Stat 100

[BarTopDancer]

It's rocket science to me! ;)

Is anyone here good at Stats (and seriously patient) who would be willing to help/look over my work? I am seriously math challenged and need all the help I can get.

My friend who was supposed to help me with my Stats has gone MIA.

It's an online class out of Penn State Uni, so the whole go to office hours and get help from the teacher thing is kinda out of the picture. It's also my only option to take this class.

Thanks!

[Kevy Baby]

Yeah, I got nothin'. I sucked at Stats.

[lashbear]

I know that 71.4% of all stats are made up on the spot......

[Kevy Baby]

I know that 71.4% of all stats are made up on the spot......No, that number is 68.9% silly with a mean average of blue.

[Alex]

In what format do you need it reviewed? If it is stats 100 I could probably review. Have an example of what is troubling you?

[BarTopDancer]

The questions are almost more critical thinking in a Stats format (analyze results in terms of what we just learned). It doesn't look like we have to memorize how to find standard deviation or other "typical" stats 100 things.

What was kicking my butt at the moment was a reference to what I think is an incorrect example. Currently awaiting instructor confirmation.

Here is one question (different then what I am waiting confirmation on).

I may just be over thinking the answer. We're studying means, outliers and standard deviations (along with bell curves, histograms and other fun stuff). I think they are all means, but they could all be outlers. Somehow I think the teacher is looking for a mix. My answers are in italics.

QUESTION: For each of the following situations, would you prefer that your data value is an outlier or the mean when compared to others’ data values? Provide reasoning.

a. your grade point average – mean - the GPA at PSU cannot be above 4.0 therefore an outlier would be extremely low.
b. age at which you graduate from college – mean (but it could be outlier).
c. monthly rent for your first apartment – an outlier towards the cheap end

[blueerica]

Sorry I missed your IM last night, the computer was on and to the side. Saw it when I flipped open the laptop.

Anyhow...

My ideas:

a. Outlier. The mean GPA at PSU isn't likely to be super high. It isn't at most colleges. You'd rather have the outlier on the high end. By assuming you'd want the mean, or average, you'd also be assuming that the GPA of PSU, or any college depending on the teacher's perspective on GPA means, is high. So yeah, I'd go with outlier.

b. Haha, could depend on this one. Wanna Doogie Howser it and graduate early? Then outlier. I'd personally prefer mean, but it would depend on your version of the mean - is it older than I imagine it is? If I could go back in time, sure, I'd take mean, but I'm also glad I was an outlier at CSULB. I'd stick with mean.

c. I'd hope monthly rent for ANY apartment would be an outlier on the low end, just as you said. Hehe, who cares if it's my first! You could reason that you'd want it to be the mean, so that you'd understand the value of blah blah blah - who cares, we wants cheap rent!

OK, time to get ready for work!

[BarTopDancer]

See, you think like me. And I'm thinking it's a trick question.

[Ghoulish Delight]

A) outlier. Most colleges grade so that the mean is a C or B, so you want to be an outlier on the high side.

B) Yeah, kinda depends. I just heard that the mean in California is now 6 years to achieve a bachelor's. But is that 6 years from the end of high school? 6 years from when you start taking college courses? Either way, odds are that the average age of graduation is significantly higher than the traditional "standard" of 22, so outlier seems the best answer.

C) I'd say "mean for the apparent quality of apartment being rented". You don't want to over pay, but if it's too cheap, you start to wonder why, that perhaps things are too good to be true.

[Alex]

Those don't sound like questions with firm answers but rather questions where you demonstrate your understanding of the terms in how your justify whichever answer you gave.

As an example, an answer to #1 (no, you really shouldn't be this wordy, I'm just having fun with it):

Well, sir, it depends on whether you want to know how I'd prefer my hypothetical GPA relate to the reality of the PSU mean GPA or how I'd prefer a hypothetical PSU GPA to relate to the reality of my GPA. If the former, since the reality is that the mean PSU GPA is likely around 3.0, and since my goal is to excel in all my studies my clear preference would be that my GPA end up being an outlier at least two standard deviations above that mean, putting me in the top 3 percent of all students.

However, if the question is how I would prefer a hypothetical mean PSU GPA to compare to my actual GPA then...

[AND HERE IT DIVERGES DEPENDING ON ACTUAL GPA]

...since my GPA is, if I may toot my own horn, a very respectable 3.7 and I am a warm-hearted person, while I suspect this would be an outlier I would prefer it were the mean, indicating that I have surrounded myself by intelligent achievers like myself.

...since my GPA is 3.1, there is little reasonable hope that this could be an outlier to the high side so would really prefer that this be very close to the mean since the alternative is likely that I am below the mean.

