Statistical Inference

Transcription

1 Statistical Iferece Statistical Iferece Statistical iferece provides us with the process of testig hypotheses uder ivestigatio from data. Procedures of statistical iferece are give for i parameter estimates ad their cofidece itervals, ad ii tests of statistical hypotheses. Some commo tests, called t-test, z-test, ad χ 2 -test, are associated respectively with t-distributio, ormal distributio, ad χ 2 -distributio. Populatio distributio ad parameters. A radom sample X 1,..., X is regarded as idepedet ad idetically distributed iid radom variables govered by pdf fx; θ. A value θ which is a vector of real umbers i geeral represets the characteristics of this uderlyig populatio distributio, ad is called a parameter. Suppose, for example, that the uderlyig distributio is the ormal distributio with µ, σ 2. The the values µ ad σ 2 are the parameters. Statistics ad poit estimates. A radom sample is viewed as a radom vector X = X 1,..., X, ad a radom variable ux costructed from the radom vector X is called a statistic. For example, the sample mea X is a statistic. A poit estimate is a statistic ux which is a best guess for the true value θ. Suppose that the uderlyig distributio is the ormal distributio with µ, σ 2. The the sample mea X is i some sese a best guess of the parameter µ. Risk fuctio ad bias. Let ux be a poit estimate for θ. The the fuctioal Rθ, u = E[uX θ 2 ] of u is called the mea square-error risk fuctio. We ca immediately observe that Rθ, u = VaruX + [EuX θ] 2 = VaruX + [bθ, u] 2, where bθ, u = EuX θ is called the bias of ux. Oe of the importat attributes of poit estimate is ubiasedess. Sice a statistic ux is a radom variable, we ca cosider the expectatio E[uX]. The the poit estimate ux of θ is called a ubiased estimator if it satisfies E[uX] = θ. For example, the sample mea X is a ubiased estimate of the mea µ, sice E X = µ. Cofidece iterval. Let X = X 1,..., X be a radom sample from fx; θ. Let u 1 X ad u 2 X be statistics satisfyig u 1 X u 2 X. If P u 1 X < θ < u 2 X = 1 α for every θ, the the radom iterval u 1 X, u 2 X is called a cofidece iterval of level 1 α. Populatio mea uder ormal assumptio. Let X 1,..., X be a radom sample from Nµ, σ. The sample mea X is a ubiased estimate of the parameter µ. The the radom variable X µ S/ has the t-distributio with 1 degrees of freedom. Thus, by usig the critical poit t α/2, 1 X µ P S/ < t α/2, 1 = P X t α/2, 1S < µ < X + t α/2, 1S Page 1 Probability ad Statistics II/April 13, 2018

2 Statistical Iferece has the probability of 1 α. This implies that the parameter µ is i the iterval X t α/2, 1S t α/2, 1 S, X + with probability 1 α. The iterval is also kow as the t-iterval. Normal assumptio is ot ecessary. Eve if a radom sample X 1,..., X is ot ormally distributed, the cetral limit theorem says that the estimate X is approximately distributed as Nµ, σ 2 / whe is large. I either case it is sesible to costruct a cofidece iterval with critical poit t α/2, 1 from t-distributio. Example 1. A radom sample of milk cotaiers is selected, ad their milk cotets are weighed. The data X 1,..., X 3.1 ca be used to ivestigate the ukow populatio mea of the milk cotaier weights. A radom sample ca be assumed to be iid ormal distributio. Suppose that we have calculated X = ad S = from the actual data with = 30. The costruct 95% cofidece iterval. Solutio. By choosig α = 0.05, we have the critical poit t 0.025,29 = 2.045, ad therefore, obtai the cofidece iterval , = 2.046, of level 0.95 or, of level 95%. Populatio proportio. Let X 1,..., X be iid Beroulli radom variables with success probability p. The sample mea X is a estimate of the parameter p, ad ubiased. By the cetral limit theorem, the radom variable X p p1 p/ has approximately N0, 1 as gets larger at least p > 5 ad 1 p > 5 by rule of thumb. Critical poit. Here we defie the critical poit z α for stadard ormal distributio by P X > z α = α with stadard ormal radom variable X. Cofidece iterval. X p P < z α/2 p1 p/ p1 p = P X z α/2 < p < X p1 p + z α/2 X1 X has the approximate probability of 1 α. Here we ca use as a estimate for Together we obtai the cofidece iterval X1 X X1 X X z α/2, X + zα/2 p1 p. Page 2 Probability ad Statistics II/April 13, 2018

