|By Gunit123 (Gunit123) on Wednesday, May 07, 2003 - 04:40 pm: Edit|
Hey. Post your thoughts about the exam here.
I know I messed up on the Part 2 question (is the test acceptable) I said yes, but it was no because you didn't know sigma and the sample size was small.
What about Part 1s that everyone had trouble with? I had a question on the one about 7 points for guessing correctly, 2 for black, and 0 for incorrect. I put that you needed to eliminate 2 but I just guessed.
|By Peterkangisfat (Peterkangisfat) on Wednesday, May 07, 2003 - 05:25 pm: Edit|
I thought the multiple choice was a little hard; part 2 was fairly easy. Do u remember any part 1 questions?
|By Pringles (Pringles) on Wednesday, May 07, 2003 - 05:55 pm: Edit|
gunit, i put eliminate 2 too. i thought part 2 was easier compared to last years. what did you guys put for 6) the last part? about the vp of the company saying coaches would be better than vans? and for the confidence interval part, did anyone get around .4 something for one of them? i didn't use a t-distribution...
|By Gunit123 (Gunit123) on Wednesday, May 07, 2003 - 06:34 pm: Edit|
Yes ... I remember a lot of m.c. questions ... uh... what about the one where it was like pick 2 something without replacement (probability) and they gave u a table that had like 0.3 0.2 0.2 0.2 and 0.1 or something. I got 0.09 for my answer I think.
The last question. I think the choice was D. It was dealing with Residuals.
A question with Roman Numerals regarding probability. I picked the choice with II and III, mutually exclusinve and P(A or B) equals P(A) + P(B) ... i think they both mean the same thing
For the question regarding how do you know if you did residuals right ... I put the mean = 0.'
I can remember more if you bring em up...
There was one where it was like what can a confidence interval give you versus a hypothesis test ... i forgot the choices ... they were all similar... I think I picked C
PART 2: The last question I said he should repeat an experiment like that in Part C. And if he finds a different p^ value, one that gives a higher confidence interval, then 0.76 is included , and anything higher, according to the graph, would mean that the coach's give more money than the vans. (for my C.I. in part C or whatever it was I got something like 0.58 to 0.74 i think, which was below 0.76)
|By Gunit123 (Gunit123) on Wednesday, May 07, 2003 - 06:37 pm: Edit|
I didn't use a t-distribution either because p^ was 0.76 Thus, all you have to do is plug into the formula: p^ +/- Z alpha/2 Radical p^q^/n
|By Winnerpl (Winnerpl) on Wednesday, May 07, 2003 - 07:15 pm: Edit|
-The last one with residuals, answer was 4 (all have to add up to 0).
-"like 0.3 0.2 0.2 0.2 and 0.1 or something. I got 0.09 for my answer I think. "
hmm I got 0.22 but I'm not sure
-you need to eliminate 2 choices for guessing to be better
-#5 required x^2 test for independece, x^2=8.9
-I got the vans to give more money anywhere when p<.75. (and part 3 of the question would suggest this)
-In both cases for type I + II errors I put that the consequences would be good for the law firm since either way they'd win the lawsuit (I'm not sure)
|By Icarus (Icarus) on Wednesday, May 07, 2003 - 07:17 pm: Edit|
i thought the whole test was really quite easy...
|By Usernamehere (Usernamehere) on Wednesday, May 07, 2003 - 08:58 pm: Edit|
Type I error occurs with the false REJECTION of a TRUE null hypothesis. The probability of this is simple the significance level of the test. Your alpha level is just defines what "unlikely" is in the statement "this is unlikely to occur solely by chance if the null hypothesis is true". Of course, unlikely doesn't mean impossible, so some of the time a very large (or very small) z or t value just is by chance. When you conclude it means something when it is just by chance, that is Type I error.
Type II error occurs with the false ACCEPTANCE of a FALSE null hypothesis. Calculating type II error is much more involved, and requires the knowledge of a specific alternative value of the population parameter.
BOTH of these represent a threat to the firm. With type I error, they falsely conclude they will be able to make their case and win. They will not be able to recover expenses. With type II error, they falsely conclude that they will not be able to win the case, so they don't take it on at all. In reality, however, they would indeed have been able to win the case, and could have made a lot of money. They lose an opportunity with type II error.
|By Smiley (Smiley) on Wednesday, May 07, 2003 - 09:01 pm: Edit|
For that problem, usernamehere, what did you put null hypothosis and the alternative hypothesis to be? I got the error right but am afraid I stated the hypotheses wrong
|By Futuredoctor (Futuredoctor) on Wednesday, May 07, 2003 - 09:32 pm: Edit|
Smiley, I have the same question.
For the null hypothesis, I said that the mean is equal to .05, while for the alternate, I said that the mean is smaller than .05. A type 1 error in this case would be that they falsely conclude that they will not be able to win the case so they don't take it on at all. A type two error would result in concluding that they will be able to make their case and win. They will not be able to recover expenses.
|By Usernamehere (Usernamehere) on Wednesday, May 07, 2003 - 10:42 pm: Edit|
Null hypothesis: p=.05
Alternative hypothesis: p>.05
I think it said that they would take the case only if more than 5% of all vehicles of the given make and model had defects. Or that if the value with defects was 5% or less, they would NOT take the case. (It said one or the other, or even both, of these)
You're really only concerned whether or not p is greater than 5%. You might try to make this (the condition you need for the firm to take the case) your null hypothesis, but that is a problem since it is an inequality. The null hypothesis must be stated in terms of a p= or mu= EQUATION (showing NO difference or NO change), not an inequality. The inequality (or difference) is left to the alternative hypothesis.