...since my GPA is 1.9 the only realistic evaluation is that I am an outlier to the low end. The coldly calculating part of me would prefer that I be at the mean, but then that would suggest that PSU kind of sucks. So the another part of me would prefer to be sucking at a good school rather than average at a bad one.

[BarTopDancer]

Thank you!!!!

[BarTopDancer]

OK, here are my answers:

your grade point average – an outlier towards the high end. Since GPA at PSU can only be 4.0, and GPA in college is usually a B or C average, I would want to my GPA to be an outlier at the high end of the GPA scale, separating myself from the mean.

b. age at which you graduate from college – this depends on when you start college, and where you attend college. Depending on when you start, how many classes you take and if you pass them on the first attempt can greatly vary the age you are when you graduate from college In CA it takes about 6 years to complete a Bachelor’s degree. Since I am 31, I am most likely already an outlier. This could be answered either way: an outlier towards the younger age, the mean, or an outlier towards an older age since I am graduating college. Since I am already an outlier, and I am ecstatic to graduate in the fall, I have to say that I prefer my value to be an outlier when I graduate.

c. monthly rent for your first apartment mean for the quality of apartment being rented. If the apartment is to cheap, it may be in an extremely unsafe area, or have underlying issues that aren’t visible upon viewing or move-in.

[BarTopDancer]

One more (for now). I may have one later but I know you guys are getting ready, if not already are camping.

8. Suppose a STAT 100 exam has 50 questions each worth 2 points. The results for this exam follow a normal distribution where the mean is 80 points with a standard deviation of 6 points.

Part A: Approximately 95% of the exam scores will be between what two values?

a. (80± 6) = (74 to 86) points
b. (80 ± 2  6) = (80 ± 12) = (68 to 92) points
c. (80 ± 3  6) = (80 ± 18) = (62 to 98) points
d. (80 ± 2) = (78 to 82) points

Approximately 95% of the exam scores will be between 62 to 98 points.

Part B: Would you expect a student to score as low as 50 points on this exam? Provide reasoning.
I would not expect a student to score as low as 50 points on this exam. I would hope students studied and absorbed knowledge to perform better. However, some students scored 62 points, and there are 5% of the students left. Based on the mean, a student could score as low as 50 points.

[Moonliner]

As stated there is nothing in the question that implies you get to choose to be either a high or low outlier. Your choice is only outlier or mean.

So using the GPA example, if you choose outlier you are taking a chance at falling on the low side: A failing grade. Whereas if you choose the mean you are guaranteed to pass but not to excel.


(Note: I am assuming that in the original statement of the problem, the part after the '-' is your initial answer and not part of the supplied question)

[BarTopDancer]

Thanks Moonie. Now I am back to being confused and overthinking. Maybe I'll add that too.

Ok,

your grade point average – The mean. an outlier could mean I’m towards the high end or the low end of the GPA scale. Since GPA at PSU can only be 4.0, and GPA in college is usually a B or C average, I would want to my GPA to be an outlier at the high end of the GPA scale, separating myself from the mean. But since I cannot pick to be at the high end of an outliner, I’d prefer my data value in this to be a mean when compared to others data values.

[Alex]

Your answers to #8 are incorrect.

Since this is homework, I am loathe to just tell you the correct answer on a factual question. But you didn't explain why you gave the answer you did for part 1.

So if you explain, I can perhaps point you into the right direction (and being shown where you went wrong will be more helpful than just being told the right answer).


For part 2, the key thing to note about your answer is that ultimately you explained that it is possible for someone to score 50 points. That is irrefutably true (it is possible for someone to have had every even numbered score between zero and 100). But that wasn't the question, the question asks whether you expect someone to score that low.

So, since you've been told that the results followed a normal distribution, the first question you need to ask yourself is this: How many standard deviations is 50 from 80?

Then using what you know about how standard deviation correlates to percentage of results, reconsider the question asked.

[BarTopDancer]

Thanks Alex.

I don't want the answers to be handed to me(thank you). I want to understand so I can get the answers on my own (important for the tests).

I used a formula in the book following the Empirical Rule for the third standard deviation. I obviously didn't understand it correctly.

My original thought was A (between 74 and 86) because it has only 1 standard deviation. But the 2 points per question threw me off.

For part 2, I over thought the question. That is one of my challenges - over thinking these things.

Ultimately I would not expect a student to score 50 points on the exam. Even with the standard deviation of 6 points, the mean is high enough to expect students will score above 50 points.

