need answer for my statistics rgarding ratio for correlated groups?
813709
What's the question?  What you are describing is a baseline phrase.
556500
50%\n\nThe mean is 200 with a deviation of 50 either way.  That means that the chances are 50 percent that the rating will be higher than 200 and 50 percen the rating will be lower than 200 because 200 is the mean.
846927
Sorry no one will be able to help you as we do not the data in #3.\n\nYou will need to figure this one out yourself...
726852
It all depends on how much different you expect the results to be between the experimental group and the control group. I am on simvastatin and have gone periods drinking more green tea. Have not watched my HDL levels closely enough to compared effects. \n\nI use statcalc in EpiInfo to run sample size calculations and find that is easy and accurate. I think that would work for your experiment and your question.
836301
Try this website for help with your statistics homework:\n\nhttp://www.statsoft.com/textbook/stathome.html
376279
100 is average, so you are above average =)  Welcome to the group!
8058
ratio is a certain value to another certain value.  say, there are 3 girls to every 5 boys.  so if there are 80 people, there will be 30 girls and 50 boys.  proportions are different. you know one of the values.  so, u find the other value using it.  so if its 30 percent girls, you will multiply 30 percent by 80 people, thus getting 24 girls.  and the rest are boys.  and as the numbers get larger, so do the proportions.
512917
13!8!/20!\n\nyour group sizes r, 2 4 6 8\n\nwithout any restriction of who goes together, u have this many choices: 20C2*18C4*14C6----A\n\nwith restriction of 7 ppl must goes together, u have this many choices: 13*12C6*6C4-----B\n\nThe prob of getting 7 ppl in a group is A/B. u do the algebra and u will get 13!8!/20!\n\n\nreasoning:\nfor 7 ppl to be in a group, the group must take at least 7 person. there is only one group that can do so. Taking the 7 ppl as one entity that must be allocated together, u have only left with 14 objects/persons to arrange. After the group for 8 take the group of 7 ppl, it can still take one and it has 13 choices from the remaining people.\n\nafter choosing the group of 8, u r left with 12 ppl to pick for the group of 6, 4 and 2 and the number of choices are, 12C6*6C4*2C2\n\nso, the total number of possibility would be 1*13*12C6*6C4*2C2.\n\n\n\n\n\n\n"....with restriction of 7 ppl must goes together, u have this many choices: 13*12C6*6C4-----B....."\n\nthis is because, in the special case of that 7 particulat person must be in a group, u group them together to form one. so, how, u only have 14 objects to allocate to groups. get it? since group of 7 can only go into the group that can fit 8 person... so, it's possibility is 1.\n\nAfter that, since the group of 8 has been filled up by the 7 person, u have one place in it left, and 13 ppl to choose from to fill it. hence, 13.\n\nthen, for the remaining of the task, u simply choose out of 12 people for your 6, 4, and 2 men groups.\n\n\n\nso, u have 1x13x12C6x6C4x2C2 to forming THIS(with 7 particular person in one group) type of group.\n\nthe prob(forming THIS group)=(number of way to form THIS type of group)/(number of way to form ANY type of group)\n\nget it?\n\n\n\n\nI'm glad u really wanna know. most ppl got the answer and they r satisfied. too bad i m not good at explaining. contact me if u need more help.\n\n\n\n\n\n\noh, sorry. u misunderstood me. yes, u r perfectly right. there r 13 ppl left after u allocate the 7 ppl to the "8" group. What i was saying was... b4 u do that, u have 14 "person".... one of them being a group of 7 people. sorry for confusing u. And from what u said, yes, u got the idea right.
278272
You'll use Z-scores and the Z-table. Z-scores convert deviations to standard deviation units. \n\nIn this case, you're looking for the range of scores that makes up the middle 88% of the distribution - 44% on either side of the mean. \n\nThe z-score is 1.56 (look in the table for the z-score that cuts off 44% of the distribution.) This tells you that 44% of the scores fall between the mean and 1.56 standard deviation units above the mean - and 44% of all scores fall between the mean and 1.56 standard deviation units below the mean. \n\nNow, since each standard deviation unit in this particular distribution is 4ml (as per your question), 1.56 standard deviation units is equal to 1.56 x 4, which is equal to 6.24ml.\n\nBack to the problem - your range is 6.24 ml above and below the mean - which is 250ml - therefore the range of volumes is from 6.24ml less than 250 to 6.24ml more than 250 - or in other words - 243.76ml to 256.24 ml.\n\nHope that helped!
793664
What's your problem - your in college - do your own homework (this son of a statistician is unimpressed).  If you don't know how to construct a confidence interval look at http://facweb.bhc.edu/EastCampus/leeb/ma108/06_01w/tsld032.htm\n\nGood Luck!!!
696159
Try this web site \n\nhttp://home.ubalt.edu/ntsbarsh/Business-stat/opre504.htm#rbosim
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you do not need to have an r^2 value to determin if two values are correlated\n\nyou just look at the sctter plot\n\nif there is a a clear trend (ie as I increase one value the other value generally decrease  or genrally decrease)\nthis can be seen in a simple scatter plot one need no statistics to to see if there is correlation\n\nif ther eis no correlation the scatterplot will be totally random and no  pattern. Correlation you are generelaly looking for a linear pattern)\n\nif you wnat to technically test and  to have a significane value you can test Ho: R^2= zero  vs   H1: R^2 ~= zero\n\nthe problem you will encounter her is when you calculate R^2 you will only be able to get a point estimate. \n\nIf you wish to test the value of R^2 equal to zero then you will need to somehow to guage the variance.\n\nTo guage the variance of R^2 you will have to utilize the jacknife variance estimator or the bootstrap variance estimator\n\nthes are methods to compute variances for thing besides the mean ie the median, the max , the min, R^2\n\n\nto jacknife you compute the R^2 n times always leaving the nth observation out of the calculation. this gives you a range of R^2's. now you can compute a variance to test or use that set to define a CI to do your test.\n\nThe bootstrap is similar in that you compute the same value multiple times, but here you will take random samples with replacement from your samples until you have chossen n observations (not with replace ment so you get a differnt sample than you started with) now compute the value of intereest (R^2) off of this sample. this will give you another estimate for the same value. Now reapeat this exercise until you have a sufficent number of estimatess (say n) then we can now compute a variance for R^2 or use the set to define a 95% CI to do our test jsut as with the jacknife method
554599
4.2\n\nfor a binomial distribution the mean is the number * the probability
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Answer is 10.2% of the sample(25) have a mean score within 3 points of the population mean score of 220.\n\nSD=10 --> 34% (normal distribution)\n   3 from mean = 3/10 x 34% = 10.2%.
251008
I'ts easy to crosntruct a covariance matrix:\nLet X be your data set.\nthen make y=X-M where M is the vector of means.\nThen the covariance matrix is given by C=1/n(y'y)\nwhere n is the number of observations
73222
Your average will be -25\nYour range will be 377\nAverage is taking the sum of the numerical vales and dividing them by the number of data points listed.\n\nie: 95 + 70 - 8 - 282 + 0 = -125/5 = -25\n\nYour range is the difference of the highest number of the group and the lowest number of the group.\n\nie: Highest = 95  Lowest = -282\n95- (-282) = 377\n\nI hope it helps.  bmsc2082yahoo.com and tell me what happens.
43772
Hi! I'm a statistician and I use http://www.statsoft.com as well as cheap textbooks from Half Priced Books in my area. Both seem to help. If you have a specific question, feel free to email me. Don't know how much good I'll be because I use fancy software for my analysis, but I'll try.