<p>CV is critical value for a confidence interval.</p>
<p>@rabbit5145</p>
<p>i was confused over whether to use a diff in proportions test but then i had little time so i skipped to 6c and d. I took the sum of all p values above 0.05 and divided it by the sum of the frequencies of each p value, because any p value below 0.05 sees the null being rejected, and the power is defined as probability of null being accepted? how did you do it?</p>
<p>Magmar, I did do a two-sample proportion test and I’m pretty sure that is correct.
As for part c, I think power is the ability to reject the incorrect null hypothesis:/ I had absolutely no idea what to do and just came up with an expected p-value. I only did part c to at least write something for d in the hope of receiving partial credit-_-##</p>
<p>haha you are right about c… cant believe i mixed up the power definition… no credit then</p>
<p>Relax… Nobody I know has gotten the correct answer. And, I’m really curious as to whether the curve is same for all forms? That would be pretty unfair.</p>
<p>Not discussing specific question obviously but, hypothetically, if I had to subtract two standard deviations, I’d have to go through the square root of those two standard deviations squared added together because I know one can’t simply combine standard deviations. I also know that, although I’m subtracting, once I square a number, the minus sign becomes positive</p>
<p>Once again, from 2008 released test or something. I know if an LSRL has correlation r and I go to z-scores, my r won’t change. Values will but all I did was standardize so my values will be smaller but have the same residuals from the LSRL.</p>
<p>Which part (c) are you guys talking about with ‘power?’ I don’t remember seeing that anywhere on the free response.</p>
<p>For the chi squared test of association, I totally blanked on how to prove normal in the conditions. At the last minute I wrote degrees of freedom which someone later told me was right, but for some reason I sketched a little box plot and wrote “slightly skewed, safe to continue” and drew an arrow to normal. Do you think they’ll ignore it?</p>
<p>Everything else about my test was correct though and so were my degrees of freedom (row-1)(column-1) which made it 2, right?</p>
<p>um degrees of freedom has nothing to do with proving the conditions for chi squared test of independence lol you just check if your expected values are greater than 1 and 80% of them are greater than 5.
btw definition of power is sensitivity of the test not the ability to reject the null</p>
<p>For the Japan one, I put the answer with “not” in it. I don’t think the small sample size matters as long as the distribution is roughly symmetric/unimodal/outlier-free.</p>
<p>I also got 350 for smallest n.</p>
<p>I put “same r as the original” for number 38. I don’t think standardizing the units changes the correlation coefficient.</p>
<p>My chi-square value was .011.</p>
<p>I got .22 for marbles and .13 for classes. I also got the graph with 1 having the highest probability and 6 the lowest for the “lowest X” MC.</p>
<p>My answers seem to be matching the majority’s. Sweet.</p>
<p>Also, I too noticed the typo on FRQ 5. What was up with that? Did everyone put “doesn’t provide any significant evidence” for that simulation? I hope they throw it out…</p>
<p>Can somebody please explain how they got 350:</p>
<p>5 (the maximum error) = 2.626(36/sqrt (413))</p>
<p>If you plug in 350 instead of 413, the error is greater than 5.</p>
<p>Does everybody agree that 2.626 is the critical value with a df of 100? And the standard deviation of the sample was 36. How is anybody getting 350? The lowest value I get is 357, which was not an option, so I chose 413. Please show me where I am making an error here…</p>
<p>for question number 6, did anyone get the one where you have to find the power of test using a graph? how do you do that?</p>
<p>Oh, seriously?! That sucks then. How much do you think that would affect my answer if I didn’t prove normalcy?</p>
<p>I thought since the cell counts were all greater than 10 that according to Central Limit Theorem it would be approximately normal.</p>
<p>@TheBombingRange I used the critical value of 2.576 rather than 2.626. If you use 2.576, 350 is below 5.</p>
<p>@thebombingrange idk where you’re getting the df=100 from lol degrees of freedom is not related to the problem (you can’t even know the degrees of freedom without knowing the sample size which is what you’re solving for)</p>
<p>99% confidence-> CV of 2.576</p>
<p>Well you know what n is, or you are trying to figure out n in this case. Either way all the values are between 100-1000 so you round down your degrees of freedom to 100 which is how he got a CV of 2.626.</p>
<p>no df in the problem use z asterisk thingy for your percent confidence and get your CV</p>
<p>Was it a chi-square test of independence or homogeneity?</p>