<p>ignore this comment =]</p>
<p>without going into specifics…since its not allowed. it definitely did NOT say association</p>
<p>I second that. ^</p>
<p>It didn’t tell you to perform a test though. I just found the expected number of men who were in party Y (around 65 I think). There were actually 40 men in party Y so I said there is an association.</p>
<p>There was no chi-square test. For that problem, the proportions for the entire population were known, which makes the use of a statistical inference procedure such a chi-square incorrect.</p>
<p>…im curious as to how you determined an expected number.</p>
<p>The second graph on number 2…what was that?</p>
<p>It was nowhere near as difficult as I expected! </p>
<p>For the one that might possibly have required a chi-squared test, I think it was actually just a simple probability problem. I just figured out if P(A) was the same as P(A given B).</p>
<p>Also the investigative task blew my mind. It was so cool how you could figure it out just with the magic of algebra and probability! It was simple but really surprising and awesome.</p>
<p>How did only 4% of highschool seniors know that APUSH question? It seemed sooo easy.</p>
<p>“Also the investigative task blew my mind. It was so cool how you could figure it out just with the magic of algebra and probability! It was simple but really surprising and awesome.”</p>
<p>LOL I’m so glad I’m not the only one who feels this way! I thought that was so fun and cool and then felt like a huge nerd.</p>
<p>The easiest AP test I’ve ever taken. 38/40 on MC, and 23/24 on FRQ.</p>
<p>shoot. is that going to be no credit though? i did the test exactly right, to a tee. Will that still be a 3/4 since I said there was an association?</p>
<p>No chi squared (: yay
I was freaking out therefor a second!</p>
<p>It makes sense that I would screw up the algebra even though I got the equation right and everything else for #6. URG. </p>
<p>Oh well. I read online that if you have the right concepts and are a little bit wrong you can still get full credit for a question, so hopefully that will get me a 4 :).</p>
<p>For the independence one there is a probability formula to determine independence. A and B are independent if they are both non-zero and P(A|B) = P(A).</p>
<p>“I also got .04 for k on last problem.
.28 = k + (1-k).25
k = .04”</p>
<p>wait…i thot it said use the interval from part (a), so you couldn’t assume k = .28.</p>
<p>for my confidence interval, i just did like from like (k-blah blah blah, k+ blah blah blah) and wrote out my statement.</p>
<p>Yeah, you shouldn’t set the equation to 0.28.</p>
<p>What was the answer to question one? I,II,III?</p>
<p>LOL stat what a joke
that was the easiest AP i have ever taken
and i thought it’d be really hard too
LOL</p>