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本课程起止时间为:2020-11-19到2021-02-22
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【作业】Chapter 1 Introduction homework 1

1、 问题:Fill in the blanks:A coin is flipped twice. Let Y =number of heads obtained, when the probability of a head for a flip equals π. a. Assuming π =0.50 , the distribution’s mean is  (            ) and standard deviation is (             ).  b. Suppose you observe y =1 and do not know π ,  the likelihood function is  (              ). c. Using the plotted likelihood function from (b),  the ML estimate of π is (             ).
评分规则: 【 (a)10分写对均值和标准差分别得5分,写错不得分。
(b)10分写对得10分,写错不得分。
(c)10分写对得10分,写错不得分。

2、 问题:Fill in the blanks:Genotypes AA, Aa, and aa occur with probabilities (π1,π2,π3). For n=3 independent observations, the observed frequencies are (n1,n2,n3). a. You can determine n3 from knowing n1 and n2 by (                  ). b. Suppose (π1, π2, π3) = (0.25, 0.50, 0.25). Find the multinomial probability that (n1,n2,n3)=(1,2,0)  is (        )c. Refer to (b), specify the values of the sample size index and parameter for that distribution, n1 has the probability distribution (              )
评分规则: 【 (a)3分写出公式得10分
(b)10分写对得10分,写错不得分。
(c)10分写对得10分,写错不得分。

3、 问题:FIl in the blanks:To collect data in an introductory statistics course, recently I gave the students a questionnaire. One question asked whether the student was a vegetarian. Of 25 students, 0 answered “yes.”They were not a random sample, but let us use these data to illustrate inference for a proportion. (You may wish to refer to Section1.4.1onmethodsofinference.) Let π denote the population proportion who would say “yes.” Consider H0:π =0.50 and Ha:π =0.50.a. Conduct the “Wald test,”  you find using  the estimated standard error is (                 ) (express it  by substituting value).b. The 95% “Wald confidence interval”  for π is    (                )c. The “score test,” for which     uses the null standard error, equals to (              )  The P-value is (               )d. The 95% score confidence interval (i.e., the set of π0 for which |z|< 1.96 in the score test) equals (                      ). (Hint: What do the z test statistic and P-value equal when you test H0:π =0.133 against Ha:π ≠ 0.133.)
评分规则: 【 (a)8分写对得满分,写错不得分。
(a)8分写对得满分,写错不得分。
(c)16分写对一个空得8分,写错不得分。
(a)8分写对得满分,写错不得分。

【作业】Chapter 2 Contingency Tables(2) homework 2

1、 问题:Fil in the balnks: A newspaper article preceding the 1994 World Cup semifinal match between Italy and Bulgaria stated that “Italy is favored 10–11 to beat Bulgaria, which is rated at 10–3 to reach the final.” Suppose this means that the odds that Italy wins are 11/10 and the odds that Bulgaria wins are 3/10. The probability that Italy wins is (             ), The probability that Bulgaria wins is (             ), 
评分规则: 【 算出两队各自赢的概率,各15分

2、 问题:Fill in the blanks:Data posted at the FBI website (www.fbi.gov) stated that of all blacks slain in 2005, 91% were slain by blacks, and of all whites slain in 2005, 83% were slain by whites. Let Y denote race of victim and X denote race of murderer.a. (Choose the true answer) Which conditional distribution do these statistics refer to?      A. Y given X       B. X given Yb. The odds ratio between X and Y is (               )c. Given that a murderer was white,  the probability that the victim was white is (                ) (Hint: How could you use Bayes’s Theorem?)
评分规则: 【 a.(B)
b.10分答对得10分,答错不得分。
c. 10分答对得10分,答错不得分。

3、 问题:Fill in the blanks:A statistical analysis that combines information from several studies is called a meta analysis. A meta analysis compared aspirin with placebo on incidence of heart attack and of stroke, separately for men and for women (J. Am. Med. Assoc., 295: 306–313, 2006). For the Women’s Health Study, heart attacks were reported for 198 of 19,934 taking aspirin and for 193 of 19,942 taking placebo.a. The odds ratio is (                  ).b. A 95% confidence interval for the population odds ratio for women is (             ).(As of 2006,results suggested that for women, aspirin was helpful for reducing risk of stroke but not necessarily risk of heart attack.)
评分规则: 【 a.20分答对得全分,答错不得分。
b.20分答对得全分,答错不得分。

【作业】Chapter 3 Generalized Linear Model(2) homework 3

1、 问题:Refer to Table 2.7 on x =mother’s alcohol consumption and Y =whether a baby has sex organ malformation.With scores (0,0.5,1.5,4.0,7.0) for alcohol consumption, ML fitting of the linear probability model has the output:a. State the prediction equation (             )   b. Use the model fit to estimate the(i) probabilities of malformation for alcohol levels 0 is (                  ), for  alcohol level 7 is  (                  ).(ii) relative risk comparing those levels is  (                  ).
评分规则: 【 a. 10s for writing out the model fomula
b. (i) 5s for each probability  (ii) 10s

2、 问题:Refer to the previous exercise1 and the solution to (b).a. The sample proportion of malformations is much higher in the highest alcohol category than the others because,although it has only one malformation, its sample size is only 38. Is the result sensitive to this single malformation observation? Re-fit the model without it (using 0 malformations in 37 observations at that level), and re-evaluate estimated probabilities of malformation at alcohol levels 0 and 7 and the relative risk.b. Is the result sensitive to the choice of scores? Re-fit the model using scores (0,1,2,3,4), and re-evaluate estimated probabilities of malformation at the lowest and highest alcohol levels and the relative risk.c. Fit a logistic regression or probit model. Report the prediction equation. Interpret the sign of the estimated effect.
评分规则: 【 a. 10s for refiting the model, 5s for each calculation of probability, 5s for relative risk, 5s for the interpretation of sensitivity
b. same scoring rule as a.
c. 4s for each model fitting result, 2s for the interpretation of parameters.

【作业】Chapter 4 Logistic Regression(2) homework 4

1、 问题:A study used logistic regression to determine characteristics associated with Y =whether a cancer patient achieved remission (1=yes). The most important explanatory variable was a labeling index (LI) that measures proliferative activity of cells after a patient receives an injection of tritiated thymidine. It represents the percentage of cells that are “labeled.” The first table shows the grouped data. Software reports the second table for a logistic regression model using LI to predict a. Conduct a Wald test for the LI effect. Interpret.b. Construct a Wald confidence interval for the odds ratio corresponding to a 1-unit increase in LI. Interpret.c. Conduct a likelihood-ratio test for the LI effect. Interpret.d. Construct the likelihood-ratio confidence interval for the odds ratio. Interpret.
评分规则: 【 a. 2s for z statistic, 1s for interpretation
b. 4s for CI, 2s for interpretation
c. 2s for statistic, 1s for result
b. 4s for CI, 2s for interpretation

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