9.2 Confidence Intervals for Means

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1 202 CHAPTER 9. ESTIMATION 9.2 Cofidece Itervals for Meas We are give X 1, X 2,..., X that are a S RS ( from a orm(mea = µ, sd = σ distributio, where µ is ukow. We kow that we may estimate µ with X, ad we have see that this estimator is the MLE. But how good is our estimate? We kow that X µ σ/ orm(mea = 0, sd = 1. (9.2.1 For a big probability 1 α, for istace, 95%, we ca calculate the quatile z α/2. The IP z α/2 X µ σ/ z α/2 = 1 α. (9.2.2 But ow cosider the followig strig of equivalet iequalities: z α/2 ( σ X z α/2 ( σ X z α/2 ( σ z α/2 X µ σ/ z α/2, X µ z α/2 ( σ, µ X + z α/2 ( σ, µ X + z α/2 ( σ. That is, σ σ IP (X z α/2 µ X + z α/2 = 1 α. (9.2.3 Defiitio The iterval [X z α/2 σ, X + z α/2 σ ] (9.2.4 is a 100(1 α% cofidece iterval for µ. The quatity 1 α is called the cofidece coefficiet. Remark The iterval is also sometimes writte more compactly as X ± z α/2 σ. (9.2.5 The iterpretatio of cofidece itervals is tricky ad ofte mistake by ovices. Whe I am teachig the cocept live durig class, I usually ask the studets to imagie that my piece of chalk represets the ukow parameter, ad I lay it dow o the desk i frot of me. Oce the chalk has bee lai, it is fixed; it does ot move. Our goal is to estimate the parameter. For the estimator I pick up a sheet of loose paper lyig earby. The estimatio procedure is to radomly drop the piece of paper from above, ad observe where it lads. If the piece of paper covers the piece of chalk, the we are successful our estimator covers the parameter. If it falls off to oe side or the other, the we are usuccessful; our iterval fails to cover the parameter.

2 9.2. CONFIDENCE INTERVALS FOR MEANS 203 The I ask them: suppose we were to repeat this procedure hudreds, thousads, millios of times. Suppose we kept track of how may times we covered ad how may times we did ot. What percetage of the time would we be successful? I the demostratio, the parameter correspods to the chalk, the sheet of paper correspods to the cofidece iterval, ad the radom experimet correspods to droppig the sheet of paper. The percetage of the time that we are successful exactly correspods to the cofidece coefficiet. That is, if we use a 95% cofidece iterval, the we ca say that, i the log ru, approximately 95% of our itervals will cover the true parameter (which is fixed, but ukow. See Figure 9.2.1, which is a graphical display of these ideas. Uder the above framework, we ca reaso that a iterval with a larger cofidece coefficiet correspods to a wider sheet of paper. Furthermore, the width of the cofidece iterval (sheet of paper should be somehow related to the amout of iformatio cotaied i the radom sample, X 1, X 2,..., X. The followig remarks makes these otios precise. Remark For a fixed cofidece coefficiet 1 α, if icreases, the the cofidece iterval gets SHORTER. (9.2.6 Remark For a fixed sample size, if 1 α icreases, the the cofidece iterval gets WIDER. (9.2.7 Example Results from a Experimet o Plat Growth. The PlatGrowth data frame gives the results of a experimet to measure plat yield (as measured by the weight of the plat. We would like to a 95% cofidece iterval for the mea weight of the plats. Suppose that we kow from prior research that the true populatio stadard deviatio of the plat weights is 0.7 g. The parameter of iterest is µ, which represets the true mea weight of the populatio of all plats of the particular species i the study. We will first take a look at a stemplot of the data: > library(aplpack > with(platgrowth, stem.leaf(weight 1 2: represets 1.2 leaf uit: 0.1 : 30 1 f 5 s * 11 5 t 3 8 f s (4 5* t f 555 s * t 3

3 204 CHAPTER 9. ESTIMATION Cofidece itervals based o z distributio Idex Cofidece Iterval Figure 9.2.1: Simulated cofidece itervals The graph was geerated by the ci.examp fuctio from the TeachigDemos package. Fifty (50 samples of size twety five (25 were geerated from a orm(mea = 100, sd = 10 distributio, ad each sample was used to fid a 95% cofidece iterval for the populatio mea usig Equatio The 50 cofidece itervals are represeted above by horizotal lies, ad the respective sample meas are deoted by vertical slashes. Cofidece itervals that cover the true mea value of 100 are plotted i black; those that fail to cover are plotted i a lighter color. I the plot we see that oly oe (1 of the simulated itervals out of the 50 failed to cover µ = 100, which is a success rate of 98%. If the umber of geerated samples were to icrease from 50 to 500 to 50000,..., the we would expect our success rate to approach the exact value of 95%.

