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Assume that out of 100 people chosen randomly in a survey in Portland 30% indicate that they vote democrat.

a) Estimate the proportion of people in Portland which vote Democrat. Give a 95% confidence interval for that proportion.

b.) Test on the 95%-level the hypothesis that there are more than 50% which vote democrat. Determine also the p-value.

Since I don't have the expectaction, variance or the standard deviation I am having trouble with parts a and b.

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    You need to estimate the proportion of people. You are given a sample proportion $\hat p$ so use that.2012-10-23
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    Hint: There is a formula with a square root of something, which comes from the central limit theorem and your $\alpha = 0.05$, so $1 - \alpha /2 = 0.975$ and you can find ${z_{0.975}}$ to compute ...2012-10-23
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    @glebovg Hm you mean $\sqrt {y_1+y_2+ ... +y_k/(k-1)}$ is it this formula?2012-10-23
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    No. How do you estimate the proportion $p$ and what is the formula for a confidence interval for a proportion?2012-10-23
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    What is 30% of 100? This gives you an estimate $\hat p$ for the proportion $p$.2012-10-23
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    @glebovg the confidence interval will be $P(−b \le N(0, 1) \le b)$ = 95% , but I am not sure what you mean by proportion p?2012-10-23
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    Your question is about proportions, correct?2012-10-23
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    @glebovg I never heard about proportions in my class. He havent taught us that yet but for your previous question it is 30 people who are democrat.2012-10-23
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    Yes. That is the estimate because we do not know the actual proportion. This looks like a typical proportions question.2012-10-23
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    So, 30 is what I will use for my expectation which will provide me the variance and standard deviation needed to solve the confidence level?2012-10-23

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Let me explain. The best estimate for the proportion $p$ is $\hat p = x/N$. Now we need to somehow tell how close this is to the actual population proportion $p$. We use confidence intervals to estimate the population proportion, for example a 95% confidence interval. To calculate the confidence interval we use $$\hat p \pm {z_{\alpha /2}}\sqrt {\frac{{\hat p\hat q}}{N}}$$ where $\hat q$ is the usual $1 - \hat p$. In your question $\alpha = 0.05$ which is the probability of type I error. So your $z$-score should be about 1.96. Now you can find the lower and the upper bound. Then we usually finish with a conclusion: We are 95% that the proportion ... is between ...

Hope this help. If you have not seen this you need to speak with your professor or look through your notes.

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    Yes it does, thank you a lot Glebovg!2012-10-24
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    Do you know how to test the hypothesis and find the p-value?2012-10-24
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    What I got was for ($1 - \hat p$) = .7 ,so $.3 \pm z_{.025}\sqrt {\frac{{.3.7}}{N}}$ is this correct? And I do not know what N is in this case?2012-10-24
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    Of course you do, what does $N$ or $n$ usually stands for in statistics? Sample size, right?2012-10-24
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    I can only help you understand the concept, I cannot do your homework for you.2012-10-24
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    That wasn't my attention, sorry if I made it look that way. But for the answer to your question is of course sample size, wow, I feel dumb. I usually see it with a lower case n. But so the answer is [(.3 - .08982), (.3 + .08982)] ?2012-10-24
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    It depends on your calculator and the $z$-table. Double check your calculations instead of asking me -- that way you will learn something.2012-10-24