6 Estimation in the One-Sample Situation

Transcription

1 6 Estimatio i the Oe-Sample Situatio SW Chapter 6 Stadard Errors ad the t Distributio We eed to add oe more small complicatio to the samplig distributio of. What we saw last time, ad i SW Chapter 5, is that if 1, 2,..., is a radom sample from a ormal populatio ad that populatio has mea µ ad stadard deviatio σ, the looks like it is a sigle umber radomly selected from a ormal distributio also with mea µ = µ but with stadard deviatio σ = σ. We get from this that µ σ = µ σ/ = Z is a stadard ormal radom variable, ad we ca use the table o the iside frot cover of SW to compute probabilities ivolvig. Ufortuately, i the cotext i which we eed to use this result, we would eed to kow σ i order to apply the result. We are samplig from a populatio i order to fid out somethig about it, so almost certaily we do ot kow what σ is. What works well is to estimate the populatio stadard deviatio σ with sample stadard deviatio S calculated from the radom sample. Our best guesses of the populatio mea ad stadard deviatio µ ad σ are the correspodig sample values ad S. While µ ad σ are costats (we do ot kow the actual values, but they are costats), ad S deped upo the actual sample radomly selected from the populatio. If we repeated the experimet ad drew a secod radom sample of observatios, we would get differet values for ad S, which is to say ad S are radom variables. If we are goig to estimate σ with S, the of course we would estimate σ = σ with S. That is exactly what we do, ad we give this quatity the ame Stadard Error of, SE. Istead of stadardizig with the expressio µ σ we use the ew expressio µ SE. Usig SE i the deomiator itroduces extra variability, though, so this o loger looks like a radom umber that came from a Z distributio. Provided our assumptios are correct (radom samplig from a ormally distributed populatio), the µ SE looks like a sigle radom umber radomly selected from a Studet s t distributio with 1 degrees of freedom (df). The amout of extra variability itroduced depeds upo the sample size ; if is very small it is a lot, but by the time is 30 or so, there is very little differece from a Z, ad i fact df = makes the t ad Z distributios the same. SW o p. 187 show how the distributio compares to the ormal it does t look like a big differece, but the probability statemets we ca make are differet eough to matter. Table 4 p. 677 i SW is a stadard table of the t distributio. It is orgaized differetly from the Normal Table, sice it gives areas uder the curve across the top ad lets you look up the critical values that geerate those areas, while the Z table gives you critical values across the side ad top ad lets you look up areas. We will go through some examples of readig this table durig the lecture. Iferece for a Populatio Mea Suppose that you have idetified a populatio of iterest where idividuals are measured o a sigle quatitative characteristic, say, weight, height or IQ. ou select a radom or represetative sample from the populatio with the goal of estimatig the (ukow) populatio mea value, idetified by µ. 29

2 This is a stadard problem i statistical iferece, ad the first iferetial problem that we will tackle. For otatioal coveiece, idetify the measuremets o the sample as 1, 2,...,, where is the sample size. Give the data, our best guess, or estimate, of µ is the sample mea: Populatio Huge set of values Ca see very little Sample 1, 2,, Mea µ Stadard Deviatio σ µ ad σ ukow Iferece i Ȳ = i = There are two mai methods that are used for ifereces o µ: cofidece itervals (CI) ad hypothesis tests. The stadard CI ad test procedures are based o the sample mea ad the sample stadard deviatio, deoted by s. We will cosider CIs i this lecture, ad hypothesis tests i the ext lecture. Let s apply the results of the precedig sectio, ad the lay out the mechaics of the procedure. The mai idea behid a CI is this: should be a pretty good guess as to what µ is, but while µ is a costat (we do t kow the value, though), is a radom variable (every possible sample gives a differet value), so most assuredly µ. Still, should ot be too far from µ, but how far away from µ do we thik could be? As a specific example, suppose we radomly sample = 9 values from a ormal populatio ad get = 22 ad S = 6. What could µ be? To aswer such a questio, apply the t distributio. looks like a sigle radom umber sampled from a t distributio with 8 df, so it should have come out somewhere i the middle of that distributio. The middle 95% of that distributio is betwee ad (from the table). So, we had a 95% chace that µ SE would fall i that rage. Substitutig the actual values of ad S we obtaied, we are 95% cofidet that 22 µ 6/ = 22 µ 9 2 is betwee ad 2.306, or equivaletly we are 95% cofidet that 22 µ is betwee ad This says that µ should be withi of 22, or i the rage to , i.e. betwee ad We still do ot kow what µ is, but to have gotte data like this µ must be somewhere betwee ad The iterval µ is referred to as a 95% cofidece iterval for µ. It is improper to say there is a 95% chace that µ is i that rage: If it is i that rage, say 25, there is a 100% chace it is i that rage, while if it is ot i that rage, say 30, there is a 0% chace it is i that rage. The 95% refers to how ofte usig this techique works (like a lifetime battig average) - this iterval either worked i capturig µ or it did ot work, ad we caot kow which is true. µ SE 30