[Alex]

A is also incorrect for Part 1. But now I'm scared I'm misremembering the % values for standard deviations (but I just confirmed I'm not). What percentage of results in a normal distribution are covered by the following:

1 standard deviation
2 standard deviations
3 standard deviations


The fact that each question on the test is worth 2 points is irrelevant to the answers. All you need is the mean score and the standard deviation. You don't need to know how many points each question was worth or how many points were possible.

The answers would be the same if each question were worth one point and there were 16,000 questions on the test.


For part 2, the key point of the question is that a score of 50 points is five standard deviations (30 point difference = 6*5) away from the mean. What percentage of results will be five or more standard deviations from the mean in a normal distribution?

[BarTopDancer]

OK, I think I understand part 1.

(80 ± 2 * 6) = (80 ± 12) = (68 to 92) points

The mean is 80. The standard deviation is 6. You have to take the Mean ± 2(SD). So 80 ± 2*6 = the answer.

Part 2.

I have no idea. I am so lost again. I don't even trust that my answer for A is correct anymore.

I think that 3 standard deviations is 18. But you already said C was incorrect for #1.

I'll look at it again in the morning. I appreciate your help.

[Alex]

The mean is 80. The standard deviation is 6. You have to take the Mean ± 2(SD). So 80 ± 2*6 = the answer.

This is correct.

The empirical rule is 68-95-99.7, right? That means that for a normal distribution, one standard deviation will include about 68% of the population. Go out to 2 standard deviations and it will include about 95% of the population. 3 standard deviations takes you up to 99.7% of the population.

So, since the question asks about covering 95% of the results, that would be two standard deviations. That's the reasoning underlying the correct formula you gave. So Answer B (68 to 92) is correct.


Part 2.

I have no idea. I am so lost again. You're correct that 3 standard deviations is ±18 points from the mean (62 to 98). And 5 standard deviations would be ±30 points.

Per the empirical rule, 99.7% of all results will be within just 3 standard deviations of the mean. So we know that only 0.3% of all results will be lower than 62 points or higher than 98 points.

I don't know if it is covered by your textbook but if you extend the empirical rule you get:

1 standard deviation = 68%
2 standard deviation = 97%
3 standard deviation = 99.7%
4 standard deviation = 99.99%
5 standard deviation = 99.9999% of all results

So, since a score of 50 is 5 standard deviations away from the mean, you would expect only 0.0001% of scores to be that far from the mean. Another way of saying that is you'd expect one student out of a million to score that low. Since it is extremely unlikely that a million students are taking the Stats 100 class you would not expect that anybody will score that low (though it remains statistically possible it is statistically improbable).

[Alex]

It's hard to lead through the logical steps asynchronously.

[BarTopDancer]

If the standard deviation was 10, then 50 would be 3 standard deviations away from the mean (10 * 3 = 30)?

Cause if that is right, then I think I understand the first part. Still trying to digest the 2nd part.

[Alex]

Yes, that is correct.

[BarTopDancer]

Thank you!

I think I understand now. Going to look at it again in the morning.

[blueerica]

Somehow, I don't remember my intro to stat course questions being this... bizarre. :) Then again, my foundation was so poor from the first round that it took second and third rounds to get it firmly cemented into my brain (thanks, o business school for requiring so many stats courses... I actually use it, even when I thought I wouldn't!)

When the teacher turned to gambling, it all made more sense...

[Alex]

Erica, did my explanations make sense (having given them I'm now paranoid about having made a silly mistake).

[BarTopDancer]

They made sense to me.

Not to speak for Erica, but I know I was expecting something along the lines of another Stats 100 class (that I tanked) where we had to memorize formulas and learn how to calculate standard deviation, etc... right off the bat. The class specifically says you don't need to learn to calculate a lot of these things since there are programs to do it for you.

In a way, this is a huge relief. In another way, it's still frustrating because my brain doesn't easily work this way and I'm overthinking the questions.

[blueerica]

Alex, your answers made sense... I was reflecting on the questions as a whole. I feel that the learning experience can be helped and hampered by using such devices as 'interesting' questions.

But, like I mentioned earlier, the minute I could apply stats to gambling, it was Straight A Central.

[BarTopDancer]

I got an 87%. Thank you very much for the help and explanations with those two questions.

Sadly, when I reviewed the work, I have no idea how I got to the conclusions I did with any of the assignments. Which means I have no idea how I am going to remember it for the test.

Do you think the teacher will accept a "Brain is full, it cannot accept any new information" message?

[blueerica]

Sadly, when I reviewed the work, I have no idea how I got to the conclusions I did with any of the assignments. Which means I have no idea how I am going to remember it for the test.


You drove there.

You were going in the right direction with some of your questions by thinking them out first, and I hope some of that comes with you during test time, though I know tests can be more pressured...