3 Statistical Iferece of level 1 α. Alterative cofidece iterval. There is a alterative ad possibly more accurate method to derive a cofidece iterval. Here we observe that X p P < z α/2 p1 p/ = P + z 2 α/2 p 2 2 X + z 2 α/2/2 p + X 2 < 0 1 α. This implies that the parameter p is i the iterval ˆp, ˆp + with probability 1 α, where X + zα/2 2 /2 ± z α/2 X1 X + zα/2 2 /4 ˆp ± =. + zα/2 2 Cocept of statistical hypotheses. Suppose that a researcher is iterested i whether the ew drug works. The process of determiig whether the outcome of the experimet poits to yes or o is called hypothesis testig. A widely used formalizatio of this process is due to Neyma ad Pearso. Our hypothesis is the the ull hypothesis that the ew drug has o effect the ull hypothesis is ofte the reverse of what we actually believe, why? Because the researcher hopes to reject the hypothesis ad aouce that the ew drug leads to sigificat improvemets. If the hypothesis is ot rejected, the researcher aouces othig ad goes o to a ew experimet. Hypothesis test for populatio mea. Hospital workers are subject to a radiatio exposure emaatig from the ski of the patiet. A researcher is iterested i the plausibility of the statemet that the populatio mea µ of radiatio level is µ 0 the researcher s hypothesis. The the ull hypothesis is H 0 : µ = µ 0. The opposite of the ull hypothesis, called a alterative hypothesis, becomes H A : µ µ 0. Thus, the hypothesis testig problem H 0 versus H A is formed. The problem here is to whether or ot to reject H 0 i favor of H A. Mechaism of rejectig H 0. To assess this hypothesis, the radiatio levels X 1,..., X are measured from patiets who had bee ijected with a radioactive tracer, ad assumed to be idepedet ad ormally distributed with the mea µ. Uder the ull hypothesis, the radom variable T = X µ 0 S/ has the t-distributio with 1 degrees of freedom. Thus, we obtai the exact probability P T t α/2, 1 = α. Whe α is chose to be a small value 0.05 or 0.01, for example, it is ulikely that the absolute value T is larger tha the critical poit t α/2, 1. The we say that the ull hypothesis H 0 is rejected with sigificace level α or, size α whe the observed value t of T satisfies t > t α/2, 1. Page 3 Probability ad Statistics II/April 13, 2018

4 Statistical Iferece Example 2. We have µ 0 = 5.4 for the hypothesis, ad decided to give a test with sigificace level α = Suppose that we have obtaied X = ad S = from the actual data with = 28. Solutio. The we ca compute T = / Sice T = 1.79 t 0.025,27 = 2.052, the ull hypothesis caot be rejected. Thus, the evidece agaist the ull hypothesis is ot persuasive. Test statistic ad p-value. The above radom variable T is called the t-statistic, or test statistic. Havig observed T = t, we ca calculate the p-value p = P Y t = 2 P Y t, where the radom variable Y has a t-distributio with 1 degrees of freedom. The we have the relatio p < α t > t α/2, 1. Thus, we reject H 0 with sigificace level α whe p < α. I the above example, we ca compute the p-value thus, we caot reject H 0. p = 2 P Y ; Oe-sided hypothesis test. I the same case of hospital workers subject to a radiatio exposure, this time the researcher is iterested i the plausibility of the statemet that the populatio mea µ is greater tha µ 0. The the hypothesis testig problem is H 0 : µ = µ 0 versus H A : µ < µ 0. The same t-statistic T = X µ 0 S/ is used as a test statistic. Ad we reject H 0 with sigificat level α whe you fid that t < t α, 1 for the observed value t of T. Oe-sided hypothesis test, cotiued. Alteratively we ca costruct the p-value p = P Y t, where the radom variable Y has a t-distributio with 1 degrees of freedom. Because of the relatio p < α t < t α, 1, we ca reject H 0 with sigificat level α whe p < α. Example 3. We use the same µ 0 = 5.4 for the hypothesis ad the same sigificace level α = 0.05, but use the oe-sided test. Recall that X = ad S = were obtaied from the data with = 28. Solutio. a The we compute T = / Sice T = 1.79 < t 0.05,27 = 1.703, the ull hypothesis H 0 is rejected. Thus, the outcome is statistically sigificat so that the populatio mea µ is smaller tha 5.4. Page 4 Probability ad Statistics II/April 13, 2018

