotiued) If takig a sample of size 5 repeatedly 0 times. Here is a histogram of 0 sample meas: Samplig Distributio of Sample Mea ( sample size = 5) Frequecy The origial data set was highly left skewed, but the set of sample meas is less skewed ad closer to bell shaped Sample Mea Observe that the empirical distributio of sample meas is more closer to bell shaped whe the size of the sample icreases. Distributio of Sample Mea I geeral, if the uderlyig populatio is closer to be bell shaped (ormally distributed), the the samplig distributio (i.e. the distributio of sample mea) will ted to be more bell shaped as well. I fact, the samplig distributio Will be ormally distributed Will have a mea equal to the mea of the populatio Will have a stadard deviatio less tha the stadard deviatio of the populatio Distributio of Sample Mea Why does it have a smaller stadard deviatio? The populatio stadard deviatio Is a measure of the distace/deviatio betwee a idividual value ad the populatio mea Is a stadard deviatio of the sample mea for = 1 The stadard deviatio of the sample mea Is a measure of the distace/deviatio betwee the sample mea ad the mea of the samplig distributio (which is the same as the populatio mea) It makes sese that the estimate of the populatio mea usig a sample mea is more accurate (closer to the populatio mea) if the sample cotais more values (a larger ) from the populatio. Therefore, the larger the sample size, the less of the deviatio of the sample mea from the true populatio mea. => stadard deviatio of sample mea is iversely related to the sample size. 4

5 Samplig Distributio of Sample Mea If a simple radom sample of size is draw from a populatio, the the samplig distributio has Mea µ = µ ad Stadard deviatio = I additio, if the populatio is ormally distributed, the The samplig distributio is ormally distributed Stadard Error The stadard deviatio of the sample mea is also called the stadard error The formula for is = This is a etremely importat formula Eample If the radom variable X has a ormal distributio with a mea of 0 ad a stadard deviatio of 1 If we choose samples of size = 4, the the sample mea will have a ormal distributio with a mea of 1 0 ad a stadard deviatio of 6 (sice 6 = ) 4 If we choose samples of size = 9, the the sample mea will have a ormal distributio with a mea of 0 ad a stadard deviatio of 4 ( sice 1 4 = ) 9 Note: if the uderlyig populatio distributio is ormal, the samplig Distributio of sample mea will be oraml regarless if the sample size is large or small. Samplig Distributio of sample Mea This is great if our radom variable X has a ormal distributio However what if uderlyig populatio distributio of the radom variable X does ot have a ormal distributio What ca we say about the samplig distributio of the sample mea? Would t it be very ice if the samplig distributio for sample mea also was ormal? This is almost true Cetral Limit Theorem The Cetral Limit Theorem states Regardless of the shape of the uderlyig populatio distributio, the samplig distributio of the sample mea becomes approimately ormal as the sample size icreases. Graphical Illustratio of the Cetral Limit Theorem Origial Populatio Distributio of : = Thus If the radom variable X is ormally distributed, the the samplig distributio of the sample mea is ormally distributed also regardless the size of the sample. For all other radom variables X, the samplig distributios are approimately ormally distributed if the size of the sample is large eough Distributio of : = Distributio of : =

6 How large is the sample? This approimatio, of the samplig distributio beig ormal, is good for large sample sizes large values of How large does have to be? A rule of thumb if is 30 or higher, this approimatio is probably pretty good Applicatios of the Cetral Limit Theorem Whe the samplig distributio of the sample mea is (eactly) ormally distributed, or approimately ormally distributed (by the CLT), we ca aswer probability questios usig the stadard ormal distributio. Eample 1 We ve bee told that the average weight of giraffes is 400 pouds with a stadard deviatio of 300 pouds We radomly picked 50 giraffes ad measured them ad foud that the sample mea was 600 pouds Is our data cosistet with what we ve bee told? ( That is, does the sample mea of 600 pouds observed support the claim that the average of populatio giraffes is 400 pouds?) Eample 1 (cotiued) Although we do ot kow the shape of the distributio of giraffe s weight. Sice the sample size is 50 which is large eough to justify the cetral limit theorem. The samplig distributio of the average weight of 50 giraffes is epected to be approimately ormal with mea 400 pouds (the same as the populatio) ad a stadard deviatio of 300 / 50 = 4.4 pouds Usig our calculatios for the geeral ormal distributio, 600 is 00 pouds over 400, ad 00 pouds is 00 / 4.4 = 4.7. That is the Z-score for the average weight of 600 is 4.7 which is a value ear the ed of the right tail of a stadard ormal distributio. From our ormal calculator, probability obtaiig a average at least this large by chace is less tha Somethig is defiitely strage we ll see what to do later i iferetial statistics Eample Or, use Z-table to solve: Eample (cotiued) Cosider a ormal populatio with µ = 50 ad =15. Suppose a sample of size 9 is selected at radom. Fid: 1) P( 45 60) ) P( 47. 5) Solutios: Sice the origial populatio is ormal, the distributio of the sample mea is also (eactly) ormal with 1) µ µ = = 50 ) 3) use TI calculator to fid the probability : ormalcdf(45,60,50,5) = ormalcdf(-e99,47.5,50,5) = = = 15 9 = 15 3 = 5 - µ z = ; z P( 45 60) = P z 5 5 = P( 1.00 z.00) = =