[BarTopDancer]

Halp - Margin of Error

Question:

A question on a youth survey was “How well do you get along with your parents?” The survey showed that 54% of the randomly selected teenagers said they get along “very well.” The reported margin of error was 5%. Is it reasonable to infer that the majority of teenagers get along “very well” with their parents? Provide reasoning.

My Answer:

According to the table in the syllabus, a 5% margin of error would mean 400 teens were surveyed. A 5% margin of error also brings down the sample percent to 49% of teens that say they get along very well with their parents. This means the majority of teens do not say they get along “very well” with their parents. However, if a larger sample was used and the margin of error shrank it is highly probable that the percentage would show that the majority of teens say they do get along very well with their parents.

----

Am I on the right track or over-thinking it again?

[BarTopDancer]

Had to turn it in. No time needed for that one.

Test this week. Scored an 80% on the practice test (it was only 5 questions) so hopefully I do ok on the actual test.

[Kevy Baby]

Test this week. Scored an 80% on the practice test (it was only 5 questions) so hopefully I do ok on the actual test.Hopefully you get more than four correct on the actual test (which presumably is longer than 5 questions)!

[BarTopDancer]

That's the idea.

[BarTopDancer]

Can someone explain p-values to me? I don't get it :(

[Kevy Baby]

Can someone explain p-values to me? I don't get it :(It is the measurement in liquid volume of the amount of urination.

[BarTopDancer]

I knew I shoulda put a disclaimer.

Seriously. Can someone explain them to me? I don't get them and they will be on the test. I don't have to know how to calculate, just interpret them.

[Alex]

I don't know how much detail you want (and to go very far I'll have to look some things up). And it would probably be better to have you try to explain it and then lead you through it but that won't work well over a message board.

Hopefully this helps, but if nothing else I suppose it gives a starting point for specific questions.




Essentially p-value is one part of a tool for determining whether a specific statistical outcome's is so far from what you'd expect to happen by random chance that it should be viewed as statistically significant.

The classic example is in flipping a coin. If it is a perfectly fair coin and you flip it 100 times, statistically you would expect to get 50 heads and 50 tails.

However, when you really perform the experiment, you end up with 57 heads and 43 tails. And statistically every possible combination from 0 heads/100 tails to 100 heads/0 tails is possible. But they are not all equally likely. Calculating the p-value is calculating how likely that result was.

The p-value here would be the specific odds of getting a result deviating 7 values from the expected (so the p-value would be the odds of getting either 57 heads or 57 tails). I'm not going to do the math -- and it can be complex so I am assuming you won't need to either -- but let's say this results in a p-value of .0713 or 7.13%.

This would mean that 7.13% of the time, when flipping a coin 100 times you'd expect to get a result of 57 (in either direction).

Now, p-value is not used by itself. It is just a statement of fact, not of importance. So it is used to help decide if a result is significant. For example, when doing a medical study, if 60% of those taking a drug are cured in 3 days, is this significant when those taking the placebo see a 55% cure rate over 3 days or was it just random deviation?

So, before doing your experiment/measurements you need to decide what your significance level will be. A very common one is 5%. This means that it was decided (and it is an arbitrary decision though different fields have professional standards) that an outcome will be viewed as statistically significant only if the observed outcome had less than a 5% chance of happening purely by chance (the p-value).

In our example, the p-value of 57 heads was 7.13%. So if we were using a 5% standard, getting 57 heads would not be viewed as statistically significant. Therefore it can not be viewed as providing support for a hypothesis that the coin is not fair. If however, you got 61 heads, the p-value for that is (again, just making up a number) 2.67%.

2.67% is less than the significance level of 5% which means it would be viewed as statistically significant. One important thing to keep in mind is that this does not prove anything, it is just evidence in favor of the hypothesis that the coin is not fair.

[Kevy Baby]

I assume you've tried the usual suspects for knowledge?

Wikipedia (http://en.wikipedia.org/wiki/P-value)
One site (http://www.statsoft.com/textbook/esc.html#What%20is%20%22statistical%20significance %22%20(p-level))
Another site (http://www.sportsci.org/resource/stats/pvalues.html)

[BarTopDancer]

Thank you Alex!!!!!!!! That does help. I'll post some more soon. The assignment is done, I missed the questions I thought I nailed. Teacher tried to explain it but it didn't make any more sense then the lesson did. I really appreciate you taking the time to help.

Kevy, no I did not check those sites. I have a book and a ton of online resources and just don't "get it". I've had pretty good luck with Alex, Moonie and GD helpin and I am so grateful that they are able to.


just need a D. just need a D. just need a D.

