=++19:’bb . G R H K X J b ; : 9 – 1. к Singh x LEC0101: 11/15 10:10-11:00amx LEC5101: 11/15 6:10-7:00pmEadie x 11/6 11:10AM-12:00 Module 4: Sampling distributionsModule 5: Data collectionModule 8: Statistical testModule 9: The effective Use of Statistical Tests=++19:’bb . G R H K X J b ; : 9 – Module 4: Sampling distributions1. Binomial model Bernoulli variable: Binomial Bernoulli n 2. Normal Model(1) Standard normal model 0, 1 ~ , n:s # of trial p: 1 1 Q 傼 N S N ~ ( , ) : : =++19:’bb . G R H K X J b ; : 9 – (2) Non standard normal 仈 僔 1. 2. 3. Z3. Poisson modelThe number of hurricane per year follows a Poisson distribution with a mean of 2. What is the probability that there are less than 2 hurricanes occurring this year Ans: 0.406P(Zb)=1-A(b) P(cb)= P(Zb)=A(a)+1- A(b) A(Z) excel/R ! P ( 2) = P (X : 0) t p N = 1 )=++19:’bb . G R H K X J b ; : 9 – (I) Sampling distributions1. Population vs sample Population (Census) Sample N n2. Central limit Theorem а sample mean variance normal а n>30 10 1 10 僔1. 傼 central limit theorem2. E(X) SD3. normal standard deviation: t distributionAs n approaches λ ξ ~ , 1 =++19:’bb . G R H K X J b ; : 9 – 仈 1 A repair shop specializes in repairing swibbles. Swibbles are big business (everyone has one!) so under normal circumstances, 200 swibbles are repaired by the shop each day. One out of every six swibbles can be repaired by replacing a broken widget. If the shop has 47 widgets in stock, what is, approximately, the probability that they run out of widgets Ans 0.0048o np = 200 ㄨ t > 1 0 n ( 1 – p) zoox 号 > 1 0 – μ = 亡 : ( 是 = 0 02 6315= 。 235P ( P > 思) = P (中寺 > 2,6=++19:’bb . G R H K X J b ; : 9 – 仈 2 A certain machine produces computer chips. The probability that the machine produces a defective chip is equal to 0.05. If the machine produces 1000 computer chips, then what is the probability that more than 900 will be non-defective Ans 1P ( P < 0 - 1) P > loh ( 1- 1》 > 1 0μ = o . 05 5=酒 命器P ( P < 0 1) = p l中寻 < ""等 )= P ( Z <7 25) ~~ 1=++19:'bb . G R H K X J b ; : 9 - Module 5: Data collection 1 Observationalstudy1. Observational study vs experiment ウ x(explanatory variable) y(response variable) Recall: explanatory variable=predictor=covariate=independent variable response variable=dependent variable=outcome Observational study Experiment Data are measurements of existing characteristics 傼 The best way we have to make causal conclusions ウ н Investigator/Scientist has no control over what might have caused the data to have certain relationships ウ Investigator/Scientist has applied a “treatment” to see how it affects the response variable2. Mechanism Causation: x y x y Reverse causation x y y x Association x y Common cause x y Confounding variable н y x confounding variable Lurking variable н 2: Experiment1.TermExperiment ウ 傼 Researcher manipulates or applies a treatment to the explanatory variables, then observes the response variable. Experimental unit: 傼 2.FactorandlevelFactor: 学 学 Level: Factor A B Placebo =++19:'bb . G R H K X J b ; : 9 - 3.Extraneousfactor x y н ウ not of interest in the current study, but may affect the responsex Extraneous factor а (Blocking)4.Blocking а bell rogers л ф 5.Experiment extraneous factor 傼 (No selection bias) 傼 Control any extraneous factors Randomly assign experimental units to treatment group Replicate, by applying each treatment to many experimental units6.CausalStatementsandExperimentalDesign7. (1) Blinding Single blind: 傼 н н ≤->conusiusf.tl=++19:’bb . G R H K X J b ; : 9 – Double blind: 傼 傼 н н ≤ (2) Placebo Effect ≤ а (3)Controlgroup а о ↓ 傼 8.MeanvsproportionMean: quantitative variable Proportion categorical variable 1 施加影响-experimentobservationdsudyi-treatmenti.uariay=++19:’bb . G R H K X J b ; : 9 – Blocked-competey.amdoniud=++19:’bb . G R H K X J b ; : 9 – 3nr-iexperiment.explanatorgaplacebo.iqvc ,3gVC.3gVCtonsetofawldrespohsedumtionof.twwmencdYesNo=++19:’bb . G R H K X J b ; : 9 – -experimentiblockchiaseedundplaeeboYes.su/No.