3 Mechaics of a CI for µ A CI for µ is a rage of plausible values for the ukow populatio mea µ, based o the observed data. To compute a CI for µ: 1. Specify the cofidece coefficiet, which is a umber betwee 0 ad 100%, i the form 100(1 α)%. Solve for α. 2. Compute the t critical value: t crit = t.5α such that the area uder the t curve (df = 1) to the right of t crit is.5α. 3. The desired CI has lower ad upper edpoits give by L = Ȳ t critse ad U = Ȳ + t crit SE, respectively, where SE = s/ is the stadard error of the sample mea. The CI is ofte writte i the form Ȳ ± t critse. I practice, the cofidece coefficiet is large, say 95% or 99%, which correspod to α =.05 ad.01, respectively. The value of α expressed as a percet is kow as the error rate of the CI. The CI is determied oce the cofidece coefficiet is specified ad the data are collected. Prior to collectig the data, the iterval is ukow ad is viewed as radom because it will deped o the actual sample selected. Differet samples give differet CIs. The cofidece i, say, the 95% CI (which has a 5% error rate) ca be iterpreted as follows. If you repeatedly sample the populatio ad costruct 95% CIs for µ, the 95% of the itervals will cotai µ, whereas 5% will ot. The iterval you costruct from your data will either cover µ, or it will ot. The legth of the CI U L = 2t crit SE depeds o the accuracy of our estimate of µ, as measured by SE = s/ the stadard error of. Less precise estimates of µ lead to wider itervals for a give level of cofidece. Assumptios for Procedures I described the classical CI. The procedure is based o the assumptios that the data are a radom sample from the populatio of iterest, ad that the populatio frequecy curve is ormal. The populatio frequecy curve ca be viewed as a smoothed histogram created from the populatio data. The ormality assumptio ca be checked usig a stem-ad-leaf display, a boxplot, or a ormal scores plot of the sample data (probably the more the better). Example Let us go through a had-calculatio of a CI, usig Miitab to geerate summary data. I will the show you how the CI is geerated automatically i Miitab. The ages (i years) at first trasplat for a sample of 11 heart trasplat patiets are as follows: Data Display AgeTra Stem-ad-Leaf Display: AgeTra 31

4 Stem-ad-leaf of AgeTra N = 11 Leaf Uit = (4) Descriptive Statistics: AgeTra Variable N N* Mea SE Mea StDev Miimum Q1 Media Q3 Maximum AgeTra The summaries for the data are: = 11, = 51.27, ad s = 8.26 so that SE = 8.26/ 11 = The degrees of freedom are df = 11 1 = 10. A ecessary first step i every problem is to defie the populatio parameter i questio. Here, let µ = mea age at time of first trasplat for populatio of patiets. Let us calculate a 95% CI for µ. The degrees of freedom are df = 11 1 = 10. For a 95% CI α =.05, so we eed to fid t crit = t.025 = Now t crit SE = = The lower limit o the CI is L = = The upper edpoit is U = = I isist that the results of every CI be summarized i words. For example, I am 95% cofidet that the populatio mea age at first trasplat is betwee 45.7 ad 56.8 years (roudig off to 1 decimal place). Miitab does all this very easily. Follow the meu path Stat > Basic Statistics > 1-Sample t (be careful that you do t select the 1-Sample Z it will treat S as if it is actually σ ad give you icorrect bouds). Uder Optios... select Cofidece Level of 95 (the default) ad Alterative: ot equal (we will uderstad that ext week). Uder Graphs check Boxplot. Do ot check Summarized data or Perform hypothesis test. ou get the followig results: Oe-Sample T: AgeTra Variable N Mea StDev SE Mea 95% CI AgeTra ( , ) We might be a little cocered about the outlier ad the possible skewess idicated i the boxplot below, sice that could be evidece we did ot sample from a ormal distributio. It will be worth tryig oe of the oparametric procedures we will lear about later, sice the assumptio of ormality is ot made there. 32

