R-Squared & Linear Regression?
830991
I'm wondering why you created the data 1-3500?  I would think you would use the set of data available and figure the standard deviation and average.  With that information, you would be able to predict tomorrows value at a certain confidence level.  I forget is that the p test or the t test.  its been a while since I took statistics.
169495
Linearly, this is fairly straightforward, or, you can use functions in Excel like trend & slope to do a regressive analysis.
368727
If you give only obscure details (B1, T/F) but tell nothing more about the problem, you are not likely to get an answer....I do a lot of statistical correlations, but this really doesn't mean much.\n\nNow, my guess on what you mean.\nYou have a series of data points (x_n, y_n). If you make a scatterplot and the points are rougly on a line you have perfect correlation, correlation coefficient is +1 or -1.\n\nIf on the other hand Y doesn't depend on X, i.e. correlation coefficient 0, then when you do a linear regression the slope of the regression line will be close to zero, as the data points (x_n, y_n) are scattered randomly.
501300
Area = Pi x R squared\nArea = Pi x (x - 6) squared\nArea = Pi x (x^2 - 12x + 36)\nArea = Pix^2 -Pi12x + Pi36\nArea = 3.14x^2 - 37.68x + 113.04\n\n(Using the approximation 3.14 for Pi)\n\nOf course, it has been a while and I'm a little rusty...
509283
In statistics we talked about residuals. a Residual is anestimate on how well your regression line fits your data. The residual is often notes as r^2. I'm not sure on exactly how to find it, but on the TI 83 plus, the residual is a flat line and the data is scattered around it. The data should be spread equally on both sides of the line. the more equal the data is spread, the better your residual line. When you find the "VarsStat1" on the TI 83, the residual value will be there noted as r^2.
870323
r=(e^20r)-1 all over 50
1065775
If a(N)=Sum(r^n, 0=>N) and -1<r<1...\n\n Is something that looks like 1/(1-r^N)
113710
Well, the y-intercept would be acceleration when the force on the object is zero.  You'd better hope it's zero!  If it's not, you might want to fix the point in your analysis (this can be done in Excel) or redo the experiment.\n\nEdit: If you're plotting acceleration vs. force, the intercept really should be zero.  Those values appear small enough to be within the error of the experiment.  They could be indicative of the effect of a systematic error, but are more likely the result of random noise.  Try fixing the intercepts at zero in Exel (right-click on the regression line and open up the options for it; there should be a checkbox for "fix intercept").  An intercept value that is not equal to zero indicates that there was acceleration when there was no force!
186891
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
196477
Root Mean Square is the square root of the sum of the squares. With that terminology is more an electric enginering term, used to measure average powwer.\nFor statistics, its more common to use minimum squares, which refers to the sum of squares of the estimation errors (residuals). Loof for regression on the web.\nGoodness of Fit is a kind of statistical test to prove (or disprove) whether a set of data comes from a given distribution. The most commonly used test for that is the Chi-square test for goodness of fit.\nAbout the third, I don't know.\nHope it Helps.
129909
You might find the following informative:\nhttp://www2.mtroyal.ab.ca/~igilborn/admn1382/regression/theory/part_b.htm
894473
Any equation explaning the relationship between two or more observations (or data points) could be called a regression. It's really an extrapolation of the underlying line (or non-linear relationship) that explains the variance.\n\nPerhaps you could do a simple one- pick a stock and use the closing price over a period (say 3 weeks, 15 observations). Then try to come up with a formula to explain the pattern.\n\nActually, that one is kind of hard without stats software. But there are others- such as average temperatures in a given area...
123406
Try Chi square testing, it starts out complicated but once you understand it, it is way faster than punnett squares! You can find all kinds of stuff about it on the internet, it's a statistical analysis.
695289
standard deviation of intercept is a quantity related to the precision of the estimated intercept of the fitted line. It can be calculated by the formula:\n\nS_a = (S_b)*(sqrt( (sum(x_j)^2)/m )) \n\nWhere, S_b is standard deviation of slope\n\nS_b = sqrt( ((m*(s_y)^^2) / [m*(sum(x_j)^2) -(sum x_j)^2] )\n\ns_y is the residual standard deviation.\n\na and b are estimated parameters of the line y = a + bx +err.\n\n[Here, by S_b, I mean S with suffix b, Sorry but its a bit difficult to type mathematics with only text.]\nI hope , I answered your question.
562312
Greg is correct for what you posted, but is that 2 after the y supposed to be superscripted as in y-squared? If that's the case:\n\ny^2 = 9.6\ny = sqrt(9.6) = 3.098
308
First load the add-in:\nTools > Add-Ins > Analysis Tool Pack\nthen find it:\nTools > Data Analysis & Choose Regression\nIf you need any more help mail me.
141538
Read the references at the end of the article or talk to an industrial engineer.\n\nhttp://en.wikipedia.org/wiki/Linear_programming