Interpretation of 99% confidence interval

#1
Hello guys,

this is my first post here in the forum and if I missed anything, then pls correct me.

My first statistic class starts this week but I have already to do some homework before the first class.

One task is the calculation and interpretation of a 99% confidence intervall.

The task was the fallowing: 79% of the Democrats are in favor of Hillary Clition for presendency. Total people asked: 649. Construct a 99% CI interval for the proportion Democrats in favor of Hillary Clinton.

My calculation so far:

n = 649

p̂ = 79:649 = 0.122

p̂ ± Z √ p̂ (1- p̂): n

0.122 ± (2.567) √ 0.122 (1-0.122): 649

0.122 + (2.567) √ 0.122 (1-0.122): 649 = 0.155

0.122 - (2.567) √ 0.122 (1-0.122): 649 = 0.89

= (0.155, 0.89)

But now I am stucked with the interpretation of the CI, even I read all chapters of my book it isn't clear to me.

It would be nice if someone could helping me with the interpretation.


Greetings
7thSlap
 

Dason

Ambassador to the humans
#2
You have quite a few issues here. Although my first question is who is assigning you to create and interpret confidence intervals BEFORE you have your first stats class? That's crazy talk.

The second is that you're misunderstanding how to get \(\hat{p}\). It's number of successes over the total sample size. In this case you don't have the number of successes - you were given \(\hat{p} = .79\) directly.

The third issue is that when it came time to calculate the end points you didn't do the computation correctly. I'm not going to dig through and figure out exactly what error your made since the p-hat wasn't correct anyways.
 
#3
Hi, thanks for answering.

My problem is that I missed my statistic course of the last year. I assumed I could repeat this course, but now the course isn't offered anymore. My professor enrolled me instead for the "statistics II" course of the fallowing year. So the professor assume that the students know the content of the first year, but I don't. I just got the homework a few days ago and the first lecture is this week...

Could you tell me how to calculate p-hat correctly?
 
#5
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Oh sorry, I am blind. I did the calculation again, is it right now?

p̂ = 0.79
p̂ ± Z √ p̂ (1- p̂): n

0.79 ± (2.567) √ 0.79 (1-0.79): 649

0. 0.79 + (2.567) √ 0. 0.79 (1-0.79): 649 = 0.83

0.79 - (2.567) √ 0. 0.79 (1-0.79): 649 = 0 .75

= (0.83, 0.75)
 

Dason

Ambassador to the humans
#6
I would flip the limits around since usually we do (lower_boundary, upper_boundary) so use (.75, .83) instead. But that looks good.
 
#8
If you repeat the experiment a number of times, 99% of the intervals will contain the proportion. Are you sure you don't want to take Stat 1? Do you know the material from Stat 1?
 
#9
Thank you. I would like to take Stat 1, but then I have to wait more than 6 months before the course is offered again. I will read the chapters of Stat 1 and maybe get some private stat lesseons.