=++19:’bb . G R H K X J b ; : 9 – ipositivehnearrdationsh.ir随 y TN , 没控制变量 ,otherpossibleconfundmgvarneshhe.IQadfamìy income=++19:’bb . G R H K X J b ; : 9 – O-e-rnnnsimplemndomsamphngfstrat.fied 村庄=++19:’bb . G R H K X J b ; : 9 – Module 8 &9 1 Hypothesis1. -傼 – ( )2. Null hypothesis Alternative hypothesis H H н Two sided example:We would like to test the average year income of a Canadian is whether $50,000 or not. Write the nullhypothesis and alternative hypothesis. : 50000 : 50000One sided example:We would like to test the average year income of a Canadian is above $50,000 or not. Write the nullhypothesis and alternative hypothesis. : 50000 : 50000We would like to test the average year income of a Canadian is below $50,000 or not. Write the nullhypothesis and alternative hypothesis. : 50000 : 500003. P value H0 ↓ 1 H0 ↓ H0 ( н H1 ) н ла н 2 儈 H0 ↓ 儈 (Fail to reject) H0 仈н accept fail to reject=++19:’bb . G R H K X J b ; : 9 – 4. Level of significance :а do not reject 0 5. Estimating a proportion doing a hypothesis test p: assume step1: hypothesisstep2: ┑ central limit theorem:Recall: central limit theorem np>10 and np(1-p) 10step3: 1 step4a: Z step4b P value significant level 。 错误可能性0test 5 状”ㄨ￥i=++19:’bb . G R H K X J b ; : 9 – 6. Critical value Z* Significance level Z Z Z Critical value One sided Two sided0.1 1.28 1.6450.05 1.645 1.960.01 2.33 2.5767. Student t distribution 仈 standard deviation 仈 僔学 Z teststep1: step2: ξ ,df=n-1df: degree of freedomstep4a: T , , *step4b P valueA四 005E = 1 . 64=++19:’bb . G R H K X J b ; : 9 – 仈 1Georgianna claims that in a small city renowned for its music school, child takes less than 5years of piano lessons. We have a random sample of 20 children from the averagecity, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years(a) Evaluate Georgianna’s claim (or that the opposite might be true) using a hypothesis test.1-1 0 : M : 5- 错的H.int#5I=4.65=z.zx-MT=-=46誉 – 0 -8 1 3 1df = n – 1 = 1 971 9 , 005 209☆ITKTq.o.os-ifailtonef.ge Ho=++19:’bb . G R H K X J b ; : 9 – 8. Error 1. 2. (Hypothesis) vs Type I error (False positive) : Type II error (False negative)9. Two erros Type I error Reject the null hypothesis when the null hypothesis is true Happens with probability Type II error Fail to reject the null hypothesis when the null hypothesis is not true Happens with probability 10. Power of test 1 P (Reject | is false) ,↓ the probability of correctly rejecting the null hypothesis.Accept Fail to Reject Reject Correct Type I error ( Type II error( Correct-→ 更严重☆-_-→ 越 越好=++19:’bb . G R H K X J b ; : 9 – Type I error Type II error Power of test 11. Increase power x x decrease variability in sampling distributionx increase sample sizex “move the true” value away from the null12. Confidence Interval 0.05, 1.961. What is the purpose of a confidence interval A. None of theseB. It provides a range of the most plausible that a theoretical can take on ~ , 1 Z , ]T distribution , , , ]☆B 1 – B’oes 随 poweroftest 增加☆ 好的影响减少不确定性增加样本数量将 t.me wue移到 轗 critìcalvaw( 10=++19:’bb . G R H K X J b ; : 9 – C. It provides the complete range of values that a statistic (such as thesample mean) can take onD. It provides a range of values that repeated measurements of a statistic(such as a sample mean) have taken on2. We have a biased coin which has a probability p of landing heads whenflipped. Suppose we perform the following experiment 100 times: In eachexperiment, we flip the coin 10 times and find the 95% confidence intervalfor p based on the proportion of heads in the 10 flips. Which of the following is TRUE A. None of theseB. 