Futuredoctor's hypotheses make sense in terms of this, but from what I remember of the problem (and I could certainly be wrong... but I hope not!), they would not take the case if p was less than or equal to .05. That can then be expressed with an equation, leaving the inequality for the >0.05. If they did say They do take it if the proportion is 0.05 or greater, however, then Futuredoctor's hypotheses are correct. (And if they're so close to being totally correct, that would probably not result in too serious a point deduction.)
|By Majinjou1 (Majinjou1) on Wednesday, May 07, 2003 - 11:57 pm: Edit|
Considering this was my first AP exam, I learned a very important concept: Time management. I went through that test so slow, spending too much time on a given problem. In the first section, I was up to question 28 when there was 10 minutes left for that section to end! LOL.
Part 2 was worst for me. The boxplots were no big deal, but the explanations, I wrote too much, needlessly wasting time. Since most of the part 2 questions were explanations, I wrote too much and didnt have time to get to the last problem. The Chi Square test was a breeze, but that problem about the ties, (chuckles) I didnt know what the was going on in that problem. I was so lost on what to do. Anyone know the answer to that problem or how they went about doing it? Thanks
Oh well, it was a great learning experience and I'll know to manage my time more effectively on the two APs I'm taking next year: Bio and Cal BC
|By Rocket1406 (Rocket1406) on Thursday, May 08, 2003 - 03:13 pm: Edit|
Hmm.. Yes the TIES PROBLEM was very strange...
Here is what I got ...
For part A, I did 2 normalcdfs and added em up:
normalcdf(-E99,14,15.7,0.7) + normalcdf(18,E99,15.7,0.7)
For part B, the 14 shirt size proportion, I did normalcdf(14,15,15.7,0.7) if the M interval was 14 to 15 I forgot, and I think I got like 0.507
I forgot what the rest of the problem asked, care to remind?
|By Rocket1406 (Rocket1406) on Thursday, May 08, 2003 - 04:56 pm: Edit|
How did I do on the TIE problem, is that anywhere close to what any of you guys got? Please post more about the Statistics exam I am curious about most of the part 2s and the harder part 1s!!!!!!
|By Winnerpl (Winnerpl) on Thursday, May 08, 2003 - 05:57 pm: Edit|
The TIE problem yes you did it correctly I believe (I got .51 for B). The last part of that question asked for the probablitlity that the number of people who choose M out of a sample of 12 is exactly 4. (No idea how to do this really, though I put some answer there).
|By Rocket1406 (Rocket1406) on Thursday, May 08, 2003 - 06:22 pm: Edit|
For choose M out of a sample of 12 is exactly 4 I did a binompdf(n,p,k)... which is binompdf(12,.51,4) = i forgot the number... is that right?
|By Usernamehere (Usernamehere) on Thursday, May 08, 2003 - 08:35 pm: Edit|
Yes, that is right, since we use binomial probability techniques when:
Each observation has two possible outcomes (medium or not)
We have a fixed n (12 in this case) and are interested in finding the probability that a certain number of "successes" occur during n observations.
The outcomes are independent (we're assuming this to be true - there isn't really any reason why it wouldn't be if they didn't tell us.)
The probability of M is the same for each observation (same thing... we're just assuming this is true - no real reason it shouldn't be).
|By Pringles (Pringles) on Thursday, May 08, 2003 - 08:44 pm: Edit|
for the coach/van problem, on one of the parts where it asked for 95% interval, was the proportion of strong support or whatever 0.35? or was that weak?
|By Rocket1406 (Rocket1406) on Thursday, May 08, 2003 - 10:34 pm: Edit|
I am not sure Pringles ...
|By Futuredoctor (Futuredoctor) on Thursday, May 08, 2003 - 10:53 pm: Edit|
Hi, can someone please answer my question about the hypothesis pair?
Also, there was a wierd question that said that a high school conducted a study of high school seniors... What type of study was this (randomized experiment, non-random experiment, random survey, non-random survey...)
Also, there was a question that asked what the study was trying to control for. I said that the answer was gpa and grade level.
Finally, one of the free-response questions said whether a control group would be necessary for the experiment question. I said no, because a matched pairs design is a form of a control.
|By Rocket1406 (Rocket1406) on Friday, May 09, 2003 - 06:10 am: Edit|
I put random survey, but I think there was an error in the question.
I put blocking GPA and Grade Level also.
I also said no, a control group would not be necessary.
|By Rocket1406 (Rocket1406) on Friday, May 09, 2003 - 09:32 pm: Edit|
Does anyone have pretty certain answers to ALL of the part 2s?
|By Rocket1406 (Rocket1406) on Saturday, May 10, 2003 - 07:02 am: Edit|
|By Rocket1406 (Rocket1406) on Saturday, May 10, 2003 - 08:34 pm: Edit|
|By Smiley (Smiley) on Saturday, May 10, 2003 - 10:36 pm: Edit|
A control group is not necessary if they are absolutely doing one of the programs and just want to find out which one to do. It is necessary if they want to know whether either program does any good. otherwise, during the 10 weeks something else might have changed that reduced stress level (weather, whatever). Control will tell you if there's even any point in using either program
**how were we supposed to get standard deviation for part b of the last one?
|By Futuredoctor (Futuredoctor) on Sunday, May 11, 2003 - 10:20 am: Edit|
A matched pairs design is in essence a control, and any other controlling would be extraneus...
I also said that because this was a randomized, comparitive experiment, causation may be implied.
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