5 The Effect of α o a Two-Sided CI A two-sided 100(1 α)% CI for µ is give by ±t crit s/. The CI is cetered at ad has legth 2t crit s/. The cofidece coefficiet 100(1 α)% is icreased by decreasig α, which icreases t crit. That is, icreasig the cofidece coefficiet makes the CI wider. This is sesible: to icrease your cofidece that the iterval captures µ you must pipoit µ with less precisio by makig the CI wider. For example, a 95% CI is wider tha a 90% CI. SW Example 6.9 page 192: Let us compute a 90% ad a 95% CI by had. Note: For large the Cetral Limit Theorem gives us the ability to treat µ σ as a Z radom variable eve without samplig from a ormal distributio. Some texts would suggest usig the 1-Sample Z procedure i this case (although that still begs the issue of ot kowig σ). I practice what we do about large is to worry a little less about lack of ormality i the populatio we sampled from (outliers ad extreme skewess are still problems, just slightly differet oes), but cotiue to use the t-procedures. Remember for large we get large df, ad for large df there is little differece betwee Z ad t. Iferece for a Populatio Proportio Assume that you are iterested i estimatig the proportio p of idividuals i a populatio with a certai characteristic or attribute based o a radom or represetative sample of size from the populatio. The sample proportio ˆp =(# with attribute i the sample)/ is the best guess for p based o the data. This is the simplest categorical data problem. Each respose falls ito oe of two exclusive ad exhaustive categories, called success ad failure. Idividuals with the attribute of iterest are i the success category. The rest fall ito the failure category. Kowledge of the populatio proportio p of successes characterizes the distributio across both categories because the populatio proportio of failures is 1 p. As a aside, ote that the probability that a radomly selected idividual has the attribute of iterest is the populatio proportio p with the attribute, so the terms populatio proportio ad probability ca be used iterchageably with radom samplig. 33

6 The diagram of this is very similar to the earlier oe. Note that a radom sample of size ow becomes just a set of S s ad F s. Populatio Huge set of S s ad F s Ca see very little Sample Proportio of S is p p is ukow #S i Sample p^ = Iferece A CI for p The derivatio of the CI follows the same basic ideas as before, except we do ot have the idea of df sice we are cosiderig as large (p 5 ad (1 p) 5). ˆp is a radom variable (it almost surely is ot p), ad it looks like a sigle umber radomly selected from a ormal distributio p(1 p) with mea µˆp = p ad stadard deviatio σˆp =, so ˆp p looks like a Z. We have the same σˆp problem as before to use this as we wish, we eed to compute the deomiator, but we eed to kow p to compute it. We estimate it istead, ad call the estimated stadard deviatio of ˆp the ˆp(1 ˆp) stadard error of ˆp, SEˆp =. Everythig proceeds as before. A two-sided CI for p is a rage of plausible values for the ukow populatio proportio p, based o the observed data. To compute a two-sided CI for p: 1. Specify the cofidece level as the percet 100(1 α)% ad solve for the error rate α of the CI. 2. Compute z crit = z.5α (i.e. area uder the stadard ormal curve to the right of z crit is.5α.) 3. The 100(1 α)% CI for p has edpoits L = ˆp z crit SE ad U = ˆp + z crit SE, respectively, where the CI stadard error is ˆp(1 ˆp) SE =. The CI is ofte writte as ˆp ± z crit SE. The CI is determied oce the cofidece level is specified ad the data are collected. Prior to collectig data, the CI is ukow ad ca be viewed as radom because it will deped o the 34

