what is the sample size needed to conduct two way ANOVA?
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.
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
647879
To run an ANOVA test, you need to make the following assumptions:\n\n1.  Samples were drawn from independent populations.\n2.  Populations are normally distributed.\n3.  Population variances are equal.\n\nNot having experience at this type of data, I can't be entirely sure that these will hold in your case, however you probably know better than I do.  My initial reaction is mixed; it could be that the variance changes after the intervention, but it may not.  The normal assumption is typically pretty loose when looking at data, and since you only have 18 observations, then it is unlikely that you would reject the null that the data is normal (translated, that means you can probably assume normallity).  The independence assumption may or may not hold since it is possible that there is some sort of seasonal effect in the quarters, but I don't know if that would happen or not.\n\nIf these assumptions do hold, then you can compute Tukey's HSD post hoc tests to see where the differences lie if you reject the null in the ANOVA.\n\nIn terms of the time series analysis, you use the ACF to determine the AR part and the PACF to determine the MA part of the ARIMA.  If there is linear decay in the ACF, then you don't have stationarity, so then differencing may be necessary.  Don't forget to take the possibilty of seasonality into account since the data is quarterly.\n\nI would get in touch with a statistical consultant to determine what tests would be best for this data.  Looking at this blind, I really can make any guarantees.
650317
If you have a computer program that can compute binomial probabilities, you can use that.\n\nIf you are doing this by hand, however, I would suggest using the normal approximation to the binomial here.\n\nn = number of trials = 200\np = prob. of success = 1/6 if we consider getting a five to be a success.\n\nThen np = 200(1/6) = 33.33 and n(1-p) = 166.67 are both >= 5 or 10 (the cutoff is different for different classes) then the sample size is large enough to use the normal.  This binomial is approximately normal with mean = np = 33.33 and SD = sqrt(np(1-p)) = sqrt(200(1/6)(5/6)) = 5.27.\n\nNow for the continuity correction.  The rectangle over 40 goes from 39.5 to 40.5.  Since we are finding the probability that the number of 5's is at least 40, then we should start at 39.5 so that all of 40 is included.\n\nz = (39.5-33.33)/5.27 = 1.17\nArea to the right of 1.17 on the standard normal is 0.121.
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.
82517
goto http://www.nal.usda.gov/fnic/foodcomp/cgi-bin/measure.pl?MSRE_NO=11134xyzxyzCassava%2C%20rawxyz
1147020
to determine the ideal sample size for a population, slovin's formula is used which is\n\nn = N / ( 1 + (N*e^2))\n\nWhere:\nn = number of samples\nN = total population\ne = margin of error\n\nSample Calculation:\nA group of students want to know the age of students in a high school but do not have the resources to survey an entire population of 2,500. If they want to use a sample with a 5% margin of error, what should their sample size be?\n\nGiven:\nN = 2,500\ne = 5% = 0.05\n\nRequired: n = ?\n\nSolution:\n\nn = N / ( 1 + (N*e^2))\nn = 2500 / ( 1 + (2500*.05^2))\nn = 2500 / 7.25\nn = 344.83   ~ 355
266644
I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
1022381
You can use a chi-square test for goodness of fit.  In the link below, O would be the observed number of times each card came up and E would be the expected.  (You would have to compute this 52 times and then add them up.).  So if you did this 500 times, the E for each card would be 500/52, which will be greater than 5, which is the typical way you determine if the sample size is large enough.  Under the assumption that the cards are at random, this would have a chi square distribution with 52-1 = 51 degrees of freedom.  You can do a test of significance using this test statistic and distribution.
611847
Depends on the sample size and population and things.  Generally, a z-table is good enough.  If the area outside is .0002, then that's around 3.7 sigmas.
