Statistics help please!!!!?
461783
This is a Two Tailed test ... in other words, you are trying to determine if you have statistical evidence to suggest that the average change was something other than zero disruptions different between weeks 1 and 2 ... just because you saw a change of 5, doesn't mean anything ... it could just be a random fluke !\n\nHo (Null Hypothesis):  Sample Aveage change in distruptions was zero -- no change, or X-Bar = 0\nHa (alternative):  Sample Average change was not 0, or X-Bar <> 0\n\nAt the 0.05 (5%) level of significance (in total), you would have 0.025 probability in the upper end of the distribution and another 0.025 probability in the other end ... that's a total of 0.05 or 5%\n\nUsing your Z-Score table, you find the Z-Score (the number of Standard Deviations in a Standardized Normal Distribution with mean of 0 and Standard Deviation of 1) associated with 0.025 or 2.5% ... this is:\n\n-1.959961082 Standard Deviations from the mean for the low tail and 1.959961082 standard deviations from the mean for the high tail.\n\nSo, the number of standard deviations from a presumed mean of zero (under the null hypothesis) for the sample at hand is:\n\n( 5 - 0 ) / 2  = 2.5\n\nThat is to say, that if the Null Hypothesis of NO CHANGE in behavior were true, a legitimate value of 5 drawn purely randomly could still happen, but that would be 2.5 Standard Deviations away.\n\nBut, your upper and lower limits were only +/- 1.959961082 !!!\n\nSo 95% of the time, you would expect to see something between +/- 1.959961082 standard deviations from zero (within this range) ... but this time, you saw a value associated with a whopping 2.5 standard deviations from the presumed mean of zero ... (we'll call 1.959961082 = 1.96 here)\n\n<< I ACTUALLY MADE A LITTLE GRAPHIC HERE, BUT IT WON'T DISPLAY PROPERLY >>\n\nSo you would reject Ho.\n\nConclusion:  Based upon the sample taken, beleieved to be truely random and under the assumptions of normality, we have statistical evidience which would tend to suggest that the observed sample difference of 5 is statistically significant at the 95% confidence level, thereby, leading us to conclude, based upon the evidence observed, that the incentive system may have played a roll in the positive inprovement of the respondents.\n\nDoes this help ?   My contact information is in my profile.\n\nRegards,\n\nAlan
188021
As you can see the answers are all over the place.  I have provided a link to a website which shows statistics on a graph from the year 2001.  Although it's not the most current of information, it is a good starting point.
1166976
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.
242624
http://en.wikipedia.org/wiki/Mean_%28statistics%29\n\nThis is a link from Wikipedia. It shows lots of examples along with explanations.
315361
Not sure what exactly type of statistics you are looking for, but here are two pages\nhttp://www.mdeda.org/info.asp\nhttp://www.unos.org/
492063
Count the total number of groups:\n9 choose 5 = 9*8*7*6/(4*3*2*1) = 9*2*7 = 126.\n\nb) Count the number of groups where 3 women where chosen and 2 men.\n(4 choose 3) * (5 choose 2) = 4 * 10 = 40.\n\nThe prob that a random committee has exactly 3 women = 40/126\n\nc) If there were no women then the number of ways this can happen is:\n(4 choose 0) * (5 choose 5).  As you could probably already see, there is only 1 way to have a committe of 5 men and 0 women.  The prob = 1/126.\n\nf) >=3 women is the same as saying 3 or 4 women as 4 women is the most you can have.\n\n(4 choose 3)*(5 choose 2) + (4 choose 4) * (5 choose 1)=\n4*10 + 1*5 = 45.  Prob = 45/126.
487348
The mean of the average of 100 randomly selected travelers would just be the same as the population mean, which is 190.\n\nThe SD of the avereage of 100 randomly selected travelers would be the SD of the population divided by the square root of the sample size, which is 20/sqrt(100) = 2.
