<p>Independence - 1 sample, 2 variables
Homogeneity - 2 sample, 2 variables</p>
<p>^Even better. Thanks :)</p>
<p>Oh thanks a lot! That probably just saved me a couple of points on the FRQ! (Won’t know until tomorrow…)</p>
<p>
</p>
<p>Finally, someone with a concrete answer haha</p>
<p>Taking it. Will Ace it. Nothing to it.</p>
<p>@mr wheezy.</p>
<p>first you find the standarddeviation of each type. so for the first one the sd would be</p>
<p>sqrt(.1*(1-.1)(25))=1.5</p>
<p>for the second it would be</p>
<p>sqrt(.45*(1-.45)(25))=2.487</p>
<p>when combining standard deviations, we take the root of the sum of squares (square the variances) so sqrt(2.487^2+1.5^2)=2.9047</p>
<p>I feel so ready for z-testing/t-testing/chi-squared testing, probability, and designing/simulating experiments.
However, I’m still shaky on least squared regression lines :(</p>
<p>Can someone help me figure this one out?</p>
<p>Independent random samples of 100 luxury cars and 250 non-luxury cars in a certain city are examined to see if they have bumper stickers. Of the 250 non-luxury cars, 125 have bumper stickers and of the 100 luxury cars,30 have bumper stickers. Which of the following is a 90 percent confidence interval for the difference in theproportion of non-luxury cars with bumper stickers and the proportion of luxury cars with bumper stickers fromthe populations of cars represented by these samples?</p>
<p>I’m between two answer choices:
They are both the same except for what’s in the radical.
The first is:
(0.5)(0.5)/250 + (0.3)(0.7)/100
The second is:
(155/350)(195/350)(1/250 + 1/100)</p>
<p>I picked the second, and the first was the right answer. How do you know when to combine the proportions and when to leave them independently?</p>
<p>Half way through the REA AP Statistics guide. I have a feeling this test is going to be really easy for some reason.</p>
<p>
</p>
<p>CC general consensus is to avoid pooling on the ap exam. There is no way to make a mistake with choice 1.</p>
<p>Ohh so you always pick choice 1? No matter what?</p>
<p>Yes, every answer on the exam is A.</p>
<p>Lol you know what I meant… You pick the one that doesn’t pool. :|</p>
<p>For the free response, don’t pool. On the MC, they won’t give two correct answers. Start with the simple looking formulas and eliminate your way to the right answer.</p>
<p>I keep forgetting all the assumptions and inferences. I really wish I had taken the class this semester instead of last.</p>
<p>Does anyone know how many points you get marked down for doing a Chi-Squared test for Homogeneity without stating it (just solving for Chi-Squared and p-value) and without stating that all expected values are greater than 5?</p>
<p>I’m definitely going to remember to do that for the test tomorrow, but I just want to know how harsh they grade (I’m assuming that would be work no credit).</p>
<p>Okay thanks An0maly.</p>
<p>Got a couple more…</p>
<p>The weights of a population of adult male gray whales are approximately normally distributed with a meanweight of 18,000 kilograms and a standard deviation of 4,000 kilograms. The weights of a population of adultmale humpback whales are approximately normally distributed with a mean weight of 30,000 kilograms and astandard deviation of 6,000 kilograms. A certain adult male gray whale weighs 24,000 kilograms. This whalewould have the same standardized weight (z-score) as an adult male humpback whale whose weight, inkilograms, is equal to which of the following?
(A) 21,000 (B) 24,000 (C) 30,000 (D) 36,000 (E) 39,000</p>
<p>I don’t understand how the answer is E… shouldn’t it be D?</p>
<p>Could someone explain this question to me?
Traffic data revealed that 35 percent of automobiles along a highway were exceeding the speed limit. It was also determined that 52 percent of all sports cars were speeding. What is the probability that a randomly selected car will be a speeding sports car?</p>
<p>I was stuck between
c.) .182 and
e.) It cannot be determined with the information given (merely because of the fact that it would be too easy otherwise)</p>
<p>Anyways, why is it choice e?</p>
<p>Oh wow never mind. I read the question wrong. :'(</p>
<p>Ohhh I was stuck on that one too.</p>
<p>It’s because you aren’t given the dependence factor (or P(B|A)). You can’t assume that they’re dependent on each other; maybe they’re completely independent events.</p>
<p>@zzxjoanw3</p>
<p>You do not pool for the confidence interval of the difference of two proportions because the formula does not require you to do so; quite straightforward. </p>
<p>YOU do, however, pool if you are using the two sample proportion test.</p>