How do you hand calculate the contrast estimate for a two-way within subject ANOVA?
479611
just use SPSS 13.0  ;)
873308
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
477523
The no of degrees of freedom for a chi-square distribution ia n-1
1073519
If you want to compute confidence intervals in this case, you will want to do a multiple regression. \n\nYou cannot combine confidence intervals in general anyway, but espcially since both variables have different units, it doesn't make sense to do, say, a two sample confidence interval.
760097
Try this: http://en.wikipedia.org/wiki/Confidence_interval
260932
it's a vector! measure of distance - magnitude, and the direction of the vector.
398063
If you have a quantitative dependent variable and a nominal independent variable, then the one-way ANOVA test is appropriate.\n\nIf you have nominal dependent and independent variables, then a chi-square test for independence is appropriate.
1150220
Bias is systematic error that favors one group over another.  There are different ways to detect bias.  The logical first step in detecting bias is to find elements where one group performs much better than the other group: such items function differentially for the two groups and thus is known as Differential Item Functioning (DIF).  DIF is a necessary but not sufficient condition for bias: bias only exists if the difference is illegitimate, i.e., if both groups should be performing equally well on the item.  The simplest DIF index is based on the differences in correct response rates between a "reference" and "focal" group.\n\nThe Mantel-Haenszel odds ratio is a non-parametric statistic using chi-square.  For every score level, it calculates the odds that the reference group will get the item correct, and that the focal group will get it correct.  It then divides reference group odds by focal group odds, showing how much more likely the reference group is to get the item correct.  You can transform results from the Mantel-Haenszel procedure into its delta\nunits to classify items into grouped results.  \n\nMantel-Haenszel has two major practical advantages: it is conceptually uncomplicated, and it doesn't require highly specialized software.  However, Mantel-Haenszel can be affected by item discrimination and it performs better with large group sizes.
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As to part one, I can not recommend any acceptable, known test for such data (perhaps what I've written below might work).\n\nAs to part two, this is a horrid approximation since the midrange as the mean is a major assumption (even more far-reaching than assuming something about the populations' probability distributions).\n\nFor part 3 and also part 1:\n\nThe first problem you run into is that you do not know anything about the nature of the population distribution (normal, uniform, etc).  Thus, a nonparametric test would be required.  I know of no nonparametric tests for any of the statistics you have (minimum, maximum or range).\n\nIf it is possible to obtain the actual values for each of the samples in each set, the Kruskal-Wallis test should suffice.  It is nonparametric, so there are no problems with the requirement of assumptions about population distributions.\n\nFurther, its job is to determine if two samples are far enough apart such that the null hypothesis that the population distributions are the same can be rejected.  This would give you what you desire.\n\nAs far as your problem is concerned, the only statistics you have are minimum, maximum and range.   Perhaps one could derive a test to test the minima or maxima of each set, but one still runs into the problem that there can be no parameter placed on the population distribution.\n\nIf you can make an assumption about the distributions of the populations of interest (normal or whatnot), then it would be possible to derive such a test.  This would be a tedious task, however, as it would require thorough knowledge of order statistics and distribution functions.\n\nOne would need to use a test statistic based of off the difference in the two statistics that you use (I would use the maxima since they are further apart than the minima and probability distributions for ranges are uncharted territory unless you're a Ph.D.).  Thus, your test statistic is based off of the difference between the maxima (Y(n) - Z(n) where the (n) are subscripts, which is 2 in this case).\n\nTo perform this test, you would need to derive the probability distribution of the difference of the two maxima and use that to find the probability of the difference being two or greater.  This is quite a task for anyone and requires a large amount of knowledge in mathematical statistics.  I believe I could perform such a task, but it would likely take some time to complete.\n\nRemember though that you must make a critical assumption about the population before you can even do this.  I would recommend using the assumption of uniformity if you are unsure of the populations' distributions.  This is because the derived test would then be least likely to give a type I error, where you reject the data improperly, because the test is less likely to reject.
440843
numerical recipes in C.  Amazon has it.
978796
Study of data with respect to different spectral responses
374562
I think that you would need at least 2 values in each cell so that the within groups SD can be estimated, so if each of the two ways has two factors then you would need a sample of at least 8, so that there is two for each cell.  Could be wrong though.
391744
This is a simple two tailed test of the equality of means.\n\nLook this up in a stats book, but the Hypothesis that you are testing is Mean1=Mean2.
173439
Sequential blood samples were evaluated from 3 subjects with AHIV (negative HIV EIA, positive HIV RNA), identified 2-3 days before peak viremia (pre-peak). HIV RNA levels and CD4 counts before antiretroviral therapy were used to estimate HIV doubling time (t(d)) and, the basic reproductive number (R(0)). HIV t(d) was calculated from the slope of ln(RNA) vs. time during rising viremia. Maximal estimates of t(d) were 1.6-3.5 days.\nhttp://www.aegis.com/conferences/croi/1998/521.html
929935
variations matter greatly in interpreting data.  Something might seem like it is returning significantly different results than the control but stat evaluation may show that the variations are minimal at best
529075
Hey seriously I dont't understand "t0.05(9)=2.26"\n\nHowever I did some of the calculation based on what I remember.\nMean of the sample = 2.2\nVariance if the sample = 9.56\n\nTo find the t-value, using the t formula,\n\nt = (2.2 - 0) / [9.56/ sq root(10)]\n= 0.7277\n\nWhy 0? Because you want to test whether the sample mean is 0. O meaning there is no effect on the B.P.\n\nSince sample size is 10, degrees of freedom is 9.\nLooking at the t-table, at 5% sig, the t-value is 2.262\nat 10% sig, t- value is 1.833.\n\nNote that we are actually looking at the t-value where the p-values are 2.5% and 5% respectively. Here we divide the level of sig by 2 because the null hypothesis is an equal sign.\nNull Hypo: sample mean = 0.\nSince it is 'equal' sign, we will divide level of significace by 2. Remember this.\n\nSo you can see that actually our t-value is smaller than the 2 t-values and we do not rej the null hypo. \n\nConclusion: We do not have enough evidence to say that the stimulus have a significant effect on the B.P. Not being able to rej null hypo doesn't mean that we accept null hypo. So we cannot claim that the stimulus doesn't have an effect on the B.P. This point is impt. \n\nHope you understand my workings. Any mistakes pls point out to me. :) I did this based on what I remember. :)