5 Statistical Iferece b Alteratively, we ca fid the p-value p = P Y < 0.05; thus, the ull hypothesis should be rejected. Oe-sided hypothesis test: Opposite case. We ca also cosider the hypothesis testig problem H 0 : µ = µ 0 versus H A : µ > µ 0. a Usig the t-statistics T = X µ 0 S/, we ca reject H 0 with sigificat level α whe the observed value t of T satisfies t > t α, 1. b Alteratively we ca costruct the p-value p = P Y t with the radom variable Y has a t-distributio with 1 degrees of freedom. Because of the relatio p < α t > t α, 1, we ca reject H 0 whe p < α. Summary. Whe the ull hypothesis H 0 is rejected, it is reasoable to fid out the cofidece iterval of the populatio mea µ. The followig table shows the cofidece iterval we ca costruct whe your ull hypothesis is rejected. Here T = X µ 0 S/ is the test statistic, ad α is the sigificace level of your choice. Test Whe to reject 1 α-level cofidece iterval H A : µ µ 0 T > t α/2, 1 X S tα/2, 1, X + S tα/2, 1 H A : µ > µ 0 T > t α, 1 X S tα, 1, H A : µ < µ 0 T < t α, 1, X + S tα, 1 Type I error: Two-sided test. We defie a fuctio Kθ of parameter θ by the probability that H 0 is rejected give µ = θ. Kθ = P Reject H 0 µ = θ is called the power fuctio. What is the probability that we icorrectly reject H 0 whe it is actually true? Such a error is called type I error, ad the probability of type I error is exactly the sigificat level α, as explaied i the followig: The probability of type I error for the two-sided hypothesis test is give by Kµ 0. The we have Kµ 0 = P T t α/2, 1 = α. Type I error: Oe-sided test. I oe-sided hypothesis test, the probability of type I error is the worst that is, largest possible probability max Kθ of type I error. Give µ = θ, the θ µ 0 radom variable X θ S/ = T θ µ 0 S/ = T δ Page 5 Probability ad Statistics II/April 13, 2018

6 Statistical Iferece has the t-distributio with 1 degrees of freedom, where δ = θ µ 0 S/. By observig that δ 0 if θ µ 0, we obtai Thus, we obtai max θ µ 0 Kθ = α. Kθ = P T t α, 1 P T δ t α, 1 = α. Power of test. What is the probability that we icorrectly accept H 0 whe it is actually false? Such probability β is called the probability of type II error. The the value 1 β is kow as the power of the test, idicatig how correctly we ca reject H 0 whe it is actually false. Agai, cosider the case of hospital workers subject to a radiatio exposure. Give the curret estimate S = s of stadard deviatio ad the curret sample size = 1, the t-statistic T = X µ 0 S/ ca be approximated by Nδ, 1 with δ = µ µ 0 s/ 1. Example 4. Suppose that the true populatio mea is µ = 5.1 versus the value µ 0 = 5.4 i our hypotheses. The we ca calculate the power of the test with δ 2.11 as follows. Solutio. a I the two-sided hypothesis testig, we reject H 0 whe T > t 0.025,27 = Therefore, the power of the test is K5.1 = P T > µ = b I the oe-sided hypothesis testig, we reject H 0 whe T < t 0.05,27 = Therefore, the power of the test is K5.1 = P T < µ = This explais why we could ot reject H 0 i the two-sided hypothesis testig. Our chace to detect the falsehood of H 0 is oly 52%, while we have 66% of the chace i the oe-sided hypothesis testig. Effect of sample size. For a fixed sigificace level α of your choice, the power of the test icreases as the sample size icreases. I the two-sided hypothesis testig discussed above, we could recommed to collect additioal data to icrease the power of the test. But how may additioal data do we eed? Here is oe possible way to calculate a desirable sample size : I the two-sided hypothesis testig, the power 1 β of the test is approximated by P T > t α/2, 1 P Y < t α/2, 1 δ + P Y > t α/2, 1 δ P Y > t α/2, 1 δ with a radom variable Y havig the t-distributio with 1 degrees of freedom. Effect of sample size, cotiued. Give the curret estimate S = s of stadard deviatio ad the curret sample size 1, we ca achieve the power 1 α/2 of the test by icreasig a total sample size ad cosequetly satisfyig δ 2t α/2,1 1. I the above example of Page 6 Probability ad Statistics II/April 13, 2018

