andrew_gelman_stats andrew_gelman_stats-2012 andrew_gelman_stats-2012-1349 knowledge-graph by maker-knowledge-mining
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Introduction: 18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introduce notation as necessary for all the information needed to solve the problem. Solution to question 17 From yesterday : 17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points? Solution: se is sqrt(.5*.5/(n/2)+.5*.5/(n/2)) = 1/sqrt(n). Solve 1/sqrt(n) = .05, you get n = (1/.05)^2 = 400.
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same-blog 1 1.0 1349 andrew gelman stats-2012-05-28-Question 18 of my final exam for Design and Analysis of Sample Surveys
Introduction: 18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introduce notation as necessary for all the information needed to solve the problem. Solution to question 17 From yesterday : 17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points? Solution: se is sqrt(.5*.5/(n/2)+.5*.5/(n/2)) = 1/sqrt(n). Solve 1/sqrt(n) = .05, you get n = (1/.05)^2 = 400.
2 0.49121362 1352 andrew gelman stats-2012-05-29-Question 19 of my final exam for Design and Analysis of Sample Surveys
Introduction: 19. A survey is taken of students in a metropolitan area. At the first stage a school is sampled at random. The schools are divided into two strata: 20 private schools and 50 public schools are sampled. At the second stage, 5 classes are sampled within each sampled school. At the third stage, 10 students are sampled within each class. What is the probability that any given student is sampled? Express this in terms of the number of students in the class, number of classes in the school, and number of schools in the area. Define appropriate notation as needed. Solution to question 18 From yesterday : 18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introd
3 0.43169364 1348 andrew gelman stats-2012-05-27-Question 17 of my final exam for Design and Analysis of Sample Surveys
Introduction: 17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points? Solution to question 16 From yesterday : 16. You are doing a survey in a war-torn country to estimate what percentage of unemployed men support the rebels in a civil war. Express this as a ratio estimation problem, where goal is to estimate Y.bar/X.bar. What are x and y here? Give the estimate and standard error for the percentage of unemployed men who support the rebels. Solution: x is 1 if the respondent is an unemployed man, 0 otherwise. y is 1 if the respondent is an unemployed man and supports the rebels, 0 otherwise. The estimate is y.bar/x.bar [typo fixed], the standard error is (1/x.bar)*(1/sqrt(n))*s.z, whe
4 0.25530908 1361 andrew gelman stats-2012-06-02-Question 23 of my final exam for Design and Analysis of Sample Surveys
Introduction: 23. Suppose you are conducting a survey in which people are asked about their health behaviors (how often they wash their hands, how often they go to the doctor, etc.). There is a concern that different interviewers will get different sorts of responses—that is, there may be important interviewer effects. Describe (in two sentences) how you could estimate the interviewer effects within your survey. Can the interviewer effects create problems of reliability of the survey responses? Explain (in one sentence). Can the interviewer effects create problems of validity of the survey responses? Explain (in one sentence). Solution to question 22 From yesterday : 22. A supermarket chain has 100 equally-sized stores. It is desired to estimate the proportion of vegetables that spoil before being sold. Three stores are selected at random and are checked: the percent of spoiled vegetables are 3%, 5%, and 10% in the three stores. Give an estimate and standard error for the percentage of spo
5 0.2471444 1333 andrew gelman stats-2012-05-20-Question 10 of my final exam for Design and Analysis of Sample Surveys
Introduction: 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution to question 9 From yesterday : 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution: estimate is p.hat=10/40=0.25. Se is sqrt(1-f)*sqrt(p.hat*(1-.hat)/n)=sqrt(1-0.4)*sqrt(0.25*0.75/40)=0.053. 95% interval is [0.25 +/- 2*0.053] = [0.14,0.36].
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same-blog 1 0.99622363 1349 andrew gelman stats-2012-05-28-Question 18 of my final exam for Design and Analysis of Sample Surveys
Introduction: 18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introduce notation as necessary for all the information needed to solve the problem. Solution to question 17 From yesterday : 17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points? Solution: se is sqrt(.5*.5/(n/2)+.5*.5/(n/2)) = 1/sqrt(n). Solve 1/sqrt(n) = .05, you get n = (1/.05)^2 = 400.
