syhuang

It's a straight forward question. Why avoid answering it? Dibs, this question is actually not that straight forward. First, in such a group, some of them are likely to get scores closer to population mean upon retaking the test. Some of them are likely to get scores higher than 140. Furthermore, the first subgroup is larger than the second subgroup. Therefore, the more accurate statement is "When retaking the test, more people in this group are likely to get scores closer to population mean than not" or "The average score of the group is likely to be closer to population mean when retaking the test". However, when talking about "people tend to score closer to the population's mean", I think it's kind of ambiguous. One side can say "Yes, they are, becuase there're more people likely to score closer to population mean or the average score is likely to be closer to population mean". The other side may say "No, because not all people are likely to score closer to population mean". Therefore, it's kind of about "semantics", does "average score" or "majority of the group" represent the whole group? Some may agree, some may disagree. I can understand what both sides are talking about, but I really don't want to join this name-calling war. So, let me try it once to see if it helps the discussion. First, the phenomenon HA described exists, but the more accurate statement is "average score of the group" instead of "people in the group". However, does the average score regress toward the "population mean"? This is another ambiguous part. I think HA's answer is yes, becuase "the average score of the group" is likely to be closer to population mean upon retaking the test. BJ's answer is no, becuase the target (not a good statistics term, but you know what I mean) of the "average score" regressing toward to is not "population mean", it's "mean of error". What's the difference? Let's say in a normally distribute population with mean of 100, one of them has a real IQ of 120 (from other more accurate tests) and he scores 140 the first time. Of course, the assumption here is normally distributed error with mean of zero. When retaking the test, (1) is he likely to score lower than 140? Yes. (2) is he likely to score closer to population mean(100)? Yes. (3) is he likely to score closer to his real IQ (120)? Yes. So what's the target of his score regress toward to? 120 or 100? I think most of you will say 120, which is his real IQ. Ok, now, go back to the example with a group of people who score 140 the first time. The assumption is the same (normally distributed error with mean of error and normally distributed population. The population mean is 100), and I simplified the example. Let's say there're three people who score 140 the first time and we know from other more accurate tests that their real IQ are Person A: 150 Person B: 125 Person C: 115 The average of their real IQ is (150+125+115)/3=130. (again, this may not be a good statistics way. Hope this can show the idea) Thus, when retaking the test, A is likely to get a score higher than 140 and be closer to his real IQ. B and C are likely to get a score lower than 140. The average score of the group is likely to be lower than 140. When retaking the test, (1) is the average score likely to be lower than 140? Yes. (2) is the average score likely to be closer to population mean(100)? Yes. (3) is the average score likely to be closer to their mean of real IQ (130)? Yes. Like the example of one person earlier, what is the target of the average score regressing toward? The population mean(100) or mean of real IQ (130)? I think most will say mean of their real IQ. Thus, the argument here is again about "semantics". You can say the statement of "regression toward mean of population" is right because the average score is likely to be closer to population mean". You can also say the statement is not right becuase the target of the regression is not population mean, which happens to be in the same direction of mean of erorr. In short, the phenomenon exists, but can the statement be called "regression toward the population mean"? it depends on what you refer to, a direction or a target.

I think many Bills fans agree there're questionable playcalling. However, what people don't agree with you is not because you question playcalling. It is because your comments like following: -------------------------------- I didn't say they would fire Fairchild, but I bet the process of decision making on game turning calls is changed. There may be a committe of Dick Jauron, the OC and the DC and Bobby April. -------------------------------- Yes, bragging rights about a change made in who makes the final decsion on game hinging calls. I wouldn't be surprized if Marv ends up jumping in on those calls. --------------------------------

Jags, OG Manuwai agree on long-term extension

Determining Draft Order • Strength of schedule for the previous season is the first tie-breaker for teams with the same winning percentage. • Divisional and conference records are the next step in the tie-breaking procedure. • As a last resort, a coin toss is used to determine the order of selection for teams with the same winning percentage.

No, I'm not talking about mean of population or population distribution. I'm only talking about error distribution and normally distributed error causing "regression toward the mean (of error)" phenomenon. You need to explain things step by step. It seems like HA agrees that not all the errors cause regression toward the meam. Although he and me may not refer to the same definition of "mean", it's not the point of my example, which is if errors always cause regression toward the mean. In my example, both "mean of population" and "mean of error" are the same, so we can focus on the effect of error distribution. It looks like the discussion now can move to the next topic (for example, mean of error vs. mean of population). Anyway, I'm done with the discussion and already let HA know what I try to show him. Now, let's go back to the usual name-calling shouting match.

Didn't I state that the "mean" in "regression toward the mean" is "the mean of error" when I brought up two questions?