95 of the intervals will contain pC. The closer p is to 0.5, the more intervals will contain pD. We expect that approximately 95 of the intervals will contain p3. Suppose that after sampling a population, we estimate some proportion pand compute a 95% confidence interval to be [0.05, 0.45]. What interpretation is correct A. With probability 0.95, the true proportion p lies between 0.05 and 0.45B. With probability 0.95, the probability that our estimate is equal to the trueproportion p is between 0.05 and 0.45C. We have sampled the same proportion repeatedly and estimate p eachtime, and 95% of the time is p^ between 0.05 and 0.45D. If we sampled the same population repeatedly and estimated p each time, 95% of the time pshould lie within the computed confidence interval4. In an experiment with two outcomes, let p be the unknown true probabilityof getting a successful outcome on any one trial. If we repeat the trial ntimes and let p^ = total number of successes / total number of trials, which of p and p^ is fixedand which is constant A. p^ is random and p is fixedB. p is random and p^ is fixedC. Both p and p^ are fixedD. Both p and p^ are random5. What is the purpose of substituting in p^ for p in the equations for theendpoints of a confidence interval for p A. p^ is a more conservative choice than pB. p is a more conservative choice than p^C. p^ is always unknown, but we can compute p following an experimentD. p is always unknown, but we can compute p^ following an experiment ifwe anticipate p^ pADDAD 错误说法: =++19:’bb . G R H K X J b ; : 9 – 1 significance level significance level reject do notreject (a) Ho 0 .2H i i p I O.2P = ~~ o . 2 9 6 95 = 1- = 0 = 00 5Z = j = 1 9 4 ※本!作’p (区 1 7 1 94 ) = 2 P ( Z 7 1 9 4)= o.0 524weakevideneeagaist.no=++19:’bb . G R H K X J b ; : 9 – 1 significance level significance level reject do notreject Hoˇ(b) H 。 : p = 0 . 2H . > 0 .2: 点4″ ㄨlP ( Z > 1 9 4) = o.oz.bz moderateevidenceagainst.it。(C) 与 前 样 Ho 0 2 H , :Pao.z.PLZ < 1 9 4) = 097 P 701 , 所以noeuidenceagainst.lt。=++19:'bb . G R H K X J b ; : 9 - 2Given significance level is 0.05.@ p = 00 5 4 P > 2所mfh.to rejeu.tt 。(b)p = o.0262 P 2所以 t.lt H 。4) P 097 > 所攻 faivtoreject.no=++19:’bb . G R H K X J b ; : 9 – 3⑧-nri(a) 5010(b) bio ( 7 7 ,o.05)4) E ( x) = 77 x 0 -0 5 = 37 53 . 7 5 7 2No=++19:’bb . G R H K X J b ; : 9 – 4…←—ōtorejeuaiypeIerroringeutHowhileltistrueplrejectHoltt.istrue)= P ( 1 , ㄨ= 2 , ㄨ= 3 , ㄨ = 5 ) 012 to.lt 0.lt 0 – 1 = 0 . 5(b) TypeElifil.tonejectttowhileltoisfnlsepliailtorejatH-IH.istlse)= P (ㄨ = 4 , ㄨ = 6 )= 0 3=++19:’bb . G R H K X J b ; : 9 – Test 2 1. If an experimental result is said to be “statistically significant”, it meansthat we conclude that the observed result contradicts the establishedmodel and/or theory due to some systematic cause. TRUE2. If the null hypothesis is rejected in favor of the alternative hypothesis, theresult is said to be statistically significant. TRUE3. A p-value tells you how unlikely the observed value of the test statistic(and more extreme values) is if the null hypothesis were true. TRUE4. A p-value can be interpreted as how likely it is that the null hypothesis istrue. FALSE5. The second step of statistical testing is to collect the data and calculatethe test statistic. TRUE6. The test statistic in statistical testing is formulated under the assumptionthat the alternative hypothesis is correct. FALSE7. In statistical tests, the legal idea of innocent until proven guilty isencapsulated by the alternative hypothesis. FALSE8. The alternative hypothesis is usually what an experiment is trying toestablish. TRUE9. TRUE or FALSE: Suppose we suspect that taking a certain medicationdaily has an effect on reducing heart attacks. An appropriate nullhypothesis in this situation is “Taking the medication daily has no effecton the incidence of heart attacks”. TRUE10. Suppose we have a coin that we suspect has been weighted so that it ismore likely to come up heads than tails and we will collect some data andcarry out a statistical test to see if this is the case. Let p denote the trueprobability of heads for this coin. The appropriate null and alternativehypotheses areA. H0: p = 0.5 vs Ha: p 0.5B. H0: p > 0.5 vs Ha: p = 0.5C. H0: p = 0.5 vs Ha: p > 0.5D. H0: p = 0.5 vs Ha: p < 0.511. Suppose we flip a coin many times and suspect that the coin is biased insome way. We perform a statistical test to evaluate this hypothesis. If thetrue probability of the coin landing heads is p, what is a reasonablealternative hypothesis statisticalysignific.atˇ 上是 错了 !reject.tl-5remrrnenrnrz Znx-我们更倾向于证明 是错误的ˇ……-=++19:'bb . G R H K X J b ; : 9 - A. p = 0.5B. p < 0.5 or p > 0.5C. p < 0.5 and p > 0.5D. None are correct12. The mean result from some experiment with 50 observations was 1.5 andthe standard deviation was 1. Let denote the true mean result. We wishto to test H0: =1 vs Ha: >1. The p-value isA. P(t50 3.5)B. P(t49 3.5)C. P(t49 3.5)D. P(t50 3.5)13. We wish to test H0: p = 0.5 vs Ha: p > 0.5 where p denotes the trueprobability of heads for a particular coin. We flip the coin 1000 times andobserve 512 heads. Without doing any calculations, the p-value and testconclusion are most likelyA. Close to 0 so we reject the null hypothesis.B. Close to 1 so we reject the null hypothesis.C. Close to 1 so we fail to reject the null hypothesis.D. Close to 0 so we fail to reject the null hypothesis.14. We wish to test H0: p = 0.5 vs Ha: p 0.5 where p denotes the trueprobability of heads for a particular coin. We flip the coin 1000 times andobserve 510 heads. Without doing any calculations, the p-value and testconclusion are most likelyA. Close to 0 so we reject the null hypothesis.B. Close to 1 so we reject the null hypothesis.C. Close to 1 so we fail to reject the null hypothesis.D. Close to 0 so we fail to reject the null hypothesis.15. The mean result of some observations was 50. Let denote the true meanresult. We wish to test the hypothesis H0: = 40 vs Ha: < 40. Withoutdoing any calculations, the p-value and test conclusion are most likelyA. Close to 0 so fail to reject the null hypothesisB. Close to 1 so fail to reject the null hypothesisC. Close to 0 so reject the null hypothesisD. Close to 1 so reject the null hypothesis16. Suppose we flip a coin many times and observe a suspiciously largenumber of heads. How does a statistical test help us assess whether or not the coin is biased A. We decide based on how likely the coin is to be unbiased given ourobservationsB. We decide based on how unlikely our observations would be if the coine.mn-9sd Tdistribut.io nt p很 ,接受"0=++19:'bb . G R H K X J b ; : 9 - were unbiasedC. We decide based on how likely our observations would be if the coin werebiasedD. We decide based on how unlikely the coin is to be unbiased given ourobservations17. Why can't we calculate the test statistic as the first step of a statisticaltest A. Because it is calculated from the p-value, which we haven't calculated yetB. Because it appears in the statement of the null hypothesis, which wehaven't formulated yetC. Because it is calculated from the sample data, which we haven't obtainedyetD. Because it can only be calculated if the null hypothesis is true18. Suppose we suspect that the average means, 1 and 2, of twopopulations are different. What is an appropriate alternative hypothesis forthis situation A. HA: -1 -2B. HA: 1 = 2C. HA: -1 -2D. None are correct19. Suppose we calculate sample means from two populations. While thesample means are not identical, the difference between them seems to besmall. How can a statistical test help here A. It can determine whether the population means are differentB. It can determine whether the difference is statistically reasonable if thepopulation means are differentC. It can determine whether the difference is statistically reasonable if thepopulation means are identicalD. It can determine whether the population means are identical20. What is the purpose of a test statistic A. All are correctB. It serves as a numerical summary of the data under the null hypothesisC. It allows us to assess how extreme our observed data is if our suspicionsare