4 9.2. CONFIDENCE INTERVALS FOR MEANS 205 The data appear to be approximately ormal with o extreme values. The data come from a desiged experimet, so it is reasoable to suppose that the observatios costitute a simple radom sample of weights 3. We kow the populatio stadard deviatio σ = 0.70 from prior research. We are goig to use the oe-sample z-iterval. > dim(platgrowth # sample size is first etry [1] 30 2 > with(platgrowth, mea(weight [1] > qorm(0.975 [1] We fid the sample mea of the data to be x = ad z α/2 = z Our iterval is therefore x ± z α/2 σ = ± , which comes out to approximately [4.823, 5.323]. I coclusio, we are 95% cofidet that the true mea weight µ of all plats of this species lies somewhere betwee g ad g, that is, we are 95% cofidet that the iterval [4.823, 5.323] covers µ. See Figure Example Give some data with X 1, X 2,..., X a S RS ( from a orm(mea = µ, sd = σ distributio. Maybe small sample? 1. What is the parameter of iterest? i the cotext of the problem. Give a poit estimate for µ. 2. What are the assumptios beig made i the problem? Do they meet the coditios of the iterval? 3. Calculate the iterval. 4. Draw the coclusio. Remark What if σ is ukow? We istead use the iterval X ± z α/2 S, (9.2.8 where S is the sample stadard deviatio. If is large, the X will have a approximately ormal distributio regardless of the uderlyig populatio (by the CLT ad S will be very close to the parameter σ (by the SLLN; thus the above iterval will have approximately 100(1 α% cofidece of coverig µ. If is small, the 3 Actually we will see later that there is reaso to believe that the observatios are simple radom samples from three distict populatios. See Sectio 10.6.

5 206 CHAPTER 9. ESTIMATION 95% Normal Cofidece Limits: σ x = 0.128, = 30 µ x f(z g( x = f(( x µ i σ x σ x x x z z shaded area Cof Level= Figure 9.2.2: Cofidece iterval plot for the PlatGrowth data The shaded portio represets 95% of the total area uder the curve, ad the upper ad lower bouds are the limits of the oe-sample 95% cofidece iterval. The graph is cetered at the observed sample mea. It was geerated by computig a z.test from the TeachigDemos package, storig the resultig htest object, ad plottig it with the ormal.ad.t.dist fuctio from the HH package. See the remarks i the How to do it with R discussio later i this sectio.

6 9.2. CONFIDENCE INTERVALS FOR MEANS 207 If the uderlyig populatio is ormal the we may replace z α/2 with t α/2 (df = 1. The resultig 100(1 α% cofidece iterval is X ± t α/2 (df = 1 S (9.2.9 if the uderlyig populatio is ot ormal, but approximately ormal, the we may use the t iterval, Equatio The iterval will have approximately 100(1 α% cofidece of coverig µ. However, if the populatio is highly skewed or the data have outliers, the we should ask a professioal statisticia for advice. The author leared of a hady acroym from AP Statistics Exam graders that summarizes the importat parts of cofidece iterval estimatio, which is PANIC: Parameter, Assumptios, Name, Iterval, ad Coclusio. Parameter: idetify the parameter of iterest with the proper symbols. Write dow what the parameter meas i the cotext of the problem. Assumptios: list ay assumptios made i the experimet. If there are ay other assumptios eeded or that were ot checked, state what they are ad why they are importat. Name: choose a statistical procedure from your bag of tricks based o the aswers to the previous two parts. The assumptios of the procedure you choose should match those of the problem; if they do ot match the either pick a differet procedure or opely admit that the results may ot be reliable. Write dow ay uderlyig formulas used. Iterval: calculate the iterval from the sample data. This ca be doe by had but will more ofte be doe with the aid of a computer. Regardless of the method, all calculatios or code should be show so that the etire process is repeatable by a subsequet reader. Coclusio: state the fial results, usig laguage i the cotext of the problem. Iclude the appropriate iterpretatio of the iterval, makig referece to the cofidece coefficiet. Remark All of the above itervals for µ were two-sided, but there are also oe-sided itervals for µ. They look like [ σ X z α, or (, X + z α σ ] ( ad satisfy ( ( σ σ IP X z α µ = 1 α ad IP X + z α µ = 1 α. ( Example Small sample, some data with X 1, X 2,..., X a S RS ( from a orm(mea = µ, sd = σ distributio. 1. PANIC

