Inference for One Numerical Population: ContinuedA Test of Hypotheses for Estimating the Median of a pdfExamples with Real DataEstimating the Median of a Count ResponsePredictionPrediction for a Normal pdfDistribution-Free PredictionWhich Method Should be Used?Some Cautionary TalesYou Need More Than a Random SampleCross-sectional Versus Longitudinal StudiesAnother Common DifficultySummaryPractice ProblemsSolutions to Practice ProblemsHomework ProblemsChapter 18Inference for On e Numerical Population:ContinuedChapter 17 did most of the heavy lifting for inference for one numerical population. By compari-son, this chapter is pretty u ser-friendly.18.1 A Test of Hypothes es for µThis section is very similar to Section 12.5, which presented a test of hypoth eses for a binomial p.Again, the i d ea is that of all of the possible values of µ, there is one value of special i nterest tothe researcher. This known special value o f interest is denoted by µ0and the nu ll hypothesis isthat µ = µ0. As in Chapter 12, the justification for the value µ0is: history; theory; or contractsor law. As in Chapter 12, the test of this section i s not terribly useful in s ci ence. Recall that thevery important McNemar’s test of Chapter 16 was a special case of the not-so-import ant test ofChapter 12. Similarly, Chapter 20 will present an important use for th e test of this section.As usual, there are three possibilities for the alternative:µ > µ0; µ < µ0; or µ 6= µ0.As in Chapter 17, we assume that our data will consi st of n i.i.d. random variables:X1, X2, X3, . . . , Xn,with summary random variables¯X and S, the mean and standard deviation of these variables. Theobserved values of these guys are:x1, x2, x3, . . . , xn, ¯x and s.Because our hypoth eses involve the mean of the popul ation, the obvious and natural choice for thetest statistic is¯X, with observed value ¯x . In order t o obtain an approximate samplin g di stributionwe standardize¯X and obtainZ =¯X − µσ/√n.469We don’t yet have our test statistic; there is a flaw inherent in this Z: we don ’t know the values µand σ. Handling σ is easy enough; w e replace it in Z with S, givingZ′=¯X − µS/√n.We will follow our approach of Chapter 17 and use Go sset’s t-curve with df = n − 1 to obtainapproximate probabilit ies for Z′. But Z′is not a test statistic because we don’t know the value of µ.Just in time, we recall that we want to know how the test statistic behaves on the assumption thatthe null hypothesis is true. Given that the null hypothesis is true, we can replace the unknown µin Z′with th e known µ0. The result is our test statist ic:T =¯X − µ0S/√n; (18.1)after data are collected, the observed value of T ist =¯x − µ0s/√n; (18.2)The three rules for finding the P-value are similar to earlier rules and are su mmarized in the fol low-ing result. The website we are using in these Course Notes gives areas to the left under a t-curve.In t he items listed b el ow, I include an equivalent area to the right rule.Result 18. 1 In the formulas below, t is given in Equation 18.2 and areas are computed under thet-curve with df = n − 1.1. For the alternative µ > µ0, the app roximate P-value equals the area to the right of t. If youprefer, the approximate P-value equals the area to the left of −t.2. For the alternative µ < µ0, the approximate P-value equals the area to the left of t. If youprefer, the approximate P-value equals the area to the right of −t.3. For t he alternative µ 6= µ0, the approximate P -val ue equals twice the area to the right of |t|.If you prefer, the approxim ate P-value equals twice the area to the left of −|t|.I will illu strate the use of these rules.Suppose that we haveµ0= 20 , n = 16, ¯x = 23.00 and s = 8.00.First, we use Equation 18.2 to obtain the observed value of the test statistic:t =¯x − µ0s/√n=23.00 − 208.00/√16= 3/ 2 = 1.50.Using t he website introduced in Chapter 17:470http://stattrek.com/online-calculator/t-distribution.aspxand the rules above:• For the alternative >: enter n − 1 = 16 − 1 = 15 for the degrees of freedom; enter −t =−1.50 in the t score box; and click on Calculate. The approximate P-value, 0.0772, appearsin the Cumulative probability box.• For the alternative <: leave 15 for the degrees of freedom; enter t = 1.50 in the t scorebox; and click o n Calculat e. The approximate P-value, 0.9228, appears in the Cumulativeprobability box.• For the alternative 6=: the approximate P-value equals twice the area to the left of −|t| =−1.50. From the above, we know that th is area equals 0.0772. Thus, the approximate P-value equals 2(0.0772) = 0.1544.If you believe that the population i s symm et ri c or app ro x imately symmetric, then the approxi-mate P-values given above should be reasonably accurate, even for relatively small values o f n.If you suspect that the population is strongly skewed and your alternative is two-sided (6=),my advice is to use the above rules if your n is very large. Of course, very l arge is vague; theguidelines we had in Chapter 17 —i.e., how large depends on how skewed—are fine here too.If, however, you suspect that the population i s strongly skewed and your alternative is one-sided (> or <), th en my advice is to never use the rules above. I don’t have the time to explainwhy, but it’s related to the fact that for a population that is strongly skewed to the ri ght [left] theincorrect confidence intervals are too small [large] much more often than they are t oo large [small].This translates into the approximate P-value being either much too large or much too small.18.2 Estimatin g the Median of a pdfRecall that a fundamental feature of a pdf is that the to tal area und er it is equal to 1 . It thus followsthat there exists a number ν (pronounced new) with the following property.The area under the pdf to the left (and right) of ν is equal to 0.5.The number ν is called the median of the pdf, for rather obvious reasons. Note my use of thedefinite article: the median. I am being a bit dishonest here. Let me explain. For every pdf wehave seen, including all the families of pdfs mentioned in Chapter 17, there is a unique number, ν,with th e property that :the area under th e pdf to the left (and righ t) of ν is equal to 0.5.It is possible mathematically, however, for there to be an interval of numbers with this property.Figure 18.1 presents a pdf (a combination of two
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