Help...Quebec Separation? Oui ou Non

daniel7lu

New Member
Since the 1960s, there has been an ongoing campaign among Quebecers to separate from Canada and form an independent nation. Should Quebec separate, the ramifications for the rest of Canada, American states that border Quebec, the North American Free Trade Agreement, and numerous multinational corporations would be enormous. In the 1993 federal election, the pro-sovereigntist Bloc Quebecois won 54 of Quebec’s 75 seats in the House of Commons. In 1994, the separatist Parti Quebecois formed the provincial government in Quebec and promised to hold a referendum on separation. As with most political issues, polling plays an important role in trying to influence voters and to predict the outcome of the referendum vote. Shortly after the 1993 federal election, The Financial Post Magazine, in cooperation with several polling companies, conducted a survey of Quebecers.
A total of 641 adult Quebecers were interviewed. They were asked the following question. (Francophones were asked the questions in French.) The pollsters also recorded the language (English or French) in which the respondent answered.
If a referendum were held today on Quebec’s sovereignty with the following question, “Do you want Quebec to separate from Canada and become an independent country?” would you vote yes or no?
1. No
2. Yes
The responses were recorded and stored in columns 1 (planned referendum vote for Francophones) and 2 (planned referendum vote for Anglophones).
Infer from the data:
1. If the referendum were held on the day of the survey, would Quebec vote to remain in Canada? (α=5%)
2. Assuming the total population of adult Quebecers is 6,500,000 on the day of survey, estimate with 95% confidence the total number of adult Quebecers who would vote for separating from Canada.

Thanks a lot...

AtlasFrysmith

New Member
Posting a homework question and the data set without even so much as an explanation of what you've tried and what difficulties you're having is not going to get you help. Please read the sticky thread about homework help.

I don't mean to be rude about this and I will happily help you understand whatever difficulties you're having, but you need to show some effort first--it will help us explain things to you if we know what you do and do not understand.

daniel7lu

New Member
Posting a homework question and the data set without even so much as an explanation of what you've tried and what difficulties you're having is not going to get you help. Please read the sticky thread about homework help.

I don't mean to be rude about this and I will happily help you understand whatever difficulties you're having, but you need to show some effort first--it will help us explain things to you if we know what you do and do not understand.

I am sorry ,i already did 3 of 4 questions ,but i dont know whether if my job is good or not..so i didn't post..

daniel7lu

New Member
Posting a homework question and the data set without even so much as an explanation of what you've tried and what difficulties you're having is not going to get you help. Please read the sticky thread about homework help.

I don't mean to be rude about this and I will happily help you understand whatever difficulties you're having, but you need to show some effort first--it will help us explain things to you if we know what you do and do not understand.
Case #1 An Analysis of Golfers in the United States (25 marks)

How many golfers are there in the United States? A survey of American adults (age 18 and above) asked each whether they had played golf at least once a month during the summer. The responses were 2 = yes and 1 = no. The survey also asked respondents to indicate which of the following household income category they fell into:
1. Under $15,000 2.$15,000 to $24,999 3.$25,000 to $34,999 4.$35,000 to $49,999 5.$50,000 to $75,000 6. Over$75,000
The latest census reveals that the number of American households in each of the income categories is as follows:
1. 75.7 million
2. 36.9 million
3. 28.3 million
4. 27.8 million
5. 21.9 million
6. 17.1 million
(Source: Statistical Abstract of the United States, 2006, Tables 685 and 1238)
*The data file is posted on the website ‘Assignment 2 Data file.xls’.
Required: An analysis of the data
1. Estimate with 95% confidence the total number of golfers.
2. Estimate with 95% confidence the number of golfers who earn at least $75,000. these are two questions i need to do .. and what i have done are as follow: ase #1 An Analysis of Golfers in the United States 1. Estimate with 95% confidence the total number of golfers. Total Sample n=1116 Golfers from sample 177 16%(0.16) = Golfers from sample/Total Sample=177/1116=16% Standard deviation= = =0.01097 =0.16 0.01097*1.96=0.16 0.0215 the 95% confidence the total number of golfers is 179+ or -24 2. Estimate with 95% confidence the number of golfers who earn at least$75,000.
Total Sample n=1116
the number of golfers who earn at least $75,000 is 68 6.1%(0.061) = Golfers from sample/Total Sample=68/1116=0.061 Standard deviation= =0.00716 =0.061 0.00716*1.96=0.061 0.014 the 95% confidence the total number of golfers who earn at least$75,000 is
69+ or -16