7 Statistical Iferece radiatio exposure of hospital workers, such size ca be calculated as 2 2tα/2,1 1s µ µ 0 2 2t0.025, = = Compariso of two populatios. We ofte wat to compare two populatios o the basis of experimet. For example, a researcher wats to test the effect of his drug o blood pressure. I ay treatmet, a improvemet could have bee due to the placebo effect whe the subject believes that he or she has bee give a effective treatmet. To protect agaist such biases, the study should cosider i the use of a cotrol group i which the subjects are give a placebo, ad a experimetal group i which the subjects are treated with the ew drug, ii the radomizatio by assigig the subjects betwee the cotrol ad the exprimetal groups radomly, ad iii a double-blid experimet by cocealig the ature of treatmet from the subjects ad the perso takig measuremets. Hypothesis test. This becomes the hypothesis testig problem H 0 : µ 1 = µ 2 versus H A : µ 1 µ 2. where µ 1 ad µ 2 are the respective populatio meas of the cotrol ad the experimetal groups Normal assumptio. As a result of experimet, we typically obtai the measuremets X 1,..., X of the subjects from the cotrol group, ad the measuremets Y 1,..., Y m of the subjects from the experimetal group. The it is usually assumed that X 1,..., X ad Y 1,..., Y m are idepedet ad ormally distributed with µ 1, σ 2 1 ad µ 2, σ 2 2, respectively. Eve whe they are ot ormally distributed, large sample sizes, m 30 esure that the tests are appropriate via the cetral limit theorem. Pooled variace procedure. Let S x ad S y be the sample stadard deviatios costructed from X 1,..., X ad Y 1,..., Y m, respectively. Whe it is reasoable to assume σ 2 1 = σ 2 2, we ca costruct the pooled sample variace S 2 p = 1S2 x + m 1S 2 y + m 2 The test statistic T = X Ȳ S p m Page 7 Probability ad Statistics II/April 13, 2018

8 Statistical Iferece has the t-distributio with + m 2 degrees of freedom uder the ull hypothesis H 0. Thus, we reject the ull hypothesis H 0 with sigificat level α whe the observed value t of T satisfies t > t α/2,+m 2. Pooled variace procedure, cotiued. Alteratively we ca compute the p-value p = 2 P Y t with Y havig a t-distributio with + m 2 degrees of freedom, ad reject H 0 whe p < α. Cofidece iterval. The followig table shows the correspodig cofidece iterval of the populatio mea differece µ 1 µ 2, whe your ull hypothesis H 0 is rejected. Test H A : µ 1 µ 2. H A : µ 1 > µ 2. H A : µ 1 < µ 2. 1 α-level cofidece iterval X 1 Ȳ t α/2,+m 2 S p + 1, m X Ȳ + t 1 α/2,+m 2S p + 1 m X 1 Ȳ t α,+m 2 S p + 1, m, X 1 Ȳ + t α,+m 2 S p + 1 m Example 5. Suppose that we cosider the sigificat level α = 0.01, ad that we have obtaied X = ad S x = from the cotrol group of size = 13, ad Ȳ = ad S y = from the experimetal group of size m = 8. Here we have assumed that σ 2 1 = σ 2 2. Solutio. The we ca compute the square root S p = of the pooled sample variace Sp, 2 ad the test statistic T = Thus, we ca obtai p = 2 P Y < 0.01, ad reject H 0. We coclude that the two populatio meas are sigificatly differet. Ad the 99% cofidece iterval for the mea differece is 0.006, Geeral procedure. Whe σ 2 1 σ 2 2, uder the ull hypothesis H 0 the test statistic T = X Ȳ Sx 2 + S2 y m has approximately the t-distributio with ν degree of freedom, where ν is the earest iteger to S 2 x 2 + S2 y m Sx 4 + S4 y 2 1 m 2 m 1 Page 8 Probability ad Statistics II/April 13, 2018.

9 Statistical Iferece Thus, we reject the ull hypothesis H 0 with sigificat level α whe the observed value t of T satisfies t > t α/2,ν. Geeral procedure, cotiued. Alteratively we ca compute the p-value p = 2 P Y t with Y havig a t-distributio with ν degrees of freedom, ad reject H 0 whe p < α. Cofidece iterval. The followig table shows the correspodig cofidece iterval of the populatio mea differece µ 1 µ 2, whe your ull hypothesis H 0 is rejected. Test H A : µ 1 µ 2. H A : µ 1 > µ 2. H A : µ 1 < µ 2. 1 α-level cofidece iterval X Ȳ t Sx 2 α/2,ν + S2 y, X m Ȳ + t α/2,ν X Ȳ t Sx 2 α/2,ν + S2 y, m, X Sx Ȳ + t 2 α/2,ν + S2 y m S 2 x + S2 y m Example 6. Suppose that we cosider the sigificat level α = 0.01, ad that we have obtaied X = ad S x = from the cotrol group of size = 13, ad Ȳ = ad S y = from the experimetal group of size m = 8 as before. Solutio. The test statistic T 3.12, ad ; thus, we obtai ν = 12 ad t 0.005, Sice T > t 0.005,12, we still reject H 0. Alteratively we ca obtai p = 2 P Y < 0.01, ad coclude that the differece is sigificat. Iferece o proportios. I experimets o pea breedig, Medel observed the differet kids of seeds obtaied by crosses from plats with roud yellow seeds ad plats with wrikled gree seeds. Possible types of progey were: roud yellow, wrikled yellow, roud gree, ad wrikled gree. Whe the data values recorded x 1,..., x takes several types, or categories, we call them the categorical data. Poit estimate. Let X be the umber of observatios for a particular type i categorical data of size, ad let p be the populatio proportio of this type that is, the probability of occurrece of this type. The the radom variable X has the biomial distributio with parameter, p. Ad the poit estimate of the populatio proportio p is ˆp = X. Page 9 Probability ad Statistics II/April 13, 2018

