Ok good, part E was the only part people had different from me, hopefully I’m right.
@Gager292 for #6 part E, I think it’s method 2. Part E doesn’t ask anything about the mean being close to 6, and also, the population size is irrelevant, only the sample size is relevant because it’s asking about the sampling distribution, so the graph that would be created if you graphed the 200 sample means. For method 2, since the sample is only coming from one production line, I feel like the sample means will be closer to eachother, which would mean less variation when compared to the sample means caused by method 1. Your logic of comparing it to a mean of 6 makes sense for part F though, because that part asks which will produce a sample mean close to 6 inches, which would definitely be method 1. I hope this makes sense lol
6e is definitely Method 1 (I am an AP Stats teacher and dozens of AP Stats teachers agree on this one; not one of whom I am aware disagrees):
With Method 1 almost all of the 365 sample means will be within three s.d. (=3X0.0077=0.0233) of the mean of the sampling distribution = 6.0; that is, almost all of those 365 sample means will lie between 5.9767 and 6.0233 inches with Method 1. On the other hand, with method II probably none of the 365 sample means will be that close to the mean of the sampling distribution (6.0) but instead about half will be clustered tightly around 5.9 and the other approximate half of the 365 will be clustered tightly around 6.1.
If you measure the variability of the two methods by the standard deviation of the sampling means, then method I’s st.dev. is 0.0077 but method II’s is about 0.1 (more than 10 times as big). If you measure by range, then method I’s range is going to be approximately 6 sd so about 0.0466, but method II’s range will be over 0.2.
DarachHD states: “Part E doesn’t ask anything about the mean being close to 6.” I don’t think that DarachHD understands what variability means. Variability is the typical distance of values from their centre. The centre (i.e. mean) of the distributions of sample means obtained by both Method I and Method II is 6. If variability is small, then the 365 means will be close to 6; if variability is large, then the 365 means will be relatively farther away from 6. So while #6e does not specifically mention “anything about the mean being close to 6,” any superficial understanding of variability recognizes that we are to comment on the relative closeness to 6 of the 365 sample means obtained by the two methods.
DarachHD is absolutely correct in stating that the size of the population is irrelevant for the purposes of answering #6e.
Thank you! This is exactly what I’ve been trying to say the entire time!
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@violindad are you talking about the question that asked which method would be better for 365 samples( one sample a day)? I agree that method one would be better because it would consistently average the two distribution to result in a mean close to 6. I also said that because method two takes a random sample of either production lines, it would vary more around the true mean of 6 because unless it got an even samples of both production lines, the distribution might be skewed. Also, I said that for a single sample, method two would be better because it provides data for a single distribution that will not vary much around its own mean, I think that was one sample of the other questions for six. Does this sound similar or credit worthy?
These are the correct answers I believe:
Ok these were my answers for #6
A. No
B. Method 1
C. Method 2
D. Normal distribution with Mean = 6, SD = .11/Sqrt(200) = .0078
E. Method 1 - I said that method one would provide less variability because I looked at it from the view of standard deviations with bigger sample sizes… If I were to use 200,000 tortillas rather than 100,000 with Method 2, I would have a lower standard deviation with Method 1 than Method 2… I also incorporated the inference that with more values, you will get closer and closer to the true mean diameter of the tortillas, so I said it was Method 1 (I don’t know If this is correct)
F. Method 1 - Basically almost the same reasoning as above.
Thanks @catullus101 and @Kyuutoryuu. I got that half of number 3 very wrong then, but if I were to miss one question, it would be on probability. It never made sense to me.
