what are the assumptions for linear regression?
402618
I have no idea about the math, but I'm thinking the assumption is that the data are relatively correlated linearly.  There's a lot of math behind the wiki link.
141538
Read the references at the end of the article or talk to an industrial engineer.\n\nhttp://en.wikipedia.org/wiki/Linear_programming
352533
Generally speaking, you call "linear" a function F (or an operator whatsoever) which satisfies the following conditions:\n\n1) F(a*X)=a*F(X) for every real number a (X is the argument of the function, whatever it be: it can be a number, a vector, a complex, a function... it doesn't matter)\n\n2) F(X+Y) = F(X)+F(Y) for every X and Y which can be taken as an argument of the function or the operator.\n\nThese two conditions can be sinthesized in the following single one:\n\nCONDITION FOR LINEARITY:\nF(a*X+b*Y) = a*F(X) + b*F(Y) for every a and b reals and for every X and Y in the domain of the function.\n\n----\n\nIn the ring of the polynomials of n variables x1, x2 ... xn, it is easily verified that the only polynomials which can be defined "linear" according to the above conditions are those in which the variables appear at the first degree only, so they are in the form:\n\nP(n) = a1*x1 + a2*x2 + ... + an*xn + a(n+1)\n\nLinear equations in n variables are obtained by setting a linear polynomial of the above type equal to 0.\n\nNon-linear equations are all other possible kinds of equations: the ones obteined by setting to 0 non-linear polynomials (you call them algebraic non-linear equations), or the ones obtained by setting ot 0 expressions which involve logarithms, exponentials etc (in that case you are dealing with transcendant equations)
1082511
I think your question have a typo.  There are three variables and I am not sure which one are you trying to solve for.
827005
It's linear alright. x = - 3/7. A little tricky because it's a line parallel to the y axis.
621346
y=mx+b (this is known as the standard form of linear equations)\n\nm=slope\nx=x values\ny=y values\nb= y-intercept\n\nso the answer for you question is:\n\nslope = -1\ny-intercept = -3\n\nEXPLANATION:\n\ny = mx + b\ny = -x-3\n\nm= -1   and    b = -3\n \nslope = m    and   y-intercept = b\n\nthus, the slope is -1 and the y-intercept is -3\n\n\nNote: the slope is rise/run (rise over run) -- to make a graph use the y-intercept and the slope -- once you have the slope and the y-intercept, making this graph should be easy because it's a linear graph -- but if you still don't get it ask your teacher\n\n\nthus, the slope is not a-1 -----there is no "a"
608881
where "Y est" equals the estimated value on dependent variable, "a" equals the intercept (supplied by SPSS),  "b" equals the slope (supplied by SPSS), and "X" equals the given value on the independent variable.
527923
Draw a scatter-plot. In short draw a graph of y against x.\nIf the graph show linear relationship, u may continue. If the graph is not linear(or curvilinear) ignore the questions as regression model is  only for linear model.\n\nSince u didn't include datas, i can only gives u equations\n\n\nQ1....\nthe mean for the number of years a teacher has taught \n    =  (∑ x)/n\n\n \nthe mean for the students' average course grade \n    =  (∑ y)/n\n\nQ2....\nthe variance for the number of years a teacher has taught \n     s²=[ ∑ (xi – mean for x value) ² ]/(n-1)\n\nthe variance for the students' average course grade\n     s²=[ ∑ (yi – mean for y value) ² ]/(n-1)\n\n\nQ3....\nthe standard deviation(s)\ns= √variance for the number of years a teacher has taught\n\ns= √variance for the students' average course grade\n\nQ4....\nthe z-score for the number of years a teacher has taught\n= (observed value of x-mean for x value)/ the standard deviation of x value\n\nthe z-score for the students' average course grade\n= (observed value of y-mean for y value)/ the standard deviation of y value.\n\nobserved value means u must calculate for each value one by one.\n\nQ5.....\nThe correlation coefficent(r)\n=1/(n-1) X  ∑(xi-mean value for x / s for x value)(yi- mean value for y)\n\ns stands for standard deviation and X is times and n is sample size\n\nThe equation looks quite tough isn't it? Normally we calculated r².\n\nr²= (SSTO-SSE)/SSTO\n\nwhere SSE = sum of squared errors\n          = ∑(yi-predicted value of y)²\n\nthe predicted value is base on the least square line( u have to draw a graph that has the sum of ( yi-predicted value of y) is smaller than any other line.) \n\n  \n     SSTO = sum of squares total\n          = ∑(yi-the mean for y value)²\n\nAfter u find r², find r by √r²\n      \nThe value of r always between -1 and +1\n#If u get value for r outside that range, means u have made some mistakes in calculation.\n#If u get r=1 or -1, it means that the relationship is perfectly linear and all data points fall on the same straight line.\n#r=+something means that when two varibles tend to increase together.(x increase, y increase)\n#r=0 means the best straight line is exactly horizontal\n\nQ6....\ni dont know about the coefficient of determination. Never heard such thing. Sorry.\n\nQ7....\nthe regression equation\n\ny-hat=b-nought  -  b-one   x\n\n  where    \nb-one = [∑(xi-the mean for x value)(yi-the mean for y value)]/  ∑(xi-the mean for x value)²\n\nb-nought = the mean for y value  -  b-one(the mean of x value)\n\nQ8....\n From y-hat=(b-nought)  -  (b-one) x\nJust simply substitude x=x1, therefore\n   predicted y = b-nought)  -  (b-one)x1\n\nQ9....\nI'm not sure about standard error of the estimate for predicting Y from X. But if if i'm not mistaken, the formula should be\n\nstandard error(s.e)= s/ [∑(yi-the mean for y value)²][b-one)\n\nwhere b-one=r(the standard deviation for x value)/(the standard deviation for y value)\n\nand s= √[SSE/(n-2)]\n\n\n\n\nGood luck
79585
If it is anything like the TI-83 plus, you need to turn on DiagnosticOn. For the 83 hit [2nd][0] which is the catalog of all commands and scroll down to DiagnosticOn hit enter twice and then run the linear regression. I assuming they didn't change the calculator that much and that this wont work for the TI-84 plus.