2 0.90291351 1348 andrew gelman stats-2012-05-27-Question 17 of my final exam for Design and Analysis of Sample Surveys
Introduction: 17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points? Solution to question 16 From yesterday : 16. You are doing a survey in a war-torn country to estimate what percentage of unemployed men support the rebels in a civil war. Express this as a ratio estimation problem, where goal is to estimate Y.bar/X.bar. What are x and y here? Give the estimate and standard error for the percentage of unemployed men who support the rebels. Solution: x is 1 if the respondent is an unemployed man, 0 otherwise. y is 1 if the respondent is an unemployed man and supports the rebels, 0 otherwise. The estimate is y.bar/x.bar [typo fixed], the standard error is (1/x.bar)*(1/sqrt(n))*s.z, whe
3 0.8928771 1361 andrew gelman stats-2012-06-02-Question 23 of my final exam for Design and Analysis of Sample Surveys
Introduction: 23. Suppose you are conducting a survey in which people are asked about their health behaviors (how often they wash their hands, how often they go to the doctor, etc.). There is a concern that different interviewers will get different sorts of responses—that is, there may be important interviewer effects. Describe (in two sentences) how you could estimate the interviewer effects within your survey. Can the interviewer effects create problems of reliability of the survey responses? Explain (in one sentence). Can the interviewer effects create problems of validity of the survey responses? Explain (in one sentence). Solution to question 22 From yesterday : 22. A supermarket chain has 100 equally-sized stores. It is desired to estimate the proportion of vegetables that spoil before being sold. Three stores are selected at random and are checked: the percent of spoiled vegetables are 3%, 5%, and 10% in the three stores. Give an estimate and standard error for the percentage of spo
4 0.87996012 1352 andrew gelman stats-2012-05-29-Question 19 of my final exam for Design and Analysis of Sample Surveys
Introduction: 19. A survey is taken of students in a metropolitan area. At the first stage a school is sampled at random. The schools are divided into two strata: 20 private schools and 50 public schools are sampled. At the second stage, 5 classes are sampled within each sampled school. At the third stage, 10 students are sampled within each class. What is the probability that any given student is sampled? Express this in terms of the number of students in the class, number of classes in the school, and number of schools in the area. Define appropriate notation as needed. Solution to question 18 From yesterday : 18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introd
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Introduction: 22. A supermarket chain has 100 equally-sized stores. It is desired to estimate the proportion of vegetables that spoil before being sold. Three stores are selected at random and are checked: the percent of spoiled vegetables are 3%, 5%, and 10% in the three stores. Give an estimate and standard error for the percentage of spoiled vegetables for the entire chain. Solution to question 21 From yesterday : 21. A country is divided into three regions with populations of 2 million, 2 million, and 0.5 million, respectively. A survey is done asking about foreign policy opinions. Somebody proposes taking a sample of 50 people from each reason. Give a reason why this non-proportional sample would not usually be done, and also a reason why it might actually be a good idea. Solution: Nonproportional sampling is usually avoided because it makes the analysis more complicated and it results in a higher standard error for estimates of the general population. It might be a good idea her
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same-blog 1 0.98060882 1349 andrew gelman stats-2012-05-28-Question 18 of my final exam for Design and Analysis of Sample Surveys
Introduction: 18. A survey is taken of 100 undergraduates, 100 graduate students, and 100 continuing education students at a university. Assume a simple random sample within each group. Each student is asked to rate his or her satisfaction (on a 1–10 scale) with his or her experiences. Write the estimate and standard error of the average satisfaction of all the students at the university. Introduce notation as necessary for all the information needed to solve the problem. Solution to question 17 From yesterday : 17. In a survey of n people, half are asked if they support “the health care law recently passed by Congress” and half are asked if they support “the law known as Obamacare.” The goal is to estimate the effect of the wording on the proportion of Yes responses. How large must n be for the effect to be estimated within a standard error of 5 percentage points? Solution: se is sqrt(.5*.5/(n/2)+.5*.5/(n/2)) = 1/sqrt(n). Solve 1/sqrt(n) = .05, you get n = (1/.05)^2 = 400.
2 0.97228676 1050 andrew gelman stats-2011-12-10-Presenting at the econ seminar
Introduction: Jim Savage saw this and pointed me to this video. I didn’t actually look at it, but given that it is labeled, “For new econ Ph.D.’s about to look for a job . . . what you might expect when you give your first talk presenting your research,” I can pretty much guess what it’ll look like.
3 0.9627887 2275 andrew gelman stats-2014-03-31-Just gave a talk
Introduction: I just gave a talk in Milan. Actually I was sitting at my desk, it was a g+ hangout which was a bit more convenient for me. The audience was a bunch of astronomers so I figured they could handle a satellite link. . . . Anyway, the talk didn’t go so well. Two reasons: first, it’s just hard to get the connection with the audience without being able to see their faces. Next time I think I’ll try to get several people in the audience to open up their laptops and connect to the hangout, so that I can see a mosaic of faces instead of just a single image from the front of the room. The second problem with the talk was the topic. I asked the people who invited me to choose a topic, and they picked Can we use Bayesian methods to resolve the current crisis of statistically-significant research findings that don’t hold up? But I don’t think this was right for this audience. I think that it would’ve been better to give them the Stan talk or the little data talk or the statistic
4 0.96272182 1143 andrew gelman stats-2012-01-29-G+ > Skype
Introduction: I spoke at the University of Kansas the other day. Kansas is far away so I gave the talk by video. We did it using a G+ hangout, and it worked really well, much much better than when I gave a talk via Skype . With G+, I could see and hear the audience clearly, and they could hear me just fine while seeing my slides (or my face, I went back and forth). Not as good as a live presentation but pretty good, considering. P.S. And here’s how to do it! Conflict of interest disclaimer: I was paid by Google last year to give a short course.
5 0.96112329 2068 andrew gelman stats-2013-10-18-G+ hangout for Bayesian Data Analysis course now! (actually, in 5 minutes)
Introduction: Here’s the link . When you’re on the hangout, please mute your own microphone! I’ll have the computer point at the blackboard. You can follow along with the slides: for the first hour for the second hour P.S. Apparently there is some limit on number of hangout participants (see comments). I didn’t know about that! Maybe next time will try “on air” hangout, I will have to learn more about this. Next week the teaching asst will do the course so no hangout, then in two weeks there is no class because it’s the day after Halloween and that’s a holiday around here. So we’ll resume this on Fri 8 Nov. See you then! P.P.S. Those of you who were able to join the hangout: Could you please let me know how the visual and sound quality were? Thanks.
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