1156399
This is a good question even for a statistician.\nThe problem is your set is small and I do not know what the deviation between samples is, however the repeatability and reproducibility information can bind the window \nthrough which the third value will fall through or perhaps even remain in.  \n\nLet me know.
303094
OK.  From any given population, you can take many, many different samples of size n.  For each of those samples, you can find the mean.  So if you take 1000 samples of size 50,  Then you would have 1000 sample means.  Now each of those sample means are probably going to be different from the overall population mean, since samples are going to look different from the population.  Also, each of the 1000 sample means are going to look different from each other because the mean is coming from different samples.\n\nSo if I were to draw a histogram (hopefully you know what that is; it is a bar graph that you can use to look at the distribution of numerical data).  Anyway, if I were to draw a histogram of my 1000 sample means,  You should notice the following three things...\n\n1.  The center of your histogram should be the same as you population mean (so if the mean of the population is 20, then the histogram of my 1000 sample means should center around 20).\n\n2.  As the sample size gets larger, the spread of the histogram should get smaller (so the spread of the means of samples of size 50 would be smaller than for the means of samples of size 25).  This is because with larger samples, you would have more information about the population, and thus you should be able to get closer to the population mean.\n\n3.  As the sample size gets larger, the shape of the histogram will look more and more like a normal curve.\n\nThis is because of the Central Limit Theorem, that states that if you take a sample of size n from any population with a finite variance, the distribution of the mean of the sample would get closer and closer to a normal curve as n gets larger.  The mean of the normal would be the same as the population mean, and the standard deviation of the normal (usually called the standard error) would  be the standard deviation of the population divided by the square root of n.  Notice that as n gets larger, the standard error gets smaller, just as described before.  So typically you can approximate these histrgrams with a normal curve if n > 25.\n\nThis phenomenon is hard to describe in words alone.  Looking at pictures of these histograms that I described would help you understand it better.
152796
try this site...\n\nhttp://www.futureaccountant.com/probability/
837986
http://www.statsoft.com/textbook/stathome.html
327897
This looks ok:\n\nhttp://www.wadsworth.com/psychology_d/templates/student_resources/workshops/stats_wrk.html\n\nor else just keep google-ing psychology+statistics.\n\nGood luck :-)
898911
Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
163979
Try the Bureau of Labor Statistics :)
907438
I could tell you the answer or I could give you an idea of what to consider when you do it yourself.\n\na) Binomial numbers and p = 0.7, q= 0.3 might help here.  You do not know the size of the population of the mailing list so you might need to make an assumption that it is large enough to assume a Normal distribution.  I think the central limit theorem might help here.\n\nb) Binomial theory, and conditional probabilities.  Look it up in your text book!  Again if the size of the population is large enough you might be able to simplify the problem.
971574
This is a proportioal problem\n\nMultiply the means\n\nmultiply the extremes\n\n- - - - - - -\n\nx = the amount of men\n\n4 / 7 = x / 28\n\nthe means = 7x\n\nThe extremes = 4 x 28\n\n- - - - - - - - - - - - - - -\n\n7x = 4 x 28\n\n7x = 112\n\n7x / 7 = 112 / 7\n\nx = 16\n\nThe answer there are 16 men\n\n- - - - - - s-
714755
A.  Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55.  So we can approximate this standard deviation by saying\n\n(55-5)/4 = 12.5\n\nB.  Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625.  Since the sample size is large, then the sample mean is approximately normal.  To be within 2 units, then you would have to be within mu-2 and mu+2.  So we can standardize.\n\nz = (mu-2-mu)/0.625 = -3.2\nz = mu+2-mu)/0.625 = 3.2\n\nThe area between these two values would then be approximately 0.9986.\n\nC. In this case, our hypothesis test would look like this.\n\nH0: mu = 35\nHA: mu < 35\n\nOur test statistic would be (31.75-35)/0.625 = -5.2\n\nWe would then reject the null hypothesis.  So we would conclude that the scientists are correct.