7 Eample (cotiued) Eample A recet report stated that the day-care cost per week i Bosto is $109. Suppose this figure is take as the mea cost per week ad that the stadard deviatio is kow to be $0. 1) Fid the probability that a sample of 50 day-care ceters would show a mea cost of $105 or less per week. - µ z = ; z P P ( 47. 5) = 5 5 = P( z. 5) = = ) Suppose the actual sample mea cost for the sample of 50 day-care ceters is $10. Is there ay evidece to refute the claim of $109 preseted i the report? Solutios: The shape of the origial distributio is ukow, but the sample size, = 50, is large. The CLT applies. The distributio of X is approimately ormal with µ = µ = 109 = = ) Use Z-table: Eample 3 (cotiued) z - µ P( ) = P z = ; 105 z. 83 = P( z 1. 41) = = Or, Use TI calculator: ormalcdf(-e99,105,109,.83) = ) Eample 3 (cotiued) To ivestigate the claim, we eed to eamie how likely a observatio is the sample mea of $10 Cosider how far out i the tail of the distributio of the sample mea is $10. z = - µ ; P( 10) = P z = P( z 389. ) = = Or usig TI calculator: ormalcdf(10,e99,10,.83) = 5.08E-5 Sice the probability is so small, this suggests the observatio of $10 is very rare (if the mea cost is really $109) There is evidece (the sample) to suggest the claim of µ = $109 is likely wrog Summary The sample mea is a radom variable with a distributio called the samplig distributio If the sample size is sufficietly large (30 or more is a good rule of thumb), the this distributio is approimately ormal The mea of the samplig distributio is equal to the mea of the uderlyig populatio The stadard error/deviatio of the samplig distributio is equal to / Distributio of the Sample Proportio 7

8 Learig Objective Describe the samplig distributio of a sample proportio Calculate probabilities of a sample proportio Sample Proportio I a electio, pollig compaies wish to estimate the percet of people who will vote for each cadidate This clearly is a situatio for samplig as it is impractical to cotact every sigle voter The desired results are proportios, for eample that 59% of the voters (a proportio of 0.59) said that they will vote for cadidate A Samplig Distributio of Sample Proportio We have the same questios for the sample proportio as we had for the sample mea What is the mea for the samplig distributio of the sample proportio? What is the stadard deviatio for the samplig distributio of the sample proportio? What is the distributio of the sample proportio? Ca we apply the Cetral Limit Theorem to approimate these with ormal distributios? The aswer is yes Sample Proportios A radom sample is take Of size Each idividual either has or does ot have a certai characteristic (dichotomous outcomes) I total, there are idividuals that have this characteristic The the sample proportio pˆ (p hat) (the proportio of idividuals with this characteristic is give by pˆ = Eample If a pollig compay polled 800 people to see if they supported a certai issue ad 475 did, the we have a sample proportio problem with = 800 = ad a sample proportio of pˆ = = Samplig Distributio of Sample proportio If the populatio proportio is p, the the distributio of the sample proportio for a sample of size Is approimately ormal if p(1-p) 10 Has a mea of µ pˆ = p Has a stadard deviatio of p( 1 p) pˆ = 8

9 Eample Assume that 80% of the people takig aerobics classes are female ad a simple radom sample of = 100 studets is take What is the probability that at most 75% of the sample studets are female? If the sample had eactly 90 female studets, would that be uusual? Eample (cotiued) The sample proportio pˆ of aerobics studets who are female Has a approimately ormal distributio Has a mea of 0.80 ad a stadard deviatio of 0.04 What is the probability that pˆ is 0.75 or less? 0.75 is 0.05 less tha the mea of is 1.5 stadard deviatios less tha the mea (i.e. the z- score is 1.5) The ormal probability P(z 1.5) =.1056 Thus P( pˆ 0.75) =.1056 Note: To obtai P( pˆ 0.75) =.1056, apply TI graphig calculator with ormalcdf(-e99,-1.5,0,1) or ormalcdf(-e99,0.75, 0.80,0.04) [or ormalcdf(0,0.75,0.80,0.04),sice probability ca t be less tha zero.] Eample (cotiued) The sample proportio pˆ of aerobics studets who are female Has a approimately ormal distributio Has a mea of 0.80 ad a stadard deviatio of 0.04 What is the probability that pˆ is 0.90 or more? Notice that istead of fidig the probability beig eactly 0.90 (which will be zero usig ormal distributio), we ca justify if 0.90 is uusual or ot (proportio is too high) by evaluatig the probability beig at least as high as 0.90, sice 0.90 is larger tha the epected value of is 0.10 more tha the mea of is.5 stadard deviatios more tha the mea (i.e. the z-score is.5) The ormal probability P(z.5) =.006 Thus P( pˆ.5) = pretty ulikely Note: To obtai P( pˆ.5) = 0.006, apply TI-calculator with ormcdf(.5,e99,0,1) = or ormcdf(0.9, E99,0.8,0.04) = [or ormcdf(0.9,1,0.8,0.04), sice probability ca oly go up to 1] Summary The sample proportio, like the sample mea, is a radom variable If the sample size is sufficietly large ad the populatio proportio p is t close to either 0 or 1, the this distributio is approimately ormal The mea of the samplig distributio is equal to the populatio proportio p The stadard deviatio of the samplig distributio is equal to p( 1 p) / Summary Summary of Samplig Distributios The sample mea ad the sample proportio ca be cosidered as radom variables The sample mea is approimately ormal with A mea equal to the populatio mea µ = µ A stadard deviatio equal to = / The sample proportio is approimately ormal with A mea equal to the populatio proportio µ pˆ = p A stadard deviatio equal to pˆ = p( 1 p) / 9