Linear Regression and Correlation Question?
339574
If both the dependent variable (Y) and the independent variable (X) are standardized, then both their means are zero and standard deviations are one.  The correlation coefficient is not effected by a change in scale, meaning that the correlation doesn't change if you add or subtract the same number from each X or Y or if you multiply X or Y by the same positive number.  An example of this is if the correlation between height and weight when height is in inches and weight is in pounds is 0.6, then the correlation if height is in centimeters and weight is in kilograms is also 0.6.  We would hope that the relationship between the two variables doesn't change if there is a change in how the variables were measured.\n\nThis is the case when standardizing, since standardizing has you subtract the mean from each value and then divide by the standard deviation.  So the correlation coefficient for the standardized values of X and Y are the same as the correlation coefficient for the original values of X and Y.\n\nNow, the formula for the slope coefficient is r(sy)/(sx), where r is the correlation coefficient and sx and sy are the standard deviations of X and Y respectively.  Since both X and Y are standardized, then both of their standard deviations are equal to one, so putting that back into the expression above, the slope coefficient is equal to r.\n\nTo summarize, if X and Y are standardized, then the slope coefficient is the same as the correlation coefficient.
662354
Yes, it shows correlation.\n1 = perfect positive correlation\n-1 = perfect negative correlation.
894092
Here is a graph of .9 which is close to what you have: http://upload.wikimedia.org/wikipedia/en...\nThe way to read this is the diagonal is all r=1 and the rest of the relations are in the opposite positions. The .9 is in the first column... and the second box down. \n\n\nHere is a general graph that explains the relationships:\nhttp://hosting.soonet.ca/eliris/remotese...\n\nJust remember that just because it has a correlation of r=0 doesn't mean there isn't some correlation: http://www.uwsp.edu/psych/stat/7/i30.gif... ...\n\nThese points are obviously related and we can come up with a model to predict the next one.
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
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.
830458
Univariate problems depend on one variable only. multivariate pepend on several variables. A log regression curve when you have two varaibles y and x, and the relation between  x and y is of the form y = k log x
913689
Add additional details to your question and I may be able to answer it.  I assume you are looking for the expected value, but some of you data is missing.\nE(X) = .7(175000) + .3(-amount of the loss)
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.
34162
I'm guessing that you are trying to graph some cell division data and are trying to figure out which variable is the independent (x axis) and dependent (y axis) variables. Let's first start with the a rule of thumb for you: I'm assuming your graphing cell divisions over time. Anytime you are graphing something over time, time is almost always the independent variable.\n\nIt is the same in this case also. The number of cells is the dependent variable. (The number of cells present is dependent on how much time has passed).
863028
Is that really the question?  Prob of a normal variable being greater than the mean?  It's just 0.5.  What's the real question, and what has Z to do with it?
450656
The statement is true.  The formula for the slope of the regression line is\n\nr(SDy)/SDx\n\nSince SDy and SDx are always positive, then the only thing that determines the sign of the slope is r, the correlation coefficient.  So if r is positive, then the slope is positive; if r is negative, then the slope is negative.
201662
As you may know if you add independent variables to a linear regression the regular R^2 will not decrease, you can actually increase the R^2 by adding more predictors, but that does not mean you have a better model or that the variable that you added is a better predictor.  So to adjust the regular R^2 taking into consideration the number of predictors the Adjusted R^2 is used.  The R^2 is alwasys between 0 and 1, the adjusted R^2 can go below 0.\n\nR^2 = 1 - SS_error / SS_total\n\nAdjusted R^2 = \n1 - (SS_error / df_error) / (SS_total/df_total)\n\n\nHere is an example with both the R^2 and Adjusted R^2\nhttp://www.ats.ucla.edu/STAT/SAS/output/reg.htm
393026
You use regression when you want to fit a model to predict some value or outcome. Multinomial regression is where you have more than one input to predict one output. \n\nHere is an example of regression:\n\nSuppose you had the following data:\n\nHeight Weight Age for 100 people. \n\nYou might want to see if you can predict