<p>^^^ I do, if you have questions PM me. No guarantee I’ll know how to do it but I think I have most of this down…</p>
<p>Which of the following can be used to show a cause and effect relationship between two variables?</p>
<p>A. A census
B. A controlled experiment
C. An observational study
D. A sample survey
E. A cross-sectional survey</p>
<p>What is the answer to that question? It seems too easy. I just want to make sure I’m right.</p>
<p>a controlled experiment.</p>
<p>@oblivion:
17 is D. This is a chi-square test for independence. There are (2-1)(2-1) = 1 degree of freedom. Note that there are two categorical variables, so imagine a 2 x 2 table in your head, one with flower color and one with stigma color. Since the chi-square statistic is 8.04 which is greater than the critical value for 5% significance and 1 degree of freedom, and the biologist also expected that the null hypothesis would be rejected (that they are not independent variables), the results are consistent with his expectations.
19 is B because the heavier one package gets, the lighter the other one will get.
23 is B because the distribution is steeper, so the standard deviation is smaller, so that means the sample size must be bigger.
38 is E, it is obvious that Y can be from 0 through 4 (just find all possible sums of 2 different or identical X variables). Then, find their respective probabilites: Y = 0 has a probability of (0.3)(0.3) = 0.09, Y = 1 is (0.4)(0.3) + (0.3)(0.4) = .24, and so on.</p>
<p>Ok, good. I started second-guessing myself for a second.</p>
<p>The distribution of the diameters of a particular variety of oranges is approx. normal with standard dev. of 0.3 inch. How does the diamter of aan oragne at the 67th percentile compare with the mean diamter?</p>
<p>a. 0.201 inch below mean
b, 0.132 inch below mean
c. 0.132 inch above mean
d. 0.201 inch above mean
e. 0.440 inch above mean</p>
<p>For anybody who has the 2007 exam, can you explain numbers 15 and 17? (i can’t write it here because they’re histograms and experiment tables…</p>
<p>^ I don’t understand 17, but 15 is D because the bulk of the data is off to the edges away from the mean and therefore the average distance (standard deviation) each unit is from the mean is going to be larger.</p>
<p>^ You find the z-score at 67%. I did invnorm (.67, 0, 1) = 0.4399. This is the number of standard deviations away from the mean so you times this by 0.3 = 0.132. Since the mean is 0 and the z-score of 0.4399 is greater than 0, it’s above the mean. So C (:</p>
<p>For 17, the critical value is 7.81 (go to the chi squared critical values chart, df = 3, tail probability = 0.05). The chi squared value you get from running the GOF-Test is 8.04, since the computed value is greater than the critical value, you reject the null hypothesis that this is a good model. Thus, the observed results are inconsistent with the expected ratio. So the answer is A.</p>
<p>^Way to go, I didn’t get that one. :)</p>
<p>why did you use 0 and 1 in invnorm (.67, 0, 1)…i thought we had to use 0.3 as std dev, not 1…</p>
<p>and for 17, is there a way to use the TI-84, instead of the chart and still get the answr you got?</p>
<p>thanks :)</p>
<p>Good luck, dudes!</p>
<p>@Charrizard: You are wrong about the degrees of freedom, it’s actually 1 since there are 2 categorical variables each with 2 types, so if you organized it into a proper data table it would be 2 rows x 2 columns (2 flower colors and 2 stigma colors)
However, you still got the right answer :)</p>
<p>^ I used 0 and 1 because you don’t know what the mean of the diameters of the oranges is. Since this is a normal distribution, the 67th percentile should be at the same z-score, so if you knew what the mean was, then you could do invnorm(.67, x, 0.3), but the z-scores would still be the same.
(Sorry, I’m not sure if this makes sense, I’m not that great at explaining… D: )</p>
<p>As for 17, I don’t really think there’s an inv chi-squared kind of thing on the calculator… so I don’t really think there’s any way to do it without using the chart…? I’m not too sure, but they do give you those charts on the exam so yeah ^^</p>
<p>Hi, quick question: I stored a lot of equations on my Ti-83 Plus calculator for the test tomorrow. Are Administrators of the exam required by the collegeboard to check that every student clear their calculator of memory before the exam? If that’s true, I just wasted a solid hour of studying.</p>
<p>Thanks in advance, and good luck all!</p>
<p>@bobtheboy: But this is a GOF-Test, not a Test for Homogeneity, so df = # of categories - 1, which in this case is 4 - 1 = 3.</p>
<p>I’m taking it tomorrow, good luck guys. and no they don’t clear calculator memories, my teacher even told us if we have trouble remembering the conditions to store them in there.</p>
<p>^Really? I thought that the College Board clearly stated that it was cheating to store notes.</p>
<p>Then if they’re so strict about it, they should clear memories.</p>
<p>Haha, I think it depends on your test center/school… since they’re the ones administering the test…</p>