1027379
Goodness of Fit\nDegrees of Freedom\nR squared\nAbsolute Sum of Squares\nSy.x\n  \nThis section seems to tell the most about the overall fit, You would want sum of squares approaching one, along with r^2, not sure about what would be a good value for goodness of fit
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
766587
(1) The correlation answers the STRENGTH of linear association between paired variables, say X and Y. On the other hand, the regression tells us the FORM of linear association that best predicts Y from the values of X.\n\n(2a) Correlation is calculated whenever:\n* both X and Y is measured in each subject and quantify how much they are linearly associated.\n* in particular the Pearson's product moment correlation coefficient is used when the assumption of both X and Y are sampled from normally-distributed populations are satisfied \n* or the Spearman's moment order correlation coefficient is used if the assumption of normality is not satisfied.\n* correlation is not used when the variables are manipulated, for example, in experiments.\n \n(2b) Linear regression is used whenever:\n* at least one of the independent variables (Xi's) is to predict the dependent variable Y. Note: Some of the Xi's are dummy variables, i.e. Xi = 0 or 1, which are used to code some nominal variables.\n* if one manipulates the X variable, e.g. in an experiment. \n\n\n(3) Linear regression are not symmetric in terms of X and Y. That is interchanging X and Y will give a different regression model (i.e. X in terms of Y) against the original Y in terms of X.\nOn the other hand, if you interchange variables X and Y in the calculation of correlation coefficient you will get the same value of this correlation coefficient.\n\n(4) The "best" linear regression model is obtained by selecting the variables (X's) with at least strong correlation to Y, i.e. >= 0.80 or <= -0.80 \n\n(5) The same underlying distribution is assumed for all variables in linear regression. Thus, linear regression will underestimate the correlation of the independent and dependent when they (X's and Y) come from different underlying distributions.
811494
linear equations are used quite frequently in real life. a person wid not a deep kmowledge of mathematics also uses them.\nfor eg:\n1.   our daily budget wid income to b spent as 'm' for 5 commodities \n      namely a,b,c----------j nd their prices p1,p2,p3,----------p10 is a linear equn.\n                  a.p1+b.p2+----------------+j.p10=m\n\n2.   also the process of cost calculation in firms require them\n      as 'w' being d wage rate of labour and a value r as raw material cost. the equn becomes-\n                  l.w+r=cost, where l is d no. of labour hired\n\n\nmany other simple examples can b quoted.
951621
1) Extrapolate the regression another day, this is the prediction.\n2) The estimated variance of the extrapolated value at point x is\n(1/S1) + (S1/D) (x- (S2/S1))^2 where\n\nD=S1 S3 - S2^2\nS1 = sum(1/s^2)\nS2 = sum (x/s^2)\nS3 = sum (x^2/s^2) \nThe sums are taken of all datapoints, i.e. I should have written s_k and x_k rather than s and x.\n\n3) 95% correspond to 1.96 standard errors to be exact.\n\n4) You probably don't want to use linear regression for prediction, as this assumes that your data are measuring points scattered around a *true* value, the regression line you find. This is clearly not the case..... In your case, it would be better to use any of the  autoregressive moving average models, like ARIMA, GARCH.\n\nTake a look at the paragraph "Time series models" on this page:\nhttp://en.wikipedia.org/wiki/Predictive_analytics
700548
So x4-5*y2*x2+4*y4=0; or (x^2-y^2)*(x^2-4y^2)=0; or (x-y)*(x+y)*(x-2y)*(x+2y)=0\nIt splits into independent linear equations: (a) y=x; (b) y=-x; (c) y=0.5*x; (d) y=-0.5*x;\nNow exclude (a) & (c); thus (x+y)*(x+2y)=0; did you mean it Don?
906303
Brute force though it may be, an x-dimensional linear regression is still your best bet for optimization.  If there is a single optimal process, the method will yield it, but you are more likely ro arrive at a number of processes, each optimized, that will yield a maximum profit.\n\nYour other alternative is to optimize each process independently, and then choose the most profitable, but you are likely to discard combinations which may yield a better end result, and you will still need  n-dimensional regressions of some sort.\n\nIn lieu of dedicated software for the problem, a spreadsheet can be used quite effectively.  For x > 7, however, you may need to apply techniques to keep numbers within your computer's ability, although this is not as problematic as it is with polynomial regressions.
493627
You can refer this book:\n\nDifferential equations : linear, nonlinear, ordinary, parial - A.C. King, J. Billingham, and S.R. Otto.