It seems like we also agree on this one. Under normal circumstances, a normally distributed error would cause "regression toward the mean". When talking about math or statistics, in theory, errors don't always cause "regression toward the mean". Thus, you can NOT simply say "errors cause regression toward the mean", since this statement is not always true. You have to be more specific and state the conditions when your statement is true. Be more scientific, for example, "In normal circumstances, errors cause regression toward the mean" or "When the error is normally distributed, error causes regression toward the mean".

As you may know, I agree that normally distributed errors cause regression toward the mean. However, here I'm trying to show you that not all the errors cause regression toward the mean by using this extremely abnormally distributed error. Please note I stated the error here is normally distributed. Let me repeat the question: If a person's real IQ is 140 (known by other more accurate tests) and scores a 160 or 120 in a test with zero-mean normally-distributed error, will he likely get a score closer to the mean when retaking the test? It seems like you agree "regression toward the mean" normally applies to individuals. Right, the person's true IQ is 140. And because of an abnormally distributed error, when this person takes this test, the only outcomes are 135 (50%) and 145 (50%). Also, as I mentioned in another post, it's common that the expected value is not one of the possible outcomes. I'm aware this case is unrealistic. However, this abnormally distributed error example shows that not all the errors cause "regression toward the mean" on individuals.

Good. We now agree this abnormally distributed error doesn't cause any individual regressed toward the mean. In other words, even with error existed, "regression toward the mean" on individuals may not happen. I think I just show you that error doesn't always cause "regression toward the mean" on individuals.

Ha, it seems like I'm good at this. Anyone also wants me to take words out of your mouth? I'm two for two today.

You said "someone", thus, it looks like you're talking about "one person". If you're talking about a group of people, please answer the questions in my previous post about if "regression toward the mean" applies to individuals. With one person taking this test, he only gets one outcome, either 135 or 145. There's nothing to average after the score is out. Also, it's called "expected value", not "average outcome". You can say, before taking the test, the expected value of the score is 140. However, once the test is taken, there is only one score, 135 or 145. Even the expected value is 140, you can not score 140 in this test. It's common that expected value is NOT one of the possible outcomes. Need more example? take the famous dice example in this thread, the expected value of one throw is 3.5, but you can never throw a 3.5. Please don't confuse expected value with possible outcomes. When talking about "regression toward the mean", it's about the possibility of the score upon retaking the test being closer to the mean. It's not about if the expected value is closer to the mean or not.

Do you imply that "regression toward the mean" only applies to a group of people and doesn't apply to individuals? Anyway, answer this question, does "regression toward the mean" apply to individuals? Or if you want an example, here is one. If a person's real IQ is 140 (known by other more accurate tests) and scores a 160 or 120 in a test with zero-mean normally-distributed error, will he likely get a score closer to the mean when retaking the test? If your answer is yes, please answer the next question. Q: If a person (not a group of people) gets a 135 in this test with abnormally distributed error, does he likely get a score (either 135 or 145) closer to the mean when retaking the test?

The statement is completely wrong. Even the average is 140, you can not score 140 in this test. You can only score 135 or 145 here. If a person gets 145 and then retakes the test, he won't get a score closer to the mean. You won't regress toward anywhere even if you keep retaking the test. Do you call a score sequence like following as "regression toward the mean"? 135, 135, 145, 135, 145,145,135,145,.... As you can see, it doesn't regress toward any value. ------ EDIT: HA, I've a feeling that you may use "average score" to reason that "regression toward the mean" still applies here. Thus, before you do that, please make sure you don't change your definition of "regression toward the mean" in past 50+ pages. If you want to use "average score of a group of people", please make sure you're not saying that "regression toward the mean" doesn't apply to individuals and only applies to the average score of a group. Please don't say that if a person's real IQ is 140 (known by other more accurate tests) and scores a 160 or 120 in a test with zero-mean normally-distributed error, he is NOT likely to get a score closer to the mean when retaking the test. If you want to use "average of all past scores", please make sure you're not changing your definition of "regression toward the mean" from "...... the next score is likely closer to the mean upon retaking the test" to "...... the average of the next score and all past scores is likely closer to the mean". Last, please don't say that a person can score 140 in a test which only has two outcomes, 135 and 145.

It's a valid assumption, but may not be a good assumption to explain what you said above. To explain the difference to him, I think one of the better ways is to show him the case where mean of error is not zero. With the assumption of normally distributed error with mean of zero, it's kind of hard to explain the difference of the two.

(HA, let me rephase the questions to see if they're more clear to you) HA, please answer the following two simplified questions: (1) Regression toward the mean I see you agree with me on this case, so I don't repeat the question here. Can we conclude the "mean" in "regression toward the mean" is "the mean of error"? (2) abnormally distributed error Assume the real IQ is 140 and the test has an abnormally distributed error as following: (Again, the real IQ is known by other more accurate tests. We then take another test with an abnormally distributed error. I change the error distribution again to make it extremely abnormal. ) 135: 50% 145: 50% Will "regression toward the mean" happen? (A) Yes (B) No My answers are (B) for both questions , what are you answers?