[Alex]

And just in case you're wondering what the real world value of this stuff is, I just had a meeting today where we discussed p-values and the topic was a in looking at how 3 different advertisements for the same product performed.

[Kevy Baby]

just need a D. just need a D. just need a D.D stands for Done.

I got one D in college - in Geology (for Life Science credit). Didn't really hurt my overall GPA (which at this point in my life is completely irrelevant).

[BarTopDancer]

D stands for Done.

I got one D in college - in Geology (for Life Science credit). Didn't really hurt my overall GPA (which at this point in my life is completely irrelevant).

BINGO

[BarTopDancer]

Halp! I'm pretty sure I did this all wrong. I read everything I have and I can't figure it out. Can someone explain where I went wrong?

3. Suppose studies have shown that approximately 20% (.2) of Americans regularly engage in exercise (at least 3 times per week).

a. If one American is randomly selected, what is the probability that he or she does not regularly engage in exercise? Set up calculation.

1 - .20 = .80

The probability that a randomly selected American does not engage in regular exercise is 80%

b. If three Americans are randomly selected, what is the probability that all three regularly engage in exercise? Set up the calculation.

1 - .20 = .80
.80 x 3 = 2.4

The probability that three randomly selected Americans engage in regular exercise is 2.4

c. What assumption is being made in part b above?

The assumption being made in part B is that the events are mutually excclusive.

d. If two Americans are randomly selected, what is the probability that the first American regularly engages in exercise while the second American does not regularly engage in exercise? Set up the calculation.

1 - .20 = .80
.80 x .20 = .16 because the two events are independent

Make sure that the values are in the correct order.

[Ghoulish Delight]

'a' is correct.

'b' is incorrect, you've made 2 errors. First, look at what it's asking. It wants odds that all three ARE regularly engaged in exercise. Second, you've used the wrong operation. Multiplication is correct, but you are not multiplying the correct thing.

'c', well there are a lot of assumptions in that. But I'm guessing that the assumption being asked for is something along the lines of you're assuming that taking removing one person from the population is not enough to have any significant effect on the percentage of the remaining population that exercises. If, for example, the population was only 5, with the same 20% chance of a person being and exerciser, if you select one person, the odds that any one of the remaining does exercise is either 25% or 0%, depending on who you picked.

d) This is correct (though I think you may have reveresed the numbers in your head). Use this answer as a model to re-evaluate b)

[Alex]

One additional comment on part b. You ran into a very handy way of knowing when you've gone astray with probabilities.

When asked for the probability of something happening the answer can never be less than 0 (or 0%, or never) or greater than 1 (or 100%, or every time). So, since you received an answer of of something having a probably of 2.4 (or 240%) you can know you went astray.

[BarTopDancer]

Thanks guys! I'm still lost on B (I had a feeling it was wrong since it was greater than 0), but I think I fixed C and D.



b. If three Americans are randomly selected, what is the probability that all three regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
.16 x 3 = .48

c. What assumption is being made in part b above?

The assumption being made in part B is that if you remove one person from the equation it will not change the outcome of the probability.

d. If two Americans are randomly selected, what is the probability that the first American regularly engages in exercise while the second American does not regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
1 - .20 = .80

[BarTopDancer]

One more for this assignment.

Refer to the article entitled [name removed] in your Supplemental References booklet. Find the value for the percent of Penn State students that said they consumed more alcohol than usual on their 21st birthday. Convert this percent to a proportion. A random sample of three Penn State students who recently turned 21 was obtained from the Penn State population. Set up the calculation for the probability that first and the third Penn State student consumed more alcohol than usual on their 21st birthday while Penn State Student 2 did not consume more alcohol than usual on their 21st birthday. Remember, the order of the values is important.

Slightly less than half – 47%
Proportion - .47

I know I'm missing figures for the second question. What type of figures do I need? I also think the proportion is incorrect.

[Alex]

For part B.

You were on the right track in your original answer for part b and I'm not sure where you've gone with your second attempt. "What is the probability that three randomly selected Americans do not exercise regularly."

You have three events there. Each of which has a probability of 0.8 of happening. So, the chance of the first one being a non-exerciser: 0.8. The same for the second. The same for the third.

0.8 0.8 0.8

You need to multiply those and I think you knew that. However, 0.8x0.8x0.8 is not the same as 0.8x3, which is what you did.

Consider the difference between 2x3 and 2x2x2.

==

As for your new question, assuming that 0.47 is the correct probability then you essentially have the exact same question as Part B.

What are the odds of the first person being a birthday drinker?
What are the odds of the second person not being a birthday drinker?
What are the odds of the third person being a birthday drinker?

