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Mathematics 19 Online
OpenStudy (anonymous):

A random sample of 140 students is chosen from a population of 3,500 students. If the mean IQ in the sample is 105 with a standard deviation of 9, what is the 95% confidence interval for the students' mean IQ score? a 100−120 b107.9−112.1 c 103.5−106.5 d108.66−111.34

OpenStudy (anonymous):

you want a 95% confidence interval, right?

OpenStudy (anonymous):

yes @abrahamglasser

OpenStudy (anonymous):

so the right side will be at .975 and the left side will be at .025 1=.95 = .05 .05 / 2 = .025 (you are taking off .025 off each side to get the interval) .95+.975 0+ .025 = .025 so the z scores are .025 and .975. do you know how to do the rest of it?

OpenStudy (anonymous):

I have no idea how to do the rest of it :/

OpenStudy (anonymous):

\[z = \frac{ x - \mu }{ \sigma }\] mu is the mean and sigma is the standard deviation now to find the interval \[.025 = \frac{ x - 105 }{ 9 }\] solve for x and that gives you the left side of the interval. now the right side \[.975 = \frac{ x - 105 }{ 9 }\] solve for x and that gives you the right side of the interval. these x's give you the answer

OpenStudy (anonymous):

tell me if you dont understand something or if you dont know where i got these values, etc

OpenStudy (anonymous):

wait, i think i made a mistake somewhere...

OpenStudy (anonymous):

yea the answers weren't coming out right for the right one I got 15.225

OpenStudy (anonymous):

105.225*

OpenStudy (anonymous):

i have to think about it...

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

ok got it

OpenStudy (anonymous):

i was thinking of a different confidence interval :( sorry about that

OpenStudy (anonymous):

ok, so the formula is..\[CI = x \pm Z _{\alpha/2}\times (\sigma/\sqrt n)\] x = Mean σ = Standard Deviation α = 1 - (Confidence Level/100) Zα/2 = Z-table value CI = Confidence Interval

OpenStudy (anonymous):

so it is \[105 \pm Z_{.025} * (9/ \sqrt 140)\]

OpenStudy (anonymous):

and the z score of .025 is 1.96

OpenStudy (anonymous):

ok so what do I do with the z score?

OpenStudy (anonymous):

did you see the formula above it?

OpenStudy (anonymous):

so it is 105±1.96∗(9/1√40)

OpenStudy (anonymous):

i mean 105±1.96∗(9/√40)

OpenStudy (anonymous):

that doesn't get me my answer what else do I have to do with that?

OpenStudy (anonymous):

it doesn't? hmm.

OpenStudy (anonymous):

none of my answer choices.

OpenStudy (anonymous):

im pretty sure the actual answer is C) 103.5 - 106.5 i dont know why i keep getting different answers...

OpenStudy (anonymous):

That's what I put down to begin with. This whole Exam is so weird. I don't understand it.

OpenStudy (anonymous):

lol neither do i :P

OpenStudy (anonymous):

although im pretty sure it's the right formula..

OpenStudy (anonymous):

whatever. statistics is weird

OpenStudy (anonymous):

Thank you so much for helping me. And I think it is. It looks really familiar.

OpenStudy (anonymous):

no problem

OpenStudy (anonymous):

Thankyou sooo muchh- this was correct!

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