<p>Also, did everyone proceed with the t-test? I just said it would be unwise to proceed and therefore we cannot make conclusions based on this sample. (I was checking assumptions and the histogram for A had a strong skew to the left. Since n was less than 15, rules say that skewness or outliers are not acceptable). Saved me a lot of time :D</p>
<p>Actually, what @anotherindiankid said freaked me out for a moment because I thought he might be right. But since the data is for the entire population, there have to be values greater than 3 SD’s away from the mean. I’ll find out for sure on Monday because my statistics teacher is going to go through the FRQ’s. </p>
<p>And @lollypop7218, I think you’re supposed to do the t-test regardless if conditions are met or not…otherwise that kinda defeats the purpose of running the tests because you can always argue that a condition isn’t met (for example, plausible independence). Though, you should state that the conditions aren’t met…which I didn’t do :(</p>
<p>how was the cluster sample for the apartments supposed to be set up? I know I got part B for that question, but just checking on part A.</p>
<p>You definitely weren’t supposed to do chi-square for independence.</p>
<p>@lollypop: I proceeded with the two sample t-test despite the slight skewness on one of the box plots I plotted, but in my conclusion I stated that the robustness of my results may be in question or something along those lines.</p>
<p>As for the chi-square of independence, I did that as well although it seems now all one had to use was a simple probability formula (<em>sigh</em>). @alittlemonster: could you explain how that is completely wrong? It seems to me that it would make sense, but if it’s completely wrong, I must be a complete failure haha.</p>
<p>anyone know when collegeboard usually release scoring guidelines?</p>
<p>@eobaggs I think you will get a good chunk of partial credit because you got the right answer, and the logic you used was fine in that you were testing for independence. I was just saying that’s not what they were looking for.</p>
<p>@ lollypop- I’m pretty sure you could still use difference of two means. If you did a normality plot you would see that the plot of the points was approximately linear, therefore it was safe to assume that it was approximately normal.</p>
<p>I felt pretty good about the Stats exam. The multiple choice was easier than any practice test I’d taken and free-response was pretty fair as well. Optimistic: 4 maybe a 5. Pessimistic: 3</p>
<p>
</p>
<p>Normality, or lack thereof, doesn’t prevent you from calculating a z-score. The z-score is just a measure of how many standard deviations away from the average score you are. Because we’re only comparing one z-score to another z-score in order to determine who is a better choice for the team, normality has no bearing on the final answer.</p>
<p>I thought the test was insanely easy.</p>
<p>Probably a 5, but pessimistically… a 5. :D</p>
<p>@joanybologna- that question was about the importance of standard deviation</p>
<ul>
<li><p>FRQ 1.
Part A: Not normal. I don’t see how these values can conclusively indicate normality. Care to contest?
Part B: Z = 2.4. — [(370-310)/25]
Part C: Player A. Compute average Z or average percentile rankings between Player A and Player B.
Kind of tripped me up at first.</p></li>
<li><p>FRQ 2. Part A: If you got this wrong…
Part B: I said no because P(M given Y) does not equal P(M). But my supposed “definition” of<br>
independence could be wrong.
Part C: Wasn’t sure; just copied it. Actually, now that I mention it, what’s the difference<br>
between the two graphical displays – if any?</p></li>
<li><p>FRQ 3. Part A: Label floors; randomly select the first two digits, skipping repeats; and select all 4<br>
apartments from the two randomly selected floors.
Part B: Ensures the subpopulation (apartments with children) is included in our sample. If<br>
cluster sampling is used, and only floors 3, 4, or 6 are selected, carpeting wear data
will not account for apartments in which children are present. </p></li>
<li><p>FRQ 4. Conditions met (make NPPs for placebo and treatment groups for normality). It is a 2-Sample<br>
T-Test. It is not Matched-Pairs because that procedure entails a before-and-after comparison
of the same experimental subject, whereas this experiment clearly does not. FTR the null that
states placebo and treatment are equally effective. And, thus, data does not provide
convincing evidence at .01 significance.</p></li>
<li><p>FRQ 5. Part A: Obvious
Part B: 2.4. 10 mph increase times the .24 coefficient.
Part C: Obvious</p></li>
</ul>
<p>-FRQ 6. Part A: (.2682, .2918)
Part B: K, .75 - .25, (.25 -.25k) - (.75-.75k), (.25+.75k)
Part D: Wrote, “Mrs. Robinson, are you trying to seduce me?” in big, fat capital letters. I had
already earned a 5 about a third of the way through the free response, so the potential
reward didn’t justify the effort on this. </p>
<p>I missed 3 or 4 multiple choice, at most. 5 for sure. Almost as easy as the AP Micro and Macro tests; What a joke!</p>
<p>^Did you take a class or self-study?</p>
<p>Shermani, I really hope you miss a 5 by like two points for being that cocky…yeah.</p>
<p>@mtv22: Yes, I took a class.</p>
<p>@Katharineclaire: Sorry you didn’t do well? Or maybe I should apologize for being informed and knowing that, by all previous accounts and parameters regarding raw score concordance, I earned a 5. By multiple choice alone, I am on the cusp of a 4.</p>
<p>6d wanted you to use the interval from a and the probability from c so i took each end of the interval and solved for k, using that as my interval. is that correct? that was the only way i could think of using both a and c in the answer.</p>
<p>using k as .04 and doing a 99 confidence interval got me something completely different but i dont think thats what you were supposed to do</p>
<p>Number 6: </p>
<p>The interval from (a) was (.2682, .2918)…We can say with 99% confidence that the true proportion of twelfth-grade students in the US who would answer this question correctly is between 0.2682 and 0.2918.</p>
<p>Part b: The probability of guessing at random is (1-k). The conditional probability of answering correctly given that one guessed is 1/4 = 0.25; the probability of answering incorrectly is thus 0.75. So, the probability of guessing at random AND answering correctly is (1-k)(0.25). The probability of guessing at random and answering incorrectly is (1-k)(0.75).</p>
<p>Part c: The probability of answering correctly and knowing the answer is k. The probability of answering correctly and not knowing the answer is (1-k)(0.25). Therefore, the probability of answering correctly is k +(1-k)(0.25) = k - 0.25k + 0.25 = 0.75k + 0.25.</p>
<p>Part d:
Okay. The interval from part a means
0.2682 < P(answering correctly) < 0.2918.
P(answering correctly) is 0.75k + 0.25, according to part (c).
0.2682 < 0.75k + 0.25 < 0.2918
I can’t find my calculator, but basically subtract 0.25 from all parts and then divide by 0.75 to get the confidence interval for k.</p>
<p>We can say with 99% confidence that the true proportion of US twelfth-grade students who know the answer to this question is between [the two numbers you would have just calculated].</p>
<p>On 2b, they exam graders may mark you down for using Chi Square. I remember doing a problem similar to number 2 from a previous AP exam in my class, and it asked us to complete a conditional probability (similar to the one in 2a), and then explain whether or not it was independent. </p>
<p>The scoring guidelines explained that we analyze Independence in the context of this problem by comparing the probabilities of the two events to look for equality or lack thereof. The guideline stated (in ridiculously capitalized and bolded letters, I felt like I was being yelled at) that if the student performed a Chi Square Test for Independence, that it would be an automatic I.</p>
<p>@Nonexistent
sweet thats what i put :D</p>
<p>anyone know when the results get back?</p>