7 208 CHAPTER 9. ESTIMATION How to do it with R We ca do Example 9.14 with the followig code. > library(teachigdemos > temp <- with(platgrowth, z.test(weight, stdev = 0.7 > temp Oe Sample z-test data: weight z = , = , Std. Dev. = 0.700, Std. Dev. of the sample mea = 0.128, p-value < 2.2e-16 alterative hypothesis: true mea is ot equal to 0 95 percet cofidece iterval: sample estimates: mea of weight The cofidece iterval bouds are show i the sixth lie dow of the output (please disregard all of the additioal output iformatio for ow we will use it i Chapter 10. We ca make the plot for Figure with > library(ipsur > plot(temp, "Cof" 9.3 Cofidece Itervals for Differeces of Meas Let X 1, X 2,..., X be a S RS ( from a orm(mea = µ X, sd = σ X distributio ad let Y 1, Y 2,..., Y m be a S RS (m from a orm(mea = µ Y, sd = σ Y distributio. Further, assume that the X 1, X 2,..., X sample is idepedet of the Y 1, Y 2,..., Y m sample. Suppose that σ X ad σ Y are kow. We would like a cofidece iterval for µ X µ Y. We kow that X Y orm mea = µ X µ Y, sd = σ 2 X + σ2 Y m. (9.3.1 Therefore, a 100(1 α% cofidece iterval for µ X µ Y is give by ( X Y ± zα/2 σ 2 X + σ2 Y m. (9.3.2 Ufortuately, most of the time the values of σ X ad σ Y are ukow. This leads us to the followig: If both sample sizes are large, the we may appeal to the CLT/SLLN (see 8.3 ad substitute S 2 X ad S Y 2 for σ2 X ad σ2 Y i the iterval The resultig cofidece iterval will have approximately 100(1 α% cofidece.

8 CONFIDENCE INTERVALS FOR DIFFERENCES OF MEANS 209 If oe or more of the sample sizes is small the we are i trouble, uless the uderlyig populatios are both ormal ad σ X = σ Y. I this case (settig σ = σ X = σ Y, 1 X Y orm mea = µ X µ Y, sd = σ + 1 m. (9.3.3 Now let U = 1 σ S 2 2 X + m 1 S σ Y. 2 ( The by Exercise 7.2 we kow that U chisq(df = + m 2 ad is ot a large leap to believe that U is idepedet of X Y; thus Z T = t(df = + m 2. (9.3.5 U/ ( + m 2 But X Y (µ X µ Y T = = σ 1 + m 1 S 2 σ 2 X + /, m 1 S 2 σ 2 Y ( + m 2 X Y (µ X µ Y (, ( 1S 2 X +(m 1S Y 2 +m 2 1 ( m t(df = + m 2. Therefore a 100(1 α% cofidece iterval for µ X µ Y is give by ( 1 X Y ± tα/2 (df = + m 2 S p + 1 m, (9.3.6 where ( 1S 2 X S p = + (m 1S Y 2 + m 2 is called the pooled estimator of σ. (9.3.7

9 210 CHAPTER 9. ESTIMATION How to do it with R The basic fuctio is t.test which has a var.equal argumet that may be set to TRUE or FALSE. The cofidece iterval is show as part of the output, although there is a lot of additioal iformatio that is ot eeded util Chapter 10. There is ot ay specific fuctioality to hadle the z-iterval for small samples, but if the samples are large the t.test with var.equal = FALSE will be essetially the same thig. The stadard deviatios are ever (? kow i advace ayway so it does ot really matter i practice. 9.4 Cofidece Itervals for Proportios We would like to kow p which is the proportio of successes. For istace, p could be: the proportio of U.S. citizes that support Obama, the proportio of smokers amog adults age 18 or over, the proportio of people worldwide ifected by the H1N1 virus. We are give a S RS ( X 1, X 2,..., X distributed biom(size = 1, prob = p. Recall from Sectio 5.3 that the commo mea of these variables is IE X = p ad the variace is IE(X p 2 = p(1 p. If we let Y = X i, the from Sectio 5.3 we kow that Y biom(size =, prob = p ad that X = Y p(1 p has IE X = p ad Var(X =. Thus if is large (here is the CLT the a approximate 100(1 α% cofidece iterval for p would be give by p(1 p X ± z α/2. (9.4.1 OOPS...! Equatio is of o use to us because the ukow parameter p is i the formula! (If we kew what p was to plug i the formula the we would ot eed a cofidece iterval i the first place. There are two solutios to this problem. 1. Replace p with ˆp = X. The a approximate 100(1 α% cofidece iterval for p is give by ˆp(1 ˆp ˆp ± z α/2. (9.4.2 This approach is called the Wald iterval ad is also kow as the asymptotic iterval because it appeals to the CLT for large sample sizes. 2. Go back to first priciples. Note that z α/2 Y/ p p(1 p/ z α/2 exactly whe the fuctio f defied by f (p = (Y/ p 2 z 2 α/2 p(1 p