HA, please answer the following two simplified questions: (1) Regression toward the mean If the real IQ is 140 and the error is normally distributed with mean of +5, what will the regression toward to? (A) 140 (B) 145 (2) abnormally distributed error Assume the real IQ is 140 and the test has an abnormally distributed error as following: 137-: 10% 138: 20% 139: 15% 140: 10% 141: 15% 142: 20% 143+: 10% Will "regression toward the mean" happen? (A) Yes (B) No My answers are (B) for both questions , what are you answers?

Correct. And I notice that in HA's post, he uses assumptions like "Assume measurement error is normally distributed with a mean of zero" to ignore that the "mean" in "regression toward the mean" is "the mean of error" and "regression toward the mean (of error)" may not occur when error is not normally distributed.

You're right, Vince Young did get a 16 on that wonderlic test when he took the exactly same test the second time. Although wonderlic score may mean nothing on his on-field performance, you still need to get the fact right.

"Regression toward the mean" is caused by error being normally distributed. If error is not normally distributed, there's no guarantee that "regression toward the mean" will happen. In other words, if error exists but is not normally distributed, "regression toward the mean" may not happen.

Instead of looking at the completion % alone, you also need to look at the opponents. One of the reasons that Young had better completion percentage in last 4 games is that he faced bad pass defenses. Young's completion % of last 4 games are: JAC: 53.3% HOU: 65.5% IND: 60% NYG: 68.6% The numbers look good. However, if I add the "average completion percentage against" of these four defenses, you'll be able to find out that Young actually only had one excellent game (NYG) on completion %. This game happened to be the only game Young has QB rating over 100, his ratings on the other three games are between 70.1 and 73.6. JAC: 53.3% (54.5%) HOU: 65.5% (63.7%) IND: 60% (65.2%) NYG: 68.6% (59.2%) Just for fun, I also list Young's completion % before the NYG game: PHI: 36.4% (59.0%) BAL: 52% (55.6%)

If I understand the post right, the 50% conclusion doesn't include two Bills games. Thus, assume Bills have 50% chance to win each of their next two games, the overall percetange of making playoff should be around 12.5% (50% * 25%).

Don't we still get a chance if Broncos lose to 49ers in week 17 and KC finishes at 9-7?

You're making things up again. I never complain which period compares to which period. What I complained is that you can't only use your method on JP and never test your method on other quarterbacks to see if your results are meaningful in all cases. Again, you're wrong. It's not about these drives, some may think these drives should count, some may not. Get the point, it's about how you massage data by using "personal judgements" to keep the stats which should be ommitted by different threshold based on your own rules. You never bring in "personal judgements" on other drives. You need to be reminded again: Everyone can create any numbers he likes by (1) Create his own rules to omit the stats he doesn't want (2) Pick the threshold to favor him most (3) Use personal judgement to retain the stats which should be ommitted by different thresholds based on his rules (4) Simplify the whole system to favor his argument (5) Get the manufactured numbers benefit his opinion These manufactured numbers are useless and show nothing more than "I think".

Tom Brady ???....... move back 1 position ..... ??? Say it ain't so!!! This should not be allowed. They can not move Brady back. Don't they know who Brady is? Sky is falling.

Do you care to show any proof that winning percentage is significantly improved with 200+ passing yards? or you just don't care about winning and only like fancy stats? For your information, the weekly passing yard leaders this season are Week 14: Weinke 423 yards Week 13: Manning 351 yards Week 12: Leinart 405 yards Week 11: Brees 510 yards Week 10: Palmer 440 yards Week 9: Roethlisberger 433 yards Week 8: Brees 383 yards Week 7: Harrington 414 yards Week 6: Delhomme 365 yards Week 5: McNabb 354 yards Week 4: Brees 349 yards Week 3: Kitna 342 yards Week 2: Manning 400 yards Week 1: Pennington 319 yards And the final scores of these games are: Week 14: Weinke 423 yards (L, 13-27) Week 13: Manning 351 yards (L, 17-20) Week 12: Leinart 405 yards (L, 26-31) Week 11: Brees 510 yards (L, 16-31) Week 10: Palmer 440 yards (L, 41-49) Week 9: Roethlisberger 433 yards (L, 20-31) Week 8: Brees 383 yards (L, 22-35) Week 7: Harrington 414 yards (L, 24-34) Week 6: Delhomme 365 yards (W, 23-21) Week 5: McNabb 354 yards (W, 38-24) Week 4: Brees 349 yards (L, 18-21) Week 3: Kitna 342 yards (L, 24-31) Week 2: Manning 400 yards (W, 43-24) Week 1: Pennington 319 yards (W, 23-16) Overall: 4W 10L