# Thread: Help...Quebec Separation? Oui ou Non

1. because out of all the golfers (2) basically all the people who voted 2, there were 68 6's. meaning that there were 68 people out of the entire vote who where golfers and who earned at least 75 000.
Does that help?

2. Originally Posted by SAG_girl
So then wat you are saying is the the standard deviation equals
N ^ 2 ( p (p-1) /n)
where ...
n = pop proportion = 0.462
N= total population = 6 500 000

you use :
T +/- 1.96*sqrt{N^2 * [p * (1-p)]/n}
to determine the intervals???
i tried that and got roughly the same answer using the p-hat formula. (n is 641, not 0.462 ).

i had a question about that too though. terzi, would that be applicable to case 1 question 1? with N-207,700,000 (total population of statscan)
also, for case 1 question2. is N limited to people that make 75+k (when using p=68/177)

3. does anyone kno if we have to draw the graphs at the end??

4. Originally Posted by terzi
Any confidence interval for an estimator E is calculated this way (assuming 95% confidence):

E +/- 1.96*sqrt[Var(E)]

If the estimator is a proportion then one should know that:

Var(p)=[p * (1-p)]/n

So, a 95% confidence interval for a proportion is:

p +/- 1.96* sqrt{[p * (1-p)]/n}

Now, when speaking of totals, it is slightly different, since the variance calculation changes:

T=N*p
Var(T)= Var(N*p) = N^2 * Var(p)
Then,
Var(T)= N^2 * [p * (1-p)]/n

So now calculate

T +/- 1.96*sqrt{N^2 * [p * (1-p)]/n}
The above formulas are useful for all the problems that involve estimating a total in a SRS. Any other method that you have posted may be incorrect, at least not as accurate. Just to be clear:

T=N*p
T +/- 1.96*sqrt{N^2 * [p * (1-p)]/n

Where:

T=Estimated total
p=Proportion of occurrence of the interest event.
n=Sample size
N=Population size

When you are only analyzing a part of the population (such as very rich people) your population of interest changes, and so your N. Anyway, be careful since sample size will also be different.

5. Originally Posted by terzi
The above formulas are useful for all the problems that involve estimating a total in a SRS. Any other method that you have posted may be incorrect, at least not as accurate. Just to be clear:

T=N*p
T +/- 1.96*sqrt{N^2 * [p * (1-p)]/n

Where:

T=Estimated total
p=Proportion of occurrence of the interest event.
n=Sample size
N=Population size

When you are only analyzing a part of the population (such as very rich people) your population of interest changes, and so your N. Anyway, be careful since sample size will also be different.
Thank you very much for explaining. But then how do we calculate the "p" is it simply #successes/sample population? Kindly elaborate please.

Thanks again for all your help.

6. Originally Posted by YorkDude
How do we calculate the "p" is it simply #successes/sample population?
Exactly, it is a simple proportion. Let's say you have the following dataset:

PERSON / SMOKING
1 / 1
2 / 1
3 / 0
4 / 0
5 / 1

The code for smoking is 1: Person does smoke and 0: Person does not smoke.

p=Proportion of smokers in sample is just:

p=3/5=0.6

Now, let's get a little more technical. The reason why binary data is usually coded as 0,1 is because that way, we can obtain the proportion with the following equation:

p=(Sum. xi)/n

As you can see, it will lead to the same result. And as you may have noticed, the above expression is the formula used for calculating the sample mean. That's the reason why the standard error of the mean and the standard error of the proportion are so alike.

7. I am a bit confused. Is income relevant to the first case 1 question 1? Do we have to divide our answers by income segments? And for case 1 question 2, what's would the changed population be?

8. Originally Posted by YorkMan
I am a bit confused. Is income relevant to the first case 1 question 1? Do we have to divide our answers by income segments? And for case 1 question 2, what's would the changed population be?
Income is irrelevant for Case 1 question 1. But it is necessary for Case 1 question 2.

In short in case 1 question 1 you will use the total population of 207.7 million. However, in Case 1 question 2 you would use 17.1 million only, as it is asking for people with income of 75K+.

I hope this helps.

Take care.

9. ## case 2 Question 2

Did everyone answer question 2 of case 2 the same way as the first case? I'm a little unsure. Please clarify.

10. Well I did answer it the same way as the Case 1 questions. It is worded similarly and it requires us to find the range in the same fashion. Also something to note here, the data is also somewhat similar.

Hope this helps.

11. Wait, but for case 1 question 2, isn't it asking for the number of golfers who earn more than 75k? Wouldn't you divide by the number of golfers then? Since you get the number of golfers from question 1, I'm not sure how you would that interval in question 2.

12. Originally Posted by YorkMan
Wait, but for case 1 question 2, isn't it asking for the number of golfers who earn more than 75k? Wouldn't you divide by the number of golfers then? Since you get the number of golfers from question 1, I'm not sure how you would that interval in question 2.
They way I did it was "Estimate with 95% confidence the number of golfers who earn over \$75,000."

1) Calculate how many people in the sample are golfers AND earn more than \$75K. = 68 people

2) calculate P-hat = 68/1116

3) Then calculate the sqrt (p-hat(1-p-hat)/n)

4) N = 17.1 million [People who earn more than 75k ]

5) now just plug in the number in the formula p-hat +/- 1.96√[p*(1-p)/n].

6) Now multiply the number you get from #5 by 17.1 million. And now you have range of golfers more than 75K.

NOTE: I may be wrong. But that is how I did it.

13. ## Graphs

Did anyone use graphs for the two Cases more specifically Case 2?

14. Originally Posted by Yorkldie
Did anyone use graphs for the two Cases more specifically Case 2?
no, i didnt.

ill see you guys again if i have any questions on the next assignment.
thanks and good luck on the quiz.

15. Wait, what graphs? There were graphs involved with this assignment? Crap...oh well... And it looks like it's just two weeks away from assignment #3. And like the "girl" above me stated, good luck and godspeed on the quiz. lol