10 Statistical Iferece We ca easily see that Eˆp = E Thus, ˆp is a ubiased estimate of p. X = 1 EX = p Poit estimate, cotiued. Recall by the cetral limit theorem that we have approximately X approx Np, p1 p whe is large. The the poit estimate ˆp is approximately distributed as the ormal distributio with parameter p, p1 p. Hypothesis test. Suppose that the vaccie ca be approved for widespread use if it ca be established that the probability p of serious adverse reactio is less tha p 0. The the hypothesis testig problem becomes H 0 : p = p 0 versus H A : p < p Let X be the umber of participats who suffer a adverse reactio amog participats. The, the radom variable X has the biomial distributio with parameter, p ad is approximated by the ormal distributio with parameter p, p1 p whe is large [that is, to satisfy p > 5 ad 1 p > 5]. Hypothesis test, cotiued. The critical poit of the stadard ormal distributio, deoted by z α, is defied as the value satisfyig P Z > z α = α where Z is a stadard ormal radom variable. Sice the ormal distributio is symmetric, it implies that P Z < z α = α. Whe p 0 > 5 ad 1 p 0 > 5, T = X p 0 p0 1 p 0 is used for the test statistic. The we ca reject H 0 i 3.2 with sigificace level α if the value t of the test statistic T satisfies t < z α. Equivaletly, we ca proceed to costruct the p-value p = P Z < t = Φt, ad reject H 0 whe p < α. Sice the cosideratio of cotiuity correctio improves the accuracy, may be used istead. T = X p p0 1 p 0 Cofidece iterval. Whe H 0 is rejected, we wat to further ivestigate the cofidece iterval for the populatio proportio p which correspods to the result of hypothesis test. We have the poit estimate ˆp = X/. The the two differet formulas X + z 2 0, α/2 + z α X X/ + z 2 α / zα 2 ˆp1 ˆp 0, ˆp + z α 3.5 ca be used for the cofidece iterval of level α. Although Formula 3.4 is kow to be more accurate, Formula 3.5 may be used i most of our problems sice it is easier to calculate. Page 10 Probability ad Statistics II/April 13,

11 Statistical Iferece Example 7. Suppose that p 0 = 0.05 is required, ad that the sigificace level α = 0.05 is chose. Ad the study shows that X = 4 adverse reactios are foud out of = 155 participats. Solutio. Note that = 7.75 > 5 ad = > 5. Thus, we have T = ad p = Φ We ca also obtai the poit estimate ˆp ad the 95% cofidece iterval 0, by usig 3.4 [we get 0, if we use 3.5]. Sice p 0.05, we caot reject the ull hypothesis. Thus, it is ot advisable that the vaccie be approved as the result of this study. Sample size calculatios. We always guaratee the possibility of icorrectly rejectig H 0 whe H 0 is true Type I error, say, to be less tha 5% of the chace. But, at the same time we sacrifice the power of detectig the falsehood of H 0 whe H 0 is false power of the test. I order for the hypothesis testig problem H 0 : p = p 0 versus H A : p < p 0, to achieve the power 1 β of the test, we eed a sample of size 2 z α p0 1 p 0 + z β p1 p. 3.6 p p 0 Sample size calculatios: Example. I the example of vaccie experimet, if the true populatio mea p is 0.025, the the power of the test is calculated as K0.025 = P Reject H 0 p = P X < To icrease the power of the test at least 0.8, we eed the sample size at least = 384. Summary. Possible ull hypotheses for the iferece o populatio proportio are H A : p p 0, H A : p > p 0, ad H A : p < p 0. I either case we ca use the test statistic Z i 3.3 if we do ot make a cotiuity correctio. The the correspodig testig procedures are summarized i the followig table. Test Whe to reject 1 α-level cofidece iterval H A : p p 0 Z > z α/2 ˆp z α/2, ˆp + z α/2 ˆp1 ˆp H A : p > p 0 Z > z α ˆp z α ˆp1 ˆp H A : p < p 0 Z < z α 0, ˆp + z α, 1 ˆp1 ˆp ˆp1 ˆp Summary, cotiued. For the sample size calculatio, use Formula 3.6 if the ull hypothesis is either H A : p > p 0 or H A : p < p 0. Whe the ull hypothesis is H A : p p 0, the sample size ca be computed as 2 zα/2 p0 1 p 0 + z β p1 p. p p 0 Page 11 Probability ad Statistics II/April 13, 2018