251100
Yes!\n\nExperimental designs are efficient strategies for designing experiments using a minimum number of runs while enabling a maximum number of factors or parameters to be tested simultaneously. They also facilitate the statistical data analysis phase and the interpretation of results.\nCovers fundamental statistical methods for the Design of Experiments (DOE): accounting for various sources of variability, tools for minimizing and controlling variability, the comparison of two groups, the comparison of more than two groups, the Student's t-test, the Analysis of Variance (ANOVA), and the techniques of multiple comparisons for locating statistical differences.\n\n\nOne Way Analysis of Variance\n\nThe one way analysis of variance allows us to compare several groups of observations, all of which are independent but possibly with a different mean for each group. A test of great importance is whether or not all the means are equal.\n\n*********************\n\nThe observations all arise from one of several different groups (or have been exposed to one of several different treatments in an experiment). We are classifying 'one-way' according to the group or treatment.\n\n\nBasically ANOVA is just one tool to use, not unlike a t-test and design of experiments we are concerned with the analysis of data generated from an experiment. It is wise to take time and effort to organise the experiment properly to ensure that the right type of data, and enough of it, is available to answer the questions of interest as clearly and efficiently as possible. \n\nThe specific questions that the experiment is intended to answer must be clearly identified before carrying out the experiment. We should also attempt to identify known or expected sources of variability in the experimental units since one of the main aims of a designed experiment is to reduce the effect of these sources of variability on the answers to questions of interest. That is, we design the experiment in order to improve the precision of our answers.\n\nTreatment\n\nIn experiments, a treatment is something that researchers administer to experimantal units . For example, a corn field is divided into four, each part is 'treated' with a different fertiliser to see which produces the most corn; a teacher practices different teaching methods on different groups in her class to see which yields the best results; a doctor treats a patient with a skin condition with different creams to see which is most effective.\n\n*********************\n\nTreatments are administered to experimental units by 'level', where level implies amount or magnitude. For example, if the experimental units were given 5mg, 10mg, 15mg of a medication, those amounts would be three levels of the treatment. 'Level' is also used for categorical variables, such as Drugs A, B, and C, where the three are different kinds of drug, not different amounts of the same thing.\n\nFactor\n\nA factor of an experiment is a controlled independent variable; a variable whose levels are set by the experimenter.\n\n*********************\n\nA factor is a general type or category of treatments. Different treatments constitute different levels of a factor. For example, three different groups of runners are subjected to different training methods. The runners are the experimental units, the training methods, the treatments, where the three types of training methods constitute three levels of the factor 'type of training'.
856461
This would be a binomial.  Assuming that p=1/2 , equal probability of male or female.\n\nThen it's like 80 trials with success (male) or failure (female)\n\nAt least 30 male means that you have to add probs of 30, 31, 32, ...,80.  It's easier to add probs of 1,2,...,29 and subtract from 1.\n\n80 is a large enough number to use the normal approximation.\n\nFor the normal, the mean would be 80/2 = 40.  So, at least 30 is\n\n1-P(X<less than 30) \n\nconvert X to Z to use standard tables.
885851
Beta is the probability of a type II error, accepting the null when the alternative is true.\n\n1-beta is called the power or ability recognize that Ho is false.  \n\nA quote from my Marx and Larsen: " ...sample size is the parameter that researchers almost invariably turn to as the mechanism for ensuring that a hypothesis test will have sufficiently high power against a given alternative"...
709491
See:\n\nThe section, Sample Size For Proportions ("Large" Populations), in the following:\nhttp://www.wku.edu/~david.neal/statistics/misc/sampsize.html\n\n...or:\n\nhttp://www.stat.auckland.ac.nz/~wild/ChanceEnc/Ch08.psampsize.pdf
258372
Oh, I've had to learn nonparametric statistics from scratch, due to my research...\nThe Dunn test is a post-hoc or 'a posteriori' test that you use _after_ a non parametric ANOVA (I assume you know what that is) indicates that there are significant differences between the groups that you're comparing. \nThe non parametric ANOVA (normally it's the Kruskal-Wallis test) can only tell you that there is an _overall_ significant difference. \nThe Dunn test then compares each pair of groups and tells you which of these pairwise differences are significant.\n \nI don't know of any statistical package that performs it (if you do, please let me know) but it's not difficult if you construct an Excel worksheet with the formulas. You can find a good account of the calculations in Zar's Biostatistical Analysis book.
522796
a) Sensitivity = P(Test + | Has disease), so out of the people who have the disease, 93.4% would test positive for it.\n\nSpecificity = P(Test - | Doesn't have disease), so out of the people who don't have the disease, 98.6% would test negative for it.\n\nb) This is again a Bayes' Rule problem\nP(T+|D) = 0.934\nP(T-|ND) = 0.968\nP(D) = 0.002\n\nP(D|T+) = P(D)P(T+|D)/(P(D)P(T+|D) + P(ND)P(T-|ND))\n= (0.002)(0.934)/(0.002(0.934) + 0.998(0.968))
842271
My rule of thumb is that the skewness should be less than 2 in absolute value.  Other people might have different rules of thumb.\n\nYou may find the Jarque-Bera test of interest.  It is a test for normally that uses both the sample skewness and sample kurtosis.  I have posted a link about it below.