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Mathematics 19 Online
OpenStudy (andijo76):

statistics question can someone check to see if this is correct i will post the question

OpenStudy (andijo76):

Express the confidence interval (0.042,0.134) in the form of p^-E<p^+E phat= (0.042+0.134)/2 phat=(0.176)2 phat=0.088 E=0.134-0,088 E=0.046 0.088<p<0.046

OpenStudy (andijo76):

is this right i am not sure

OpenStudy (marissalovescats):

@jim_thompson5910 is pretty good at this stuff I believe

OpenStudy (kirbykirby):

Is it supposed to be something of the form \( xy \pm z\) because \(0.088 \pm 0.046\) would indeed give that confidence interval (0.042, 0.134)

OpenStudy (kirbykirby):

(I'm just not sure I understand the notation given )

jimthompson5910 (jim_thompson5910):

The steps phat= (0.042+0.134)/2 phat=(0.176)/2 phat=0.088 are correct

jimthompson5910 (jim_thompson5910):

That's the point estimate

jimthompson5910 (jim_thompson5910):

Now onto the margin of error E=(0.134-0.042)/2 E=0.092/2 E=0.046

jimthompson5910 (jim_thompson5910):

So \[\Large \hat{p} - E < p < \hat{p} + E\] turns into \[\Large 0.088 - 0.046 < p < 0.088 + 0.046\]

OpenStudy (andijo76):

so that is the answer i just needed to combine the numbers together on each side the way you did ?

jimthompson5910 (jim_thompson5910):

well they want it in the form \[\Large \hat{p} - E < p < \hat{p} + E\] so I think you just start with \[\Large \hat{p} - E < p < \hat{p} + E\] and plug in the values for p-hat and E

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