12 Statistical Iferece Compariso of two proportios. A researcher is iterested i whether there is discrimiatio agaist wome i a uiversity. I terms of statistics this is the hypothesis testig problem H 0 : p A = p B versus H A : p A > p B where p A ad p B are the respective populatio proportios of me ad wome who are admitted to the uiversity. The researcher decided to collect the data for graduate program i the uiversity. Poit estimate. Let X ad Y be the respective umbers of me ad wome who are admitted to the graduate school. The test statistic is give by Me Wome Admit X Y Dey X m Y Total m Z = ˆp A ˆp B ˆp1 ˆp m where ˆp A = X/ ad ˆp B = Y/m are the poit estimates of p A ad p B, ad ˆp = X + Y + m is called a pooled estimate of the commo populatio proportio. Hypothesis test. Uder the ull hypothesis, the probability that Z > z α becomes approximately less tha α. Thus, we reject H 0 whe the observed value z of Z satisfies z > z α. Or, equivaletly we ca reject p = 1 Φz < α. Cofidece iterval. We may wat to further ivestigate the cofidece iterval for the differece p A p B. Havig costructed the hypothesis test problems H A : p A p B, H A : p A > p B, or H A : p A < p B, the followig table shows the correspodig testig procedure ad the cofidece iterval. Test Rejectio 1 α-level cofidece iterval ˆp H A : p A p B z > z α/2 ˆp A ˆp B z A 1 ˆp A α/2 + ˆp B1 ˆp B m, ˆp A ˆp B + z α/2 ˆp A 1 ˆp A + ˆp B1 ˆp B m, H A : p A > p B z > z α ˆp A ˆp B z α ˆp A 1 ˆp A + ˆp B1 ˆp B m, 1 ˆp H A : p A < p B z < z α 1, ˆp A ˆp B + z A 1 ˆp A α + ˆp B1 ˆp B m Page 12 Probability ad Statistics II/April 13, 2018

13 Statistical Iferece Example 8. The followig table classifies the applicatios for the graduate school accordig to admissio status ad sex. Me Wome Total Admit Dey Total Solutio. We have ˆp A = 97/ , ˆp B = 40/ , ad ˆp = 137/ Ad we ca obtai Z = / / ad p = 1 Φ Thus, we caot reject H 0, idicatig that there is o discrimiatio agaist wome i this particular graduate program. I fact, the alterative hypothesis H A : p A < p B will be established i this example. Goodess of fit. I the experimet o pea breedig, Medel observed the differet kids of seeds obtaied by crosses from plats with roud yellow seeds ad plats with wrikled gree seeds. Possible types of progey were: roud yellow RY, wrikled yellow WY, roud gree RG, ad wrikled gree WG. Ad Medel s theory predicted the associated probabilities of occurrece as follows. RY WY RG WG Probabilities 9/16 3/16 3/16 1/16 We wat to test whether the data from observatio is cosistet with his theory goodess of fit test, i which the statemet of ull hypothesis becomes the model is valid. Chi-square test. I geeral, each observatio is classified ito oe of k categories or cells, which results i the cell frequecies X 1,..., X k. The goodess of fit to a particular model ca be assessed by comparig the observed cell frequecies X 1,..., X k with the expected cell frequecies E 1,..., E k, which are predicted from the model. The discrepacy betwee the data ad the model ca be measured by the Pearso s chi-square statistic χ 2 = k i=1 X i E i 2 E i. Chi-square test, cotiued. Uder the ull hypothesis that is, assumig that the model is correct, the distributio of Pearso s chi-square χ 2 is approximated by the chi-square distributio with df = umber of cells 1 umber of parameters i the model Page 13 Probability ad Statistics II/April 13, 2018