falseD. It allows us to calculate the p-value21. Which of the following is TRUE about the p-value A. It indicates the likelihood of obtaining data at least as extreme as theobserved data if the null hypothesis were trueB. A large p-value indicates that the null hypothesis is falseC. A large p-value indicates that the null hypothesis is true=++19:'bb . G R H K X J b ; : 9 - D. It indicates the likelihood of obtaining data at least as extreme as theobserved data if the null hypothesis were false22. Which of the following hypotheses would be suitable for testing using astatistical test A. The average age of undergraduate students is 20B. 4 is the only solution to the function x^2í16=0C. Every undergraduate student is at least 16 years oldD. The average employee is healthy23. Suppose we suspect that University A tends to graduate a significantlydifferent proportion of students than University B does. If we assessedthis using a statistical test, how would the test statistic be formulated A. Under the assumption that the two proportions are equalB. Under the assumption that the two proportions are differentC. Under the assumption that both universities graduate the same number ofstudentsD. Under the assumption that each universities graduate a different numberof students24. You want to investigate the relationship between exercise and weight lossamong university students, but you do not know whether weight gain dueto increased muscle mass might be offset by weight loss due todecreased fat stores. You should FIRST...A. Design your study to distinguish between the two influences on weightB. Define precisely your null hypothesis and alternative hypothesisC. Use a two-tailed test to evaluate the significance of your resultsD. Use a one-tailed test to evaluate the significance of your results25. Consider the following scenario: we have a coin which we suspect to bebiased. We get to flip the coin as many times as we want to perform astatistical test. If we incorrectly conclude that the coin is biased at the endof the experiment, we must pay $100; otherwise, no transfer of moneytakes place. Which of the following is TRUE about Type I and Type IIerrors A. We should aim to reduce Type II error, because the consequences of aType I error are not particularly costlyB. We should aim to reduce Type I and Type II error equally, because theconsequences of either of them are particularly costlyC. We should aim to reduce Type I error, because the consequences of aType II error are not particularly costlyD. We should aim to reduce Type I error, because the consequences of aType I error are particularly costly1-5 TTTFT 6-10 FFTTC 11-15 BCCCB 16-20 BCCCA 21-25 AAAADb)=1-A(b) P(cb)= P(Zb)=A(a)+1- A(b) A(Z) excel/R ! P ( 2) = P (X : 0) t p N = 1 ) =++19:'bb . G R H K X J b ; : 9 - (I) Sampling distributions 1. Population vs sample Population (Census) Sample N n 2. Central limit Theorem а sample mean variance normal а n>30 10 1 10 僔 1. 傼 central limit theorem 2. E(X) SD 3. normal standard deviation: t distribution As n approaches λ ξ ~ , 1 =++19:’bb . G R H K X J b ; : 9 – 仈 1 A repair shop specializes in repairing swibbles. Swibbles are big business (everyone has one!) so under normal circumstances, 200 swibbles are repaired by the shop each day. One out of every six swibbles can be repaired by replacing a broken widget. If the shop has 47 widgets in stock, what is, approximately, the probability that they run out of widgets Ans 0.0048 o np = 200 ㄨ t > 1 0 n ( 1 – p) zoox 号 > 1 0 – μ = 亡 : ( 是 = 0 02 63 1 5 = 。 235 P ( P > 思) = P ( 中寺 > 2,6 =++19:’bb . G R H K X J b ; : 9 – 仈 2 A certain machine produces computer chips. The probability that the machine produces a defective chip is equal to 0.05. If the machine produces 1000 computer chips, then what is the probability that more than 900 will be non-defective Ans 1 P ( P < 0 - 1) P > lo h ( 1- 1》 > 1 0 μ = o . 