Since you need to know the odds of all three things happening together, you multiply those odds.

[BarTopDancer]

Thanks Alex! So B would be

.80 x .20 = .16
.8 x .8 x .8 = .51

-----

New question:

636 students answered the survey.

0.47 said they consumed more alcohol then usual

636 x .47 = 299 students who consumed more alcohol than usual

337 students did not consume more alcohol than usual. This is 53%.

(0.47 + 0.53 = 1)

So, I think the caculation would be

.47 x .53 x .47 = .12

[Alex]

I'm not sure why you're putting the 0.8 x 0.2 in the Part B answer. All you need is the second part.

But otherwise that's right.

[BarTopDancer]

Thanks Alex!

I'm not sure why I put that either. I'm sure at the time I had a good (incorrect) reason.

[Ghoulish Delight]

-----

New question:

636 students answered the survey.

0.47 said they consumed more alcohol then usual

636 x .47 = 299 students who consumed more alcohol than usual

337 students did not consume more alcohol than usual. This is 53%.

(0.47 + 0.53 = 1)

So, I think the caculation would be

.47 x .53 x .47 = .12Close, but I think this is incorrect. Notice they said that the order matters.

Remember the "what assumptions are being made" part of question b? Don't make those same assumptions for this one.

[BarTopDancer]

Close, but I think this is incorrect. Notice they said that the order matters.

Remember the "what assumptions are being made" part of question b? Don't make those same assumptions for this one.

Confused.

.47 drank more than usual (1st and 3rd students)
.53 did not consume more than usual (2nd)

If you remove either the 1st or the third student the probability doesn't change.

If I do .47 x .53 = .25

I'm so lost. :(

[Alex]

Sorry, BTD, I suspect this is going to confuse you more but now GD and I hash it out and I don't know any other way without getting explicit.


GD: I thought about that, but the resulting difference doesn't make a huge difference until 1/100th of a percent even if pool depletion is taken into account.

299/636 x 337/635 x 298/634 = 0.1172727 = 11.72727%

Not taking pool depletion into account it is:

299/636 x 337/636 x 299/636 = 0.1171119 = 11.71119%

Using BTD two-digit rounding results in:

.47 x .53 x .47 = 0.117077 = 11.7077% (ignoring the issue of significant digits).

However, the question, as set up does not draw the three students from the pool of survey respondents so it isn't actually possible to take pool depletion into account. The survey is a 636 student subset. The three students in question are drawn from the entire pool of Penn State students and the question does not give the information on that pool size.

But that is the hard part of seeing questions out of context of the class. I don't know what level of imprecision is acceptable. So if I were doing this homework I'd probably present the second equation above. But I don't think the first one (which is what I think GD is going for) is justified. Though it would be if the three students were drawn from the population of survey respondents.

[BarTopDancer]

You two can hash it out all you want. I've gotta turn it in now, so hopefully I'll understand the teacher's corrections.

Thanks for your help!

[BarTopDancer]

I got number 3 wrong.

Suppose studies have shown that approximately 20% (.2) of Americans regularly engage in exercise (at least 3 times per week).

a. If one American is randomly selected, what is the probability that he or she does not regularly engage in exercise? Set up calculation.

1 - .20 = .80

The probability that a randomly selected American does not engage in regular exercise is 80%

b. If three Americans are randomly selected, what is the probability that all three regularly engage in exercise? Set up the calculation.

.8 x .8 x .8 = .51

(-1)
(.2) x (.2) x (.2) = .008

c. What assumption is being made in part b above?

The assumption being made in part B is that if you remove one person from the equation it will not change the outcome of the probability.

(-1) sure the probability will change!!!

The answer is that we are assuming independence of the events. Otherwise, we could not just multiply the individual event probabilities together to arrive at the “joint probability” of all 3 events happening

d. If two Americans are randomly selected, what is the probability that the first American regularly engages in exercise while the second American does not regularly engage in exercise? Set up the calculation.

.80 x .20 = .16
1 - .20 = .80

(-.5)
Correct order is important and will be important on the exam. The answer is :
(.20) x (.80) = .16



Make sure that the values are in the correct order.

[Alex]

Crap. I'm partly to blame for b being wrong. GD saw that mistake on your part of using the odds on the wrong side but then I flipped them around again in helping.

Part C is the difficulty of trying to help without the class context. Odds are that was a specific point in the chapter, though I'd argue the answer you gave is a subset of the answer he was looking for. GD's answer is still mostly correct (though adding the word "significantly" to "will not change" might make it better). It just isn't the assumption being sought.