14 Statistical Iferece degrees of freedom. Therefore, if you observe that χ 2 = x ad x > χ 2 α,df, the we ca reject the ull hypothesis, castig doubt o the validity of the model. Or, by computig the p-value p = P X > x with a radom variable X havig the chi-square distributio with df degrees of freedom, equivaletly we ca reject the ull hypothesis whe p < α. Example 9. I the experimet of pea breedig, we have obtaied the data as i the followig table. RY WY RG WG Frequecies Solutio. With the total umber of observatios = 556, the expected cell frequecies from the Medel s theory ca be calculated as RY WY RG WG Expected frequecies We ca compute the Pearso s chi-square χ 2 = Sice the Medel s model has o parameter, the chi-square distributio has 3 = 4 1 degrees of freedom ad we get the p-value p = Thus, there is little reaso to doubt the Medel s theory o the basis of Pearso s chi-square test. Test of idepedece. Cosider agai the study of discrimiatio agaist wome i uiversity admissio. I the study, there are two characteristics: me or wome; admitted or deied. The researcher wated to kow whether such characteristics are liked or idepedet. For such a study, we take a radom sample of size from the populatio, which is summarized i the cotigecy table Me Wome Total Admit X 11 X 12 X 1 Dey X 21 X 22 X 2 Total X 1 X 2 = X Test of idepedece, cotiued. The statemet of ull hypothesis becomes the two characteristics are idepedet. Uder the ull hypothesis, the expected frequecies for the cotigecy table ca be give by Me Wome Total Admit p 1 q 1 p 1 q 2 p 1 Dey p 2 q 1 p 2 q 2 p 2 Total q 1 q 2 The poit estimates of p 1, p 2, q 1, ad q 2 are ˆp 1 = X 1 /, ˆp 2 = X 2 /, ˆq 1 = X 1 /, ad ˆq 2 = X 2 /. With these poit estimates, the chi-square statistic is 2 2 χ 2 X ij X i X j / 2 =, X i X j / i=1 j=1 Page 14 Probability ad Statistics II/April 13, 2018

15 Statistical Iferece ad the degree of freedom is = 1. Test of idepedece: Example. chi-square statistic By usig the same data as before, we ca obtai the χ 2 = + [ /442]2 [ /442] / /442 [ /442]2 [ /442] / / , ad the p-value p = Thus, the ull hypothesis is rejected at ay reasoable level, idicatig that the two characteristics are somewhat depedet. Histogram. The data poits x 1,..., x are typically cosidered as the observed values of radom variables X 1,..., X havig a commo probability distributio fx. To judge the quality of data, it is useful to evisage a populatio by the followig graphical represetatio, called histogram. Relative frequecy. Cosider the itervals [20.45, 21.35, [21.35, 22.25, [22.25, 23.15,..., [25.85, The the umber of observatios f i i the i-th iterval becomes a sample frequecy, ad the desity f i h i = width of the i-th iterval. forms the height of the i-th rectagle above the i-th iterval i the histogram. Whe the width of each iterval is equally chose, the width w is called badwidth ad the height h i becomes h i = f i w. Stem ad leaf plot. A stem ad leaf plot is much like a histogram except it portrays a data set itself. The leadig digits of the data values become stems, which split the trailig digits as leaves. The trailig digits are rouded dow to a sigle digit if ecessary. Page 15 Probability ad Statistics II/April 13, 2018

16 Statistical Iferece = Media ad quartiles. The media is the value of the middle data poit, ad defied by { X +1/2 if is odd; X /2 + X /2+1 /2 if is eve. For example, the media of 2, 4, ad 7 is 4. Whe there is a eve umber of umbers, the media is the mea of the two middle umbers. Thus, the media of the umbers 2, 4, 7, 12 is 4+7/2 = 5.5. The 25-percetile is the value idicatig that 25% of the observatios takes values smaller tha the value. Similarly, we ca defie 50-percetile, 75-percetile, ad so o. Note that 50-percetile is the media. We call 25-percetile the lower sample quartile ad 75-percetile the upper sample quartile. Boxplot. A box is draw stretchig from the lower sample quartile the 25-percetile to the upper quartile the 75-percetile. The media is show as a lie across the box. Therefore 1/4 of the distributio is betwee this lie ad the right of the box ad 1/4 of the distributio is betwee this lie ad the left of the box. Vertical lies dotted, called whiskers, stretch out from the eds of the box to the largest ad smallest data. Outliers. Graphical presetatios ca be used to idetify odd-lookig value which does ot fit i with the rest of the data. Such a value is called a outlier. I may cases a outlier is discovered to be a misrecorded data value, or represets some special coditio that was ot i effect whe the data were collected. I the histogram ad boxplot below, the value i far right appears to be quite separate from the rest of the data, ad ca be cosidered to be a outlier. Page 16 Probability ad Statistics II/April 13, 2018