10 For two proportios p 1 ad p 2, we may collect idepedet biom(size = 1, prob = p samples of size 1 ad 2, respectively. Let Y 1 ad Y 2 deote the umber of successes i the respective samples. We kow that Y 1 p1 (1 p 1 orm mea = p 1, sd = 1 ad Y 2 p2 (1 p 2 orm mea = p 2, sd = 2 so it stads to reaso that a approximate 100(1 α% cofidece iterval for p 1 p 2 is give by ˆp1 (1 ˆp 1 ( ˆp 1 ˆp 2 ± z α/2 + ˆp 2(1 ˆp 2, ( where ˆp 1 = Y 1 / 1 ad ˆp 2 = Y 2 / 2. Remark Whe estimatig a sigle proportio, oe-sided itervals are sometimes eeded. They take the form ˆp(1 ˆp 0, ˆp + z α/2 (9.4.5 or ˆp(1 ˆp ˆp z α/2, 1 (9.4.6 or i other words, we kow i advace that the true proportio is restricted to the iterval [0, 1], so we ca trucate our cofidece iterval to those values o either side How to do it with R > library(hmisc > bicof(x = 7, = 25, method = "asymptotic" PoitEst Lower Upper > bicof(x = 7, = 25, method = "wilso" PoitEst Lower Upper

11 212 CHAPTER 9. ESTIMATION The default value of the method argumet is wilso. A alterate way is > tab <- xtabs(~geder, data = RcmdrTestDrive > prop.test(rbid(tab, cof.level = 0.95, correct = FALSE 1-sample proportios test without cotiuity correctio data: rbid(tab, ull probability 0.5 X-squared = 2.881, df = 1, p-value = alterative hypothesis: true p is ot equal to percet cofidece iterval: sample estimates: p > A <- as.data.frame(titaic > library(reshape > B <- with(a, utable(a, Freq 9.5 Cofidece Itervals for Variaces I am thikig oe ad two sample problems here How to do it with R I am thikig about sigma.test i the TeachigDemos package ad var.test i base R here. 9.6 Fittig Distributios How to do it with R I am thikig about fitdistr from the MASS package [84]. 9.7 Sample Size ad Margi of Error Sectios 9.2 through 9.5 all bega the same way: we were give the sample size ad the cofidece coefficiet 1 α, ad our task was to fid a margi of error E so that Some examples we saw were: ˆθ ± E is a 100(1 α% cofidece iterval for θ. E = z α/2 σ/, i the oe-sample z-iterval, E = t α/2 (df = + m 2S p 1 + m 1, i the two-sample pooled t-iterval.

12 9.7. SAMPLE SIZE AND MARGIN OF ERROR 213 We already kow (we ca see i the formulas above that E decreases as icreases. Now we would like to use this iformatio to our advatage: suppose that we have a fixed margi of error E, say E = 3, ad we wat a 100(1 α% cofidece iterval for µ. The questio is: how big does have to be? For the case of a populatio mea the aswer is easy: we set up a equatio ad solve for. Example Give a situatio, give σ, give E, we would like to kow how big has to be to esure that X ± 5 is a 95% cofidece iterval for µ. Remark Always roud up ay decimal values of, o matter how small the decimal is. 2. Aother ame for E is the maximum error of the estimate. For proportios, recall that the asymptotic formula to estimate p was Reasoig as above we would wat ˆp(1 ˆp ˆp ± z α/2. ˆp(1 ˆp E = z α/2, or (9.7.1 = z 2 ˆp(1 ˆp α/2. (9.7.2 E 2 OOPS! Recall that ˆp = Y/, which would put the variable o both sides of Equatio Agai, there are two solutios to the problem. 1. If we have a good idea of what p is, say p the we ca plug it i to get = z 2 p (1 p α/2. (9.7.3 E 2 2. Eve if we have o idea what p is, we do kow from calculus that p(1 p 1/4 because the fuctio f (x = x(1 x is quadratic (so its graph is a parabola which opes dowward with maximum value attaied at x = 1/2. Therefore, regardless of our choice for p the sample size must satisfy = z 2 p (1 p α/2 z2 α/2 E 2 4E. ( The quatity z 2 α/2 /4E2 is large eough to guaratee 100(1 α% cofidece. Example Proportio example Remark For very small populatios sometimes the value of obtaied from the formula is too big. I this case we should use the hypergeometric distributio for a samplig model rather tha the biomial model. With this modificatio the formulas chage to the followig: if N deotes the populatio size the let m = z 2 p (1 p α/2 (9.7.5 E 2 ad the sample size eeded to esure 100(1 α% cofidece is achieved is m 1 + m 1 N If we do ot have a good value for the estimate p the we may use p = 1/2. =. (9.7.6

Chapter 22. Comparing Two Proportions. Copyright 2010 Pearson Education, Inc.

Chapter 22. Comparing Two Proportions. Copyright 2010 Pearson Education, Inc. Chapter 22 Comparig Two Proportios Copyright 2010 Pearso Educatio, Ic. Comparig Two Proportios Comparisos betwee two percetages are much more commo tha questios about isolated percetages. Ad they are more