questions for any statistics people

[bigworm82]

I have a question on a worksheet i just can't wrap my head around. We've moved on past confidence intervals and what not so now i can't remember how to do it .

Can anybody look at this question and help me figure it out. Can give me the answer if you want but i'd prefer a description of how you got to said answer .

I'm probably over analyzing it

Thanks guys.

http://i256.photobucket.com/albums/hh182/bigworm82/2011-04-13161226.jpg

[IPD]

damn, it's been like...13 years since i took statistics. i got lost somewhere around chi square. :( i usedtacould be able to do that problem. :mad:

p.s.

am i correct in thinking that the standard deviation is .1 micrometers?

[IPD]

ok wait, i'm not familiar with my greek symbology from statistics. .5 would be standard deviation. so the question is, in a sample size of 250...how many would fall outside of that .5 milimeters (which would be 10x the acceptable deviation--per manufacturer direction).

is that correct?

[onebadmollafolla]

Shit. Three years ago I could have solved this more than likely. I was quite good at managerial statistics when I took it. Don't recall much of it anymore.

[bigworm82]

sorry i was at a work meeting.

The standard deviation is .5mm

The question wants to know what the max confidence interval is as far as i read it.


The thing that i'm confused on it says not to exceed 250 for sample size but does that mean can use less than 250.

[19Eclipse90]

I'm like the rest of everyone here, it's been a while since I've touched this stuff and I don't reckon I was terrific with it to begin, but here's my go at it. Don't take it for truth because I could very well be wrong; it makes sense to me. :p

The mean isn't 0.05mm. The mean is unknown and, as far as I can tell, not needed. The +/- 0.05mm is equal to +/- [Z_alpha/2*(sigma/sqrt(n))]. Sigma is given as 0.5, and you know the sample size, n, as 250. Thus you can solve for Z_alpha/2 -- I calculate ~1.58. From the z-table, you find 0.9429 -- 94.29% as your level of confidence. This also means that alpha/2 is 0.571 (from 1-.9429) and it follows that alpha is 0.1142. Calculate your confidence interval 100(1-0.1142)% and you obtain ~88.58%. This checks out because I obtained a confidence interval of ~88.25% initially using linear interpolation. For reference as to what I interpolated over (this stuff seems pretty standard), a confidence interval of 80% has a Z_alpha/2 value of 1.2816; 90%, 1.6449; 95%, 1.9600; 96%, 2.0537; 98%, 2.3263; 99%, 2.5758.

What it means, as far as I understand it, is that you are 94.29% sure (statistically speaking) that 88.25% of the sample measurements will fall under management's specifications.

Good luck, and do let me know if I'm wrong! :bigthumb:

[bigworm82]

well i turn it in tmrw morning and am supposed to have it back when i leave school at 5pm so will post up the answer.. It looks right after reading through that though.... i've just been focusing on Ho and ha

[bigworm82]

well it's official i don't like my stats teacher.. she just put an X through it.. but gave me 1/2 credit. But no notes on what needed to be changed :(. There was a couple questions like that.

[19Eclipse90]

So I was wrong? Bugger....

[bigworm82]

not completely.. may have given to much info and that's why got partial credit i don't know.. irritated didn't actually show me where i got things wrong on some of the questions so i can study for the test.

[UTRacerX9]

Well, it asked for a confidence interval. So your answer should have been something like +/- 3.1 or something, where that is the range that you would expect the product to meet specifications. I think you may have had the right idea as posted above by Eclipse, but phrased it a bit wrong.

[bigworm82]

yeah i don't know..i'm not happy with the teacher just X'ing stuff out and not explaining what i did wrong though.. More irate now knowing i didn't need this stats class and could have taken any other 4 credit class :(. I'll bring it up tuesday after the test though cause i don't think it's right that she didn't say what was wrong.