05 5 =酒 命器 P ( P < 0 1) = p l 中寻 < ""等 ) = P ( Z <7 25) ~~ 1 =++19:'bb . G R H K X J b ; : 9 - Module 5: Data collection 1 Observationalstudy 1. Observational study vs experiment ウ x(explanatory variable) y(response variable) Recall: explanatory variable=predictor=covariate=independent variable response variable=dependent variable=outcome Observational study Experiment Data are measurements of existing characteristics 傼 The best way we have to make causal conclusions ウ н Investigator/Scientist has no control over what might have caused the data to have certain relationships ウ Investigator/Scientist has applied a “treatment” to see how it affects the response variable 2. Mechanism Causation: x y x y Reverse causation x y y x Association x y Common cause x y Confounding variable н y x confounding variable Lurking variable н 2: Experiment 1.TermExperiment ウ 傼 Researcher manipulates or applies a treatment to the explanatory variables, then observes the response variable. Experimental unit: 傼 2.FactorandlevelFactor: 学 学 Level: Factor A B Placebo =++19:'bb . G R H K X J b ; : 9 - 3.Extraneousfactor x y н ウ not of interest in the current study, but may affect the response x Extraneous factor а (Blocking) 4.Blocking а bell rogers л ф 5.Experiment extraneous factor 傼 (No selection bias) 傼 Control any extraneous factors Randomly assign experimental units to treatment group Replicate, by applying each treatment to many experimental units 6.CausalStatementsandExperimentalDesign 7. (1) Blinding Single blind: 傼 н н ≤ - > conusiusf.tl =++19:’bb . G R H K X J b ; : 9 – Double blind: 傼 傼 н н ≤ (2) Placebo Effect ≤ а (3)Controlgroup а о ↓ 傼 8.MeanvsproportionMean: quantitative variable Proportion categorical variable 1 施加影响 – experimentobservationdsudyi-treatmenti.ua riay =++19:’bb . G R H K X J b ; : 9 – Blocked-competey.amdoniud =++19:’bb . G R H K X J b ; : 9 – 3 nr-iexperiment.explanatorgaplacebo.iq vc ,3gVC.3gVCtonsetofawldrespohsedumtionof.twwmencd Yes No =++19:’bb . G R H K X J b ; : 9 – – experimentiblockchiaseedundplaeeboYes.su/No. =++19:’bb . G R H K X J b ; : 9 – ipositivehnearrdationsh.ir 随 y T N , 没控制变量 ,otherpossibleconfundmgvarneshhe.IQadfamìy income =++19:’bb . G R H K X J b ; : 9 – O-e-rnnnsimplemndomsamphngfstrat.fied 村庄 =++19:’bb . G R H K X J b ; : 9 – Module 8 &9 1 Hypothesis 1. -傼 – ( ) 2. Null hypothesis Alternative hypothesis H H н Two sided example: We would like to test the average year income of a Canadian is whether $50,000 or not. Write the null hypothesis and alternative hypothesis. : 50000 : 50000 One sided example: We would like to test the average year income of a Canadian is above $50,000 or not. Write the null hypothesis and alternative hypothesis. : 50000 : 50000 We would like to test the average year income of a Canadian is below $50,000 or not. Write the null hypothesis and alternative hypothesis. : 50000 : 50000 3. P value H0 ↓ 1 H0 ↓ H0 ( н H1 ) н ла н 2 儈 H0 ↓ 儈 (Fail to reject) H0 仈н accept fail to reject =++19:’bb . G R H K X J b ; : 9 – 4. Level of significance :а do not reject 0 5. Estimating a proportion doing a hypothesis test p: assume step1: hypothesis step2: ┑ central limit theorem: Recall: central limit theorem np>10 and np(1-p) 10 step3: 1 step4a: Z step4b P value significant level 。 错误可能性 0 test 5 状 ” ㄨ ￥i =++19:’bb . G R H K X J b ; : 9 – 6. Critical value Z* Significance level Z Z Z Critical value One sided Two sided 0.1 1.28 1.645 0.05 1.645 1.96 0.01 2.33 2.576 7. Student t distribution 仈 standard deviation 仈 僔 学 Z test step1: step2: ξ ,df=n-1 df: degree of freedom step4a: T , , *step4b P value A四 005 E = 1 . 64 =++19:’bb . G R H K X J b ; : 9 – 仈 1 Georgianna claims that in a small city renowned for its music school, child takes less than 5 years of piano lessons. We have a random sample of 20 children from the average city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years (a) Evaluate Georgianna’s claim (or that the opposite might be true) using a hypothesis test. 1-1 0 : M : 5 – 错的 H.int#5I=4.65=z.zx-MT=-=46誉 – 0 -8 1 3 1 df = n – 1 = 1 9 7 1 9 , 005 209 ☆ ITKTq.o.os-ifailtonef.ge Ho =++19:’bb . G R H K X J b ; : 9 – 8. Error 1. 