kapkomet Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Balta1701 @ Apr 15, 2009 -> 05:17 PM) 2.4 to 23.1, I believe. ^^ Link to comment Share on other sites More sharing options...
Balta1701 Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (shipps @ Apr 15, 2009 -> 03:21 PM) ha,its so cute when balta is being modest. You know you are correct you smart son of a b****. Actually there could be a bit of error in my answer, I just discovered that neither Excel nor Maple have the inverse error function as one of their standard functions, so I had to go to a table that I'm not totally sure how to read. Link to comment Share on other sites More sharing options...
kapkomet Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Balta1701 @ Apr 15, 2009 -> 05:25 PM) Actually there could be a bit of error in my answer, I just discovered that neither Excel nor Maple have the inverse error function as one of their standard functions, so I had to go to a table that I'm not totally sure how to read. It's just reading a std. deviation chart and confidence intervals, right? Link to comment Share on other sites More sharing options...
shipps Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Balta1701 @ Apr 15, 2009 -> 05:25 PM) Actually there could be a bit of error in my answer, I just discovered that neither Excel nor Maple have the inverse error function as one of their standard functions, so I had to go to a table that I'm not totally sure how to read. I have a belly button. Link to comment Share on other sites More sharing options...
Balta1701 Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (kapkomet @ Apr 15, 2009 -> 03:27 PM) It's just reading a std. deviation chart and confidence intervals, right? Ah, that's the 2nd decimal place. Figured it out now. No, it's an x-y chart of areas under the normal distribution curve. Just an odd way of presenting that data. The first decimal place is the y axis, the 2nd is the x axis. Link to comment Share on other sites More sharing options...
kapkomet Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Balta1701 @ Apr 15, 2009 -> 05:32 PM) Ah, that's the 2nd decimal place. Figured it out now. No, it's an x-y chart of areas under the normal distribution curve. Just an odd way of presenting that data. The first decimal place is the y axis, the 2nd is the x axis. The area under the curve covered by 2 std. deviations... that's what I was thinking... right? Link to comment Share on other sites More sharing options...
Balta1701 Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (kapkomet @ Apr 15, 2009 -> 03:43 PM) The area under the curve covered by 2 std. deviations... that's what I was thinking... right? 2 standard deviations is 95%. 98% is something like 2.33 S.D. Link to comment Share on other sites More sharing options...
Goober Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (shipps @ Apr 15, 2009 -> 05:28 PM) I have a belly button. OMG, I have one too! Link to comment Share on other sites More sharing options...
lostfan Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (shipps @ Apr 15, 2009 -> 06:28 PM) I have a belly button. I just farted. Link to comment Share on other sites More sharing options...
dasox24 Posted April 15, 2009 Share Posted April 15, 2009 I hated statistics. Link to comment Share on other sites More sharing options...
Buehrle>Wood Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Balta1701 @ Apr 15, 2009 -> 10:17 PM) 2.4 to 23.1, I believe. Is this the range? I need the "mean +- somethingsomething". Obviously the mean is easy since they give it to you. I can't figure out the somethingsomething part though. I have been doing 2.33(sd/sqrt of sample). Does not seem to be working. Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Buehrle>Wood @ Apr 15, 2009 -> 06:08 PM) Is this the range? I need the "mean +- somethingsomething". Obviously the mean is easy since they give it to you. I can't figure out the somethingsomething part though. I have been doing 2.33(sd/sqrt of sample). Does not seem to be working. ECON 202? Link to comment Share on other sites More sharing options...
Buehrle>Wood Posted April 15, 2009 Share Posted April 15, 2009 Yes. Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Buehrle>Wood @ Apr 15, 2009 -> 04:55 PM) A random sample of 20 college students was taken and it was found that they drank an average of 12.75 sodas per week and the sample standard deviation was 4.44. If we know that the number of sodas drank per week by college students is normally distributed, compute the 98% confidence interval estimate of the population mean. Anyone know the answer to that? 12.75 + or - 2.3133 so (10.4367, 15.0633) Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Buehrle>Wood @ Apr 15, 2009 -> 06:10 PM) Yes. Use pages 134 and 135 in your course book, if you ahve the notes for those pages it basically spells out what to do. Hope it helps, just ask if you need more help. Link to comment Share on other sites More sharing options...