Part D was just a continuation of the same mistake in Part B of having flipped the odds of the two outcomes.

[BarTopDancer]

It's ok Alex. I still ended up with a C on the assignment. I did get the last question correct (the 2nd one I posted about).

I just wanted to understand the discrepancies.

Only 2 more assignments, and 2 tests. If I can hold this C I will be super happy. If I get a D in the class I'll still be super happy, cause I passed!

[BarTopDancer]

I need a total of 365 points to pass. I have 255 points. There are 220 points left in the class. I need to get 110 of them.

2 assignments left, one 50 points, one 70 points and one test worth 100 points.

I'm starting to freak out.

[Kevy Baby]

Statistically, it would appear that you have a good shot.

[Ghoulish Delight]

So all you need on 3 assignments is 50%? Doesn't sound hard.

[Kevy Baby]

So all you need on 3 assignments is 50%? Doesn't sound hard.Or 92% on the two assignments and she can skip the exam!

[BarTopDancer]

No, in theory it doesn't. Still starting to freak out. It is a little hard seeing a D as a passing grade when I pull As and Bs on everything else.

I also emailed the teacher to make sure I can pass even if I tank the tests. Some classes, regardless of their point structure have a "fail the tests fail the class" rule.

The assignment for this week is back to more of a Psych stats, variables, experiments. I should do well on it.

[Kevy Baby]

Well, you didn't say anything about GRADES - you just mentioned PASSING.

Silly!

Besides: D stands for Done!

[BarTopDancer]

The D is passing.

My old roommates gf used to say "C your way to your degree". Then again she was a 4.+ PhD student.

So I think I understand...

Researchers asked a sample of 50 1st grade teachers and a sample of 50 12th grade teachers how much of their own money they spent on school supplies in the previous school year. They wanted to see if teachers at one grade level spend more than teachers at the other grade level.

a. What type of study is found—observational or randomized experiment? Explain. – This study is an observational experiment. The subjects are the teachers, and the amount of personal money they spend on school supplies is being observed.
b. What is the experimental unit? – The experimental unit is the teachers
c. What is the response variable? – The response variable is the money spent
d. What is the explanatory variable? – The explanatory variable is the teachers from different grade levels.
e. Do we have to worry about confounding variables in this instance? Why? If so identify a possible confounding variable? Yes, we have to worry about confounding variables in this example. Confounding variables are found in observational experiments. The confounding variable may be the amount of discretionary money each teacher has to spend.
f. Are either of the terms retrospective study or prospective study relevant? Explain. This study would be a retrospective study – the teachers are asked to recall how much of their own money they spent in the previous school year.

[BarTopDancer]

One more

2. A research team compared two methods of measuring treadwear on tires. Eleven tires were each measured for treadwear by two different methods: one method was based on weight while the other method was based on groove wear. For convenience, each tire was measured first by weight method and then second by the groove wear.

a. The two samples are which of the following: two independent or two dependent (matched pairs)? Explain. The two samples used are independent samples. The tires were measured for weight and then by groove. The weight of the tire may be affected by the amount of groove wear on the tire.
b. What type of study is found—observational or randomized experiment? Explain. This study is a randomized experiment. The measurements are randomly assigned to the tire.
c. What is the experimental unit? The experimental unit is the tire.
d. What is the response variable? The response variable is the treadwear.
e. What is the explanatory variable? The explanatory variable is the measurements (weight or groove wear).

[BarTopDancer]

Teacher says I get the grade that equates to the point I earn. He also said the last test is much easier than the previous test. That gives me hope, and a C *is* obtainable.

[BarTopDancer]

Would anyone [who understands stats] be willing to check my assignment? It's not due until tomorrow night. PM me your email if you are.

Thanks

[Kevy Baby]

Would anyone [who understands stats] be willing to check my ass!

[Tref]

!

?

[BarTopDancer]

Can you guys wait 2 weeks to have fun with this thread? Please?

I'm totally stressing over this class :(

[Tref]

Can you guys wait 2 weeks to have fun with this thread? Please?

I'm totally stressing over this class :(

Umm, no.

[Alex]

I'm hesitant to provide assistance on these since they are essentially vocabulary questions and it has been a long time since I've worried about the precise definitions of the terms being asked about.

