Stat 100

[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]

Originally Posted by BarTopDancer

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]

Originally Posted by Kevy Baby

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