Case 2 Quebec Separation? Oui ou Non
If the referendum were held on the day of the survey, would Quebec vote to remain in Canada? (α=5%)
The total number of Quebecers is 641
The number of Quebecers who would vote Quebec to separate from Canada and become an independent country is 296
The number of Quebecers who would not vote Quebec to separate from Canada and become an independent country is 345
=number of Quebecers who would not vote Quebec to separate from Canada and become an independent country/Total number of Quebecers=345/641=0.538
Standard deviation= =0.01969
=0.538 1.96*0.01969=0.538 0.0386
the 95% confidence Quebecers vote Quebec to remain in Canada is
345+ or -25

daniel7lu

New Member
Posting a homework question and the data set without even so much as an explanation of what you've tried and what difficulties you're having is not going to get you help. Please read the sticky thread about homework help.

I don't mean to be rude about this and I will happily help you understand whatever difficulties you're having, but you need to show some effort first--it will help us explain things to you if we know what you do and do not understand.
Could u help me to check it..

and i don't know how to do the 2rd question of case2...

Thanks ..

terzi

TS Contributor
What you have done seems correct for a Simple Random Sampling. Since you don't seem to have studied weights or anything like it, we will keep it that way. Still you are only estimating totals for your sample.

Try estimating the totals with

T=N*p

Where p is the proportion you obtained and N is the size of the population.

To obtain the confidence interval, consider that

Var(T)= Var(N*p)= (N^2)*V(p)

Hope that helps

daniel7lu

New Member
What you have done seems correct for a Simple Random Sampling. Since you don't seem to have studied weights or anything like it, we will keep it that way. Still you are only estimating totals for your sample.

Try estimating the totals with

T=N*p

Where p is the proportion you obtained and N is the size of the population.

To obtain the confidence interval, consider that

Var(T)= Var(N*p)= (N^2)*V(p)

Hope that helps
.

Thanks a lot for helping...

But I am still confused about
Var(T)= Var(N*p)= (N^2)*V(p)

I never saw it before..

I don't know how to use it to do with last question in detail...

terzi

TS Contributor
Well, a confidence interval for the total is defined as (with a 95% confidence)

T + / - 1.96*sqrt(Var(T))

So you need to calculate Var(T).

It is also equivalent to use the following:

Var(T)=N^2 * Var(x bar)

Try reviewing how you did calculate your confidence intervals

daniel7lu

New Member
Well, a confidence interval for the total is defined as (with a 95% confidence)

T + / - 1.96*sqrt(Var(T))

So you need to calculate Var(T).

It is also equivalent to use the following:

Var(T)=N^2 * Var(x bar)

Try reviewing how you did calculate your confidence intervals

Hi~~Could I do it in this way?
2.Assuming the total population of adult Quebecers is 6,500,000 on the day of survey, estimate with 95% confidence the total number of adult Quebecers who would vote for separating from Canada.