FRQs done by a teacher I know:
- a.) Both corporations roughly symmetric/same median salary. Corporation A has greater IQR and Range than Corporation B. Corporation A has two high end outliers and Corporation B doesn't.
b.) Might choose Corporation A due to their being a greater opportunity to earn a higher salary. 3 of the 30 at Corporation A make over 60k a year and none at Corporation B make this. But with this comes also the chance of making a little lower than Corporation B as some of Corporation A’s employees are indeed lower in terms of salary.
c.) Accept job at Corp. B because after 5 years I will be making between 40k and 60k because this is the middle 50 percentile of the distribution. Will most likely receive a modest raise, in Corp. A some are still at starting salaries after 5 years.
a(I). No, .20 is included in the interval.
a(II). No, there are many other values besides .20 in the interval .09 to .21 and .20 is on the upper end.
b.) ME = .06/sqrt(4) = .03
c.) Program is not working as intended because .20 isn’t included in the interval .12 to .18 using new ME.
a.) 0.85
b.) 1.73
c.) .2824
d.) Increases due to the 0.15 probability being taken away, has to be distributed to other probabilities now to equal everything to 1.
Ho = P(aspirin) - P(no aspirin) = 0 ; Ha = P(aspirin) - P(no aspirin) < 0
State Conditions…
Test: 2 Proportion Z-Test
z = -1.75 and p = .0385 using calculator… Reject Ho, evidence to suggest taking aspirin does in fact reduce colon cancer.
a.) There is a moderately strong, positive, linear relationship between the arm span and height of the 12 seniors.
b.) Graph 2
c.) 62.05 inches || 11.74 + 0.8247(61) = y
a.) No
b.) Method 1
c.) Method 2
d.) Mean = 6 ; SD = .0078 according to Central Limit Theorem.
e.) Method 1
f.) Method 1
Hope you guys did well!
On the 2 proportion z test question, would I lose a point if I just said
Because .03<.05 we reject Ho that “” & accept the alternative that “”
I never said “p-value” anywhere in the problem. I already know I lost a point because I didn’t calculate the test statistic (I’m stupid). Would I lose another for this, or would the grader know that’s the pvalue?
Also for #2
If I got B wrong could I still get full credit
for C?
Depending on the grader would be the decider for your situation with the p-value…
If you got B wrong on Number 2, you would get credit for C if the answer you put for B still makes your explanation for C True (provided your answer for C is “No” with an explanation for why)
For example, say you said the ME = .02 and you got CI (.13,.17), and then you said no based off of that answer yes you will still receive full credit.
BUT, if you said this ME = .05 and had CI = (.10,.20) and then you said no based off that answer you probably wouldn’t get credit for that.
@Gager292
I put .28 as the new ME & said no…Would that be wrong?
This was such an easy exam but I made a lot of stupid mistakes. 
What does it take to get a 5?
Wait you said .28, did you mean .028?
If you put .28 I don’t think you would receive full credit because .15 + or - .28 is (-.13, .43) respectively which includes .20 so that would contradict the answer for C if you put this.
@Gager292
I put .28. That would have such a large interval so I thought that even though .20 was in it it would still show that the machine isn’t working as intended…oh well.
Okay I have a serious question about #4 on the AP exam. I dd everything correctly, but in the midst of taking the exam, I forgot how to find the p-value in the calculator, so instead I took the Z-score at the 95% confidence interval (1.96) and compared it to my z-score (it was a negative value, but I don’t remember exactly what it was). I rejected the null hypothesis because I said that the calculated z score fell into the rejection zone of the confidence level. My Princeton Review book said I could do that and AP gives you full credit, but it was my first time actually trying it out. I did reject the null hypothesis and in my conditions I stated that if the z-score is lower than z*=1.96 at the 95% confidence level, we would reject it. Also, for the test I only stated it was a Z-test, but I followed it with the two proportion z test equation. Is it still possible to get 4/4? I know for a fact that I passed (only need a 3 to max out the offered college credit), but this FRQ could throw me into the higher peaks of a 4.
did anyone take the make-up exam today? if so, PM me 
magic125,
I heard that the multiple choice was tough and the free response was easy. Can’t say anymore.
I just took our final exam and it was harder than the AP itself ,_,
@Maadii05 You have a final exam? At our school, the AP exam counts as our final exams so we don’t have any finals.