2. (Hypothesis) vs Type I error (False positive) : Type II error (False negative) 9. Two erros Type I error Reject the null hypothesis when the null hypothesis is true Happens with probability Type II error Fail to reject the null hypothesis when the null hypothesis is not true Happens with probability 10. Power of test 1 P (Reject | is false) ,↓ the probability of correctly rejecting the null hypothesis. Accept Fail to Reject Reject Correct Type I error ( Type II error( Correct – → 更严重 ☆ -_- → 越 越好 =++19:’bb . G R H K X J b ; : 9 – Type I error Type II error Power of test 11. Increase power x x decrease variability in sampling distribution x increase sample size x “move the true” value away from the null 12. Confidence Interval 0.05, 1.96 1. What is the purpose of a confidence interval A. None of these B. It provides a range of the most plausible that a theoretical can take on ~ , 1 Z , ] T distribution , , , ] ☆ B 1 – B ‘ o es 随 poweroftest 增加 ☆ 好的影响 减少不确定性 增加样本数量 将 t.me wue移到 轗 critìcalvaw ( 1 0 =++19:’bb . G R H K X J b ; : 9 – C. It provides the complete range of values that a statistic (such as the sample mean) can take on D. It provides a range of values that repeated measurements of a statistic (such as a sample mean) have taken on 2. We have a biased coin which has a probability p of landing heads when flipped. Suppose we perform the following experiment 100 times: In each experiment, we flip the coin 10 times and find the 95% confidence interval for p based on the proportion of heads in the 10 flips. Which of the following is TRUE A. None of these B. 95 of the intervals will contain p C. The closer p is to 0.5, the more intervals will contain p D. We expect that approximately 95 of the intervals will contain p 3. Suppose that after sampling a population, we estimate some proportion p and compute a 95% confidence interval to be [0.05, 0.45]. What interpretation is correct A. With probability 0.95, the true proportion p lies between 0.05 and 0.45 B. With probability 0.95, the probability that our estimate is equal to the true proportion p is between 0.05 and 0.45 C. We have sampled the same proportion repeatedly and estimate p each time, and 95% of the time is p^ between 0.05 and 0.45 D. If we sampled the same population repeatedly and estimated p each time, 95% of the time p should lie within the computed confidence interval 4. In an experiment with two outcomes, let p be the unknown true probability of getting a successful outcome on any one trial. If we repeat the trial n times and let p^ = total number of successes / total number of trials, which of p and p^ is fixed and which is constant A. p^ is random and p is fixed B. p is random and p^ is fixed C. Both p and p^ are fixed D. Both p and p^ are random 5. What is the purpose of substituting in p^ for p in the equations for the endpoints of a confidence interval for p A. p^ is a more conservative choice than p B. p is a more conservative choice than p^ C. p^ is always unknown, but we can compute p following an experiment D. p is always unknown, but we can compute p^ following an experiment if we anticipate p^ p ADDAD 错误说法 : =++19:’bb . G R H K X J b ; : 9 – 1 significance level significance level reject do not reject (a) Ho 0 .2 H i i p I O . 2 P = ~~ o . 2 9 6 9 5 = 1- = 0 = 00 5 Z = j = 1 9 4 ※本!作’ p (区 1 7 1 94 ) = 2 P ( Z 7 1 9 4) = o.0 524 weakevideneeagaist.no =++19:’bb . G R H K X J b ; : 9 – 1 significance level significance level reject do not reject Ho ˇ (b) H 。 : p = 0 . 2 H . > 0 . 2 : 点 4 ” ㄨ l P ( Z > 1 9 4) = o.oz.bz moderateevidenceagainst.it 。 (C) 与 前 样 Ho 0 2 H , ao.z.PL Z < 1 9 4) = 097 P 701 , 所以noeuidenceagainst.lt。 =++19:'bb . 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