Answer: =number of Quebecers who would vote Quebec to separate from Canada and become an independent country/Total number of Quebecers=296/641=0.462
Standard deviation= =0.0196855{sqrt[(1-0.462)*0.462)/n(641)]}
confidenceinterval=0.462+ or -0.0386
When the total population is 6，500，000， then
=6，500，000*（0.462+ or -0.0386）=3003000 + or -250900

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yorkguy

Guest
hi there...
i am really stuck on this assignment...can you show me step by step how you came up with the answer for case 1? i really appreciate it if you can help me out. Thanks

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lynxter

New Member
I understand what Daniel is doing in part one...he is using the following formula to answer the total number of golfers :
xbar (+and-) z a/2 (standard deviation/ square root of n).

z a/2 = z.025=1.96 (95% confidence level)
and xbar = 0.16
n= 1116
so square root of n= 33.40658618

The only thing I do not understand is how Daniel got the standard deviation?

daniel7lu

New Member
I understand what Daniel is doing in part one...he is using the following formula to answer the total number of golfers :
xbar (+and-) z a/2 (standard deviation/ square root of n).

z a/2 = z.025=1.96 (95% confidence level)
and xbar = 0.16
n= 1116
so square root of n= 33.40658618

The only thing I do not understand is how Daniel got the standard deviation?
standard deviation is equal to
square root of p(1-p)/n=square root of (1-0.16)*0.16=square root of 0.1344/1116=square root of 0.00012043=0.01097407

daniel7lu

New Member
hi there...
i am really stuck on this assignment...can you show me step by step how you came up with the answer for case 1? i really appreciate it if you can help me out. Thanks
there are many formulas it could not showing here becaure they are icons...
i am also looking for helps here, but no one tell me what i done is exactly correct or not...

if you still could not understand you can send me a email.
but i am really not sure whether if my job is right or not.

Y

yorkguy

Guest
i am still confused about how you got 179+/- 24 for final answer.

are you supposed to multiply 0.16 by the total sample population: n or Total Population: N?

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lynxter

New Member
Daniel thank you for helping me understand how you got standard deviation.

I too have to ask how you end up with the total number of golfer between:
179+ or -24....the reason I ask is, are you allowed to have a negative number of golfer when we already know the total is 177 who answered they do golf.

I am not totally sure but, would it work to use the confidence interval estimator of p, since the case is dealing with nominal data?

the formula is: p hat (+and-) z a/2 square root of [p-hat(1-p-hat)/n]

Like I said I do not know for sure because when did the formula I got:
LCL: 0.1372, UCL: 0.1800
So, now I am not sure if you add 14 and 18 to 177 or 0.16?

terzi

TS Contributor
Wow, this post has become interesting

When the total population is 6,500,000, then
=6,500,000*(0.462+ or -0.0386)=3003000 + or -250900
This is incorrect, because you are multiplying the confidence interval, not creating a new one.

Let me try to explain the correct solution.

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}

Voila

NOTE: THE ABOVE IS ONLY POSSIBLE FOR SIMPLE RANDOM SAMPLING, SINCE FOR OTHER SAMPLING PROCEDURES THE VARIANCE ESTIMATORS MAY CHANGE. SINCE THIS IS A CLASS PROJECT, I THINK YOU SHOULD ASSUME S.R.S.

Y

yorkguy

Guest
hi lynxter...
Daniel got the 179+/- 24 by...(0.16x1116)=179 , (0.0215x1116)=24
i think daniel is trying to say is the total number of golfer is between 155 and 203.

i am not sure if you multiply 0.16 by 1116(sample) or by (207.7 million) total population to get the intervals...

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daniel7lu

New Member
i am still confused about how you got 179+/- 24 for final answer.

are you supposed to multiply 0.16 by the total sample population: n or Total Population: N?
JUST TIMES THE TOTAL SAMPLE WHICH IS EQUALED 1116(BUT NOW I FOUND IT SHOUD BE TIMES THE TOTAL population )

daniel7lu

New Member
hi lynxter...
Daniel got the 179+/- 24 by...(0.16x1116)=179 , (0.0215x1116)=24
i think daniel is trying to say is the total number of golfer is between 155 and 203.

i am not sure if you multiply 0.16 by 1116(sample) or by (207.7 million) total population to get the intervals...
You are right. later yeterday i found it should be times by total population(207.7 million)

SAG_girl

New Member
so then the total number of golfers would equal
42838265.15 + - 4438774.3
lcl 38399490.7 golfers
ucl 47277039.3 golfers
????