That said:

Researchers asked a sample of 50 1st grade teachers and a sample of 50 12th grade teachers how much of their own money they spent on school supplies in the previous school year. They wanted to see if teachers at one grade level spend more than teachers at the other grade level.

a. What type of study is found—observational or randomized experiment? Explain. – This study is an observational experiment. The subjects are the teachers, and the amount of personal money they spend on school supplies is being observed.
b. What is the experimental unit? – The experimental unit is the teachers
c. What is the response variable? – The response variable is the money spent
d. What is the explanatory variable? – The explanatory variable is the teachers from different grade levels.
e. Do we have to worry about confounding variables in this instance? Why? If so identify a possible confounding variable? Yes, we have to worry about confounding variables in this example. Confounding variables are found in observational experiments. The confounding variable may be the amount of discretionary money each teacher has to spend.
f. Are either of the terms retrospective study or prospective study relevant? Explain. This study would be a retrospective study – the teachers are asked to recall how much of their own money they spent in the previous school year.A. I think you're correct that it is an observational study, but the reason isn't so much that you're observing something (you're always observing something) but rather because the experimental units (teachers) are not randomly assigned to the explanatory variable (grade level of teacher).
B. Correct.
C. Correct.
D. Correct.
E. Correct. This type of question screams for an expectation that something will be quoted from the text book, so I wouldn't know if you have. However, I don't think you can state that "confounding variables are found in observational experiments." They are more common in them but they can exist in randomized experiments as well and it is possible for an observational experiment to not have them. But you do always have to try and consider them.
F. Correct.

For your second question I must be missing something because I don't see how enough information is provided to answer most of the questions about it. I don't see where an experiment is being described, just methods for gathering data.

[BarTopDancer]

Thanks Alex. I understand your hesitance too. I think I understand enough to pass this assignment - it is mostly vocabulary and understanding how to interpret results.

I think A should be randomized experiment, not observational study. As I progressed through the assignment I realized I misunderstood the difference. The confounding variables are found more in randomized experiments - and this fits the situation now that I fixed it.

For the 2nd question, that is all that is provided.

[BarTopDancer]

I submitted my last assignment this afternoon. I need 55 points between this assignment and my test to maintain my D is for DONE grade. There is a slim chance I'll get a C.

To those who had feedback and assistance, thank you so much for your help, I really couldn't have done it without it.

[blueerica]

Thanks to all that helped her in completing this course.

I really couldn't stand to hear any more whining and complaining...

And really, no, thank you, and kudos to BTD for finally.. <gasp> finally... making it through Stat. I am happy, and proud of you. And will be ever more proud once December comes. That's it, and you'll finally have the pot of gold you've been seeking!

[BarTopDancer]

I passed!

YAY!!!!!!!!!

Can we burn this thread?

[Moonliner]

Sure.

http://www.riordanlawgroup.com/images/explosion.jpg

[Pirate Bill]

Can we burn this thread?

Would you prefer...

meteors (http://www.netdisaster.com/go.php?mode=meteor&destruction=massive&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
slow burn (http://www.netdisaster.com/go.php?mode=burn&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
dog poop (http://www.netdisaster.com/go.php?mode=dog&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
or
cream pie (http://www.netdisaster.com/go.php?mode=creampie&destruction=massive&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)?

[Kevy Baby]

Would you prefer...

meteors (http://www.netdisaster.com/go.php?mode=meteor&destruction=massive&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
slow burn (http://www.netdisaster.com/go.php?mode=burn&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
dog poop (http://www.netdisaster.com/go.php?mode=dog&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
or
cream pie (http://www.netdisaster.com/go.php?mode=creampie&destruction=massive&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)?That's cool!

[katiesue]

Yay - Congratulations!!

[BarTopDancer]

They're all blocked at work!

[Pirate Bill]

They're all blocked at work!

Time to start looking for a new job.

[Kevy Baby]

They're all blocked at work!Talk to your IT person about that.

Oh, wait...

[BarTopDancer]

Time to start looking for a new job.

Talk to your IT person about that.

Oh, wait...

:evil: :evil: :evil: ;)

[BarTopDancer]

Would you prefer...

meteors (http://www.netdisaster.com/go.php?mode=meteor&destruction=massive&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
slow burn (http://www.netdisaster.com/go.php?mode=burn&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
dog poop (http://www.netdisaster.com/go.php?mode=dog&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)
or
cream pie (http://www.netdisaster.com/go.php?mode=creampie&destruction=massive&url=http%3A//www.loungeoftomorrow.com/LoT/showthread.php%3Ft%3D8002)?

Those are awesome, though now I want a cigarette.

[Pirate Bill]

Maybe your prof should use The Manga Guide to Statistics (http://nostarch.com/mg_statistics.htm) as the text next year. It's sure to be a page turner (if you're into manga...or a teen boy...or a dirty old man).

[Kevy Baby]

(if you're into manga...or a teen boy...or a dirty old man).What if the dirty old man is in the teen boy (and by teen boy I mean 18 or 19, so as to keep the perversion legal).