7 actual sample selected. Differet samples give differet CIs. The cofidece i, say, the 95% CI (which has a.05 or 5% error rate) ca be iterpreted as follows. If you repeatedly sample the populatio ad costruct 95% CIs for p, the 95% of the itervals will cotai p, whereas 5% (the error rate) will ot. The CI you get from your data either covers p, or it does ot. The legth of the CI U L = 2z crit SE depeds o the accuracy of the estimate ˆp, as measured by the stadard error SE. For a give ˆp, this stadard error decreases as the sample size icreases, yieldig a arrower CI. For a fixed sample size, this stadard error is maximized at ˆp =.5, ad decreases as ˆp moves towards either 0 or 1. I essece, sample proportios ear 0 or 1 give arrower CIs for p. However, the ormal approximatio used i the CI costructio is less reliable for extreme values of ˆp. Example: The 1983 Tyleol poisoig episode highlighted the desirability of usig tamper-resistat packagig. The article Tamper Resistat Packagig: Is it Really? (Packagig Egieerig, Jue 1983) reported the results of a survey o cosumer attitudes towards tamper-resistat packagig. A sample of 270 cosumers was asked the questio: Would you be willig to pay extra for tamper resistat packagig? The umber of yes respodets was 189. Costruct a 95% CI for the proportio p of all cosumers who were willig i 1983 to pay extra for such packagig. Here = 270 ad ˆp = 189/270 =.700. The critical value for a 95% CI for p is z.025 = The CI stadard error is give by SE = =.028, so z crit SE = =.055. The 95% CI for p is.700 ±.055. ou are 95% cofidet that the proportio of cosumers willig to pay extra for better packagig is betwee.645 ad.755. (How much extra?). Appropriateess of the CI The stadard CI is based o a large sample stadard ormal approximatio to z = ˆp p SE. A simple rule of thumb requires p 5 ad (1 p) 5 for the method to be suitable. The populatio proportio p is ukow so you should use ˆp i these formulae to check the suitability of the CI. Give that ˆp ad (1 ˆp) are the observed umbers of successes ad failures, you should have at least 5 of each to apply the large sample CI. I the packagig example, ˆp = 270 (.700) = 189 (the umber who support the ew packagig) ad (1 ˆp) = 270 (.300) = 81 (the umber who oppose) both exceed 5. The ormal approximatio is appropriate here. 35

8 More Accurate Cofidece Itervals Large sample CIs for p should be iterpreted with cautio i small sized samples because the true error rate usually exceeds the assumed (omial) value. For example, a assumed 95% CI, with a omial error rate of 5%, may be oly a 80% CI, with a 20% error rate. The large sample CIs are usually overly optimistic (i.e. too arrow) whe the sample size is too small to use the ormal approximatio. SW use the followig method, origially suggested by Ala Agresti, for a 95% CI. The stadard method computes the sample proportio as ˆp = y/ where y is the umber of idividuals i the sample with the characteristic of iterest, ad is the sample size. Agresti suggested estimatig the proportio with p = (y + 2)/( + 4), with a stadard error of SE = p(1 p) + 4, ad usig the usual iterval with these ew summaries: p ± 1.96SE. This appears odd, but just amouts to addig two successes ad two failures to the observed data, ad the computig the stadard CI. This adjustmet has little effect whe is large ad ˆp is ot close to either 0 or 1, as i the Tyleol example. Let us do examples usig SW s proposed CI. SW Examples 6.16 ad 6.17, page Miitab Implemetatio A CI for p ca be obtaied i Miitab from summary data from the meu path Stat > Basic Statistics > 1 Proportio, check Summarized data, eter Number of trials () ad Number of evets (# Successes), click Optios, eter Cofidece level i percet (95.0 usually), igore Test proportio for ow, select Alterative: ot equal, ad check Use test ad iterval based o ormal distributio. The above choices produce a CI based upo ˆp. I order to use SW s CI based o p, add 4 to ad 2 to # Successes. Fially, to get the best iterval (arguably the correct oe), do ot check Use test ad iterval based o ormal distributio. This third choice produces what is kow as a exact iterval it is a lot harder to explai how we get it (I ll idicate where it comes from ext week), but the cofidece level ad error rate are correct ad ot subject to approximatio like the other two itervals. Miitab is a little uique i providig this. Let us examie Miitab results from two examples: 36

9 The Tyleol Example: Usig ˆp: Sample X N Sample p 95% CI Z-Value P-Value ( , ) Usig p: Sample X N Sample p 95% CI Z-Value P-Value ( , ) Usig exact iterval: Exact Sample X N Sample p 95% CI P-Value ( , ) Igore the Z-Value ad P-Value etries for ow. ou ca see that the itervals all agree for ay practical iterpretatio. Example 6.17 p. 209 of SW Usig ˆp: Sample X N Sample p CI Z-Value P-Value (*, *) * NOTE * The ormal approximatio may be iaccurate for small samples. Usig p: Sample X N Sample p 95% CI Z-Value P-Value ( , ) * NOTE * The ormal approximatio may be iaccurate for small samples. Usig exact iterval: Exact Sample X N Sample p 95% CI P-Value ( , ) The oly oe of these I would trust is the exact oe. The oe based o p is surprisigly iformative, though. Miitab s warig o the other two should ot be igored. 37