Balta1701 Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (bigruss22 @ Apr 15, 2009 -> 04:12 PM) 12.75 + or - 2.3133 so (10.4367, 15.0633) You need to multiply the 2.3133 by the standard deviation of the data. Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 (edited) QUOTE (Balta1701 @ Apr 15, 2009 -> 06:17 PM) You need to multiply the 2.3133 by the standard deviation of the data. 12.75 + or minus 2.33(4.44/sqroot of 20) is the equation. Edited April 15, 2009 by bigruss22 Link to comment Share on other sites More sharing options...
Buehrle>Wood Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (bigruss22 @ Apr 15, 2009 -> 11:18 PM) 12.75 + or minus 2.33(4.44/sqroot of 20) is the equation. I took the quiz over 30 times and I kept getting it wrong using that same formula. Never did get it right, even put the same answer as the one you gave. Maybe my Mallard is screwed up or something. I even did it just now using that equation and got it wrong(I knw you can't submit past 5, but it'll tell you if you're right or wrong). I'm going to talk to the TA I guess. Link to comment Share on other sites More sharing options...
Balta1701 Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (bigruss22 @ Apr 15, 2009 -> 04:18 PM) 12.75 + or minus 2.33(4.44/sqroot of 20) is the equation. If the standard deviation (1 sigma) is 4.44, does it make sense that the 98% confidence limit can be less than 1 sigma away from the mean? Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Buehrle>Wood @ Apr 15, 2009 -> 06:24 PM) I took the quiz over 30 times and I kept getting it wrong using that same formula. Never did get it right, even put the same answer as the one you gave. Maybe my Mallard is screwed up or something. I even did it just now using that equation and got it wrong(I knw you can't submit past 5, but it'll tell you if you're right or wrong). I'm going to talk to the TA I guess. Which quiz is this? The third one of the set for this midterm? I had problems with that one, but it was all stupid mistakes in calculations. Maybe you had trouble with that (like you were doing it wrong but didnt realize it). I always divide the stdev. by the sqroot of n first, then I multiply that number by the z value. Basically, that method just cuts down on errors for me for when i put it in the calculator. Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 QUOTE (Balta1701 @ Apr 15, 2009 -> 06:26 PM) If the standard deviation (1 sigma) is 4.44, does it make sense that the 98% confidence limit can be less than 1 sigma away from the mean? It all depends on n, the sample size. Since sqroot of 20 is 4.4721 then it will be less than 1 sigma. But if n was larger than it would increase, all part of statistics. Link to comment Share on other sites More sharing options...
Buehrle>Wood Posted April 15, 2009 Share Posted April 15, 2009 The first question on the last quiz was the problem. Link to comment Share on other sites More sharing options...
bigruss Posted April 15, 2009 Share Posted April 15, 2009 (edited) QUOTE (Buehrle>Wood @ Apr 15, 2009 -> 06:33 PM) The first question on the last quiz was the problem. Ahh, then you have to use t chart. So you have mean + or - tvalue(stdev./sqroot of n) To find the t value you need to look at the confidence value, which for that problem is 98% so you split 2% in half (100-98=2) so you look at the first row (because .2/2= .1, so its the first row) of the t chart and go down to 20, since thats your n. So you would have 1.325 as your tvalue. then just plug in the rest. Edited April 15, 2009 by bigruss22 Link to comment Share on other sites More sharing options...
SnB Posted April 16, 2009 Share Posted April 16, 2009 QUOTE (Buehrle>Wood @ Apr 15, 2009 -> 06:10 PM) Yes. colleen schultz still teach it? that class sucked. Link to comment Share on other sites More sharing options...
bigruss Posted April 16, 2009 Share Posted April 16, 2009 QUOTE (SnB @ Apr 15, 2009 -> 07:01 PM) colleen schultz still teach it? that class sucked. I dont like her, but the class is pretty easy to me, but then again I think stats is a subject that just comes to some people and not to others. Then again, I was AWFUL at Calc, so I get to make up a little bit here. Link to comment Share on other sites More sharing options...
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