17 Statistical Iferece Page 17 Probability ad Statistics II/April 13, 2018

18 Exercises Exercises Problem 1. A experimeter is iterested i the hypothesis testig problem H 0 : µ = 3.0mm versus H A : µ 3.0mm, where µ is the populatio mea of thickess of glass sheets. Suppose that a sample of = 21 glass sheets is obtaied ad their thickesses are measured. a For what values of the t-statistic does the experimeter accept the ull hypothesis with a size α = 0.10? b For what values of the t-statistic does the experimeter reject the ull hypothesis with a size α = 0.01? Suppose that the sample mea X = 3.04mm ad the sample stadard deviatio is S = 0.124mm. Is the ull hypothesis accepted or rejected with α = 0.10? With α = 0.01? Problem 2. A machie is set to cut metal plates to a legth of mm. The legth of a radom sample of 24 metal plates have a sample mea of X = mm ad a sample stadard deviatio of S = 0.019mm. Is there ay evidece that the machie is miscalibrated? Problem 3. A experimeter is iterested i the hypothesis testig problem H 0 : µ = versus H A : µ > where µ is the populatio mea of the desity of a chemical solutio. Suppose that a sample of = 31 bottles of the chemical solutio is obtaied ad their desities are measured. a For what values of the t-statistics does the experimeter accept the ull hypothesis with a size α = 0.10? b For what values of the t-statistics does the experimeter reject the ull hypothesis with a size α = 0.01? Suppose that the sample mea X = ad the sample stadard deviatio is S = Is the ull hypothesis accepted or rejected with α = 0.10? With α = 0.01? Problem 4. A chocolate bar maufacturer claims that at the time of purchase by a cosumer the average age of its product is o more tha 120 days. I a experimet to test this claim a radom sample of 26 chocolate bars are foud to have ages at the time of purchase with a sample mea of X = days ad a sample stadard deviatio of S = 13.4 days. With this iformatio how do you feel about the maufacturer s claim? Page 18 Probability ad Statistics II/April 13, 2018

19 Exercises Problem 5. Measuremet of suspeded particles i µg/m 3 ca be made for air quality moitorig. Let µ 1 ad µ 2 be the average cocetratio of suspeded particles i the city ceter of Melboure ad Housto, respectively. Usig 1 = 13 observatios for Melboure ad 2 = 16 observatios for Housto, we test H 0 : µ 1 = µ 2 versus H A : µ 1 < µ 2 a Assumig the equal variaces, fid the critical regio with α = b Suppose that X = 72.9, S 1 = 25.6, Ȳ = 81.7 ad S 2 = 28.3 for Melboure ad Housto, respectively. The calculate the test statistic, ad state the coclusio. Problem 6. 14% of drivers used a seat belt, ad a advertisig campaig was coducted to icrease this proportio. Two moths after the campaig, 104 drivers ouf of a radom sample of 590 drivers were wearig a seat belt. a Defie the ull hypothesis ad alterative hypothesis usig the proportio p of driviers usig a seat belt after the campaig. b Fid the critical regio with α = c Calculate the p-value, ad state the coclusio. Page 19 Probability ad Statistics II/April 13, 2018

20 Aswers to exercises Aswers to exercises Problem 1. a For the t-statistic T, we fail to reject the ull hypothesis if T t 0.05,20 = b For the t-statistic T, we reject the ull hypothesis if T > t 0.005,20 = We obtai T = / 1.478, ad therefore, we fail to reject the ull hypothesis with 21 α = Sice the p-value must be greater tha 0.10, we clearly fail to reject the ull hypothesis with α = Problem 2. We ca set the hypotheses H 0 : µ = versus H A : µ Sice T = / 3.61 > t ,23 = 2.807, we reject the ull hypothesis with α = 0.01, ad fid evidece that the machie is miscalibrated. Problem 3. a For the t-statistic T, we fail to reject the ull hypothesis if T t 0.1,30 = b For the t-statistic T, we reject the ull hypothesis if T > t 0.01,30 = We obtai T = / 2.844, ad therefore, we reject the ull hypothesis with α = Sice the p-value must be less tha 0.01, we are clearly able to reject the ull hypothesis with α = Problem 4. We ca set the hypotheses H 0 : µ = 120 versus H A : µ > 120 Sice T = / < t 0.1,25 = 1.316, we fail to reject the ull hypothesis with α = 0.1, ad could ot fid sufficiet evidece agaist the maufacturer s claim. Problem 5. a For the test statistic T, we ca form the critical regio T < t 0.05,27 = b We ca calculate Sp 2 = T = /13 + 1/16 Thus, we fail to reject H 0, ad therefore, we caot fid sufficiet evidece that Melboure has the lower cocetratio of suspeded particles tha Housto. Problem 6. a We defie the hypotheses H 0 : p = 0.14 versus H A : p > 0.14 b For the test statistic Z we ca form the critical regio Z > z 0.01 = c We obtai Z = , ad therefore, calculate p-value by 1 Φ2.54 = Thus, we reject H 0, ad therefore, we fid evidece that the campaig was successful. Page 20 Probability ad Statistics II/April 13, 2018