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Mathematics 17 Online
OpenStudy (howard-wolowitz):

5. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. 0.438 < p < 0.505 0.444 < p < 0.500 0.471 < p < 0.472 0.435 < p < 0.508

OpenStudy (michele_laino):

here a 95% confidence level corresponds to a 1.96*sigma, where sigma is the standard deviation

OpenStudy (kirbykirby):

\[ C.I=p\pm z\cdot \sqrt{\frac{p(1-p)}{n}}\]

OpenStudy (howard-wolowitz):

Ok, guys I"m not sure how to use that to get the answer?

OpenStudy (kirbykirby):

The sample proportion \(p\) is the proportion of voters in favour. Do you think you can find this?

OpenStudy (howard-wolowitz):

honestly no.. but I can try.. ok whats C

OpenStudy (kirbykirby):

oh C.I. i just wrote to mean "confidence interval", which is the formula for the confidence interval for a proportion

OpenStudy (kirbykirby):

(Hmm maybe you saw the \(p\) written as \(\hat{p}\)). The formula still works the same, although technically \(\hat{p}\) means the calculated proportion (found in your data), and \(p\) is the true population proportion

OpenStudy (michele_laino):

I got sigma = 0.5 @kirbykirby

OpenStudy (kirbykirby):

Almost. I see where you get that number, but don't forget the divided by "n" in the formula, where \(n\) is the sample size (how many voters there are in total).

OpenStudy (michele_laino):

ok!

OpenStudy (howard-wolowitz):

this stuff is hard for me

OpenStudy (michele_laino):

the mean value is: 408/865 = 0.472

OpenStudy (kirbykirby):

did you manage to find your proportion \(p\)? Basically it is: \[ \frac{number ~of~people~in~favour}{total~number~of~people}\]

OpenStudy (howard-wolowitz):

im working on it

OpenStudy (howard-wolowitz):

well hes saying he got 0.471 < 0 < 0.472

OpenStudy (howard-wolowitz):

or would it be 0.43

OpenStudy (howard-wolowitz):

because I got n divided to get 0.471 ... is that right

OpenStudy (kirbykirby):

I'm not sure I understand what you mean :S But \(p=\dfrac{408}{865}=0.472\) as @Michele_Laino mentioned . That it just the value of p. Then, z=1.96 (the z-value for a 95% confidence interval when using the normal distribution)

OpenStudy (howard-wolowitz):

oh see I didnt realize that srry

OpenStudy (howard-wolowitz):

so tell me what I need to be trying to figure out?

OpenStudy (kirbykirby):

\[C.I=p\pm z\cdot \sqrt{\frac{p(1-p)}{n}}\] You have p, you have z, and \(n=865\). You have all the ingredients to use the formula

OpenStudy (howard-wolowitz):

ok give me one sec dont leave.. so u can tel me if im right ok

OpenStudy (howard-wolowitz):

p=? z=?

OpenStudy (howard-wolowitz):

what is p ans z

OpenStudy (howard-wolowitz):

ok so is it C or not?

OpenStudy (howard-wolowitz):

I got C

OpenStudy (kirbykirby):

\[C.I=\frac{408}{865}\pm 1.96\cdot \sqrt{\frac{\frac{408}{865}(1-\frac{408}{865})}{865}}\]

OpenStudy (howard-wolowitz):

ok give me a sec to solve that

OpenStudy (kirbykirby):

ok

OpenStudy (howard-wolowitz):

how do i type that =- sign

OpenStudy (howard-wolowitz):

+-

OpenStudy (kirbykirby):

oh that means you need to do 2 operations. First you do + then do -. But it is really saying you have 2 equations to type

OpenStudy (howard-wolowitz):

first one i got 0.472

OpenStudy (kirbykirby):

\[\frac{408}{865}+ 1.96\cdot \sqrt{\frac{\frac{408}{865}(1-\frac{408}{865})}{865}}\] \[\frac{408}{865}- 1.96\cdot \sqrt{\frac{\frac{408}{865}(1-\frac{408}{865})}{865}}\]

OpenStudy (howard-wolowitz):

second one i got 0.47054517

OpenStudy (howard-wolowitz):

so the answer is C

OpenStudy (kirbykirby):

hm I don't get that :s

OpenStudy (howard-wolowitz):

how dont u??????????????

OpenStudy (howard-wolowitz):

round it to 0.471

OpenStudy (kirbykirby):

oh I think I know why you get that. The 865 has to be in the square root sign

OpenStudy (kirbykirby):

If you have trouble typing it on the calculator you could try doing this: (which is the same)

OpenStudy (howard-wolowitz):

i got 0.438 and i dont use a cal

OpenStudy (kirbykirby):

yeah I get 0.438 too

OpenStudy (howard-wolowitz):

then i got 0.505

OpenStudy (kirbykirby):

yeah :)

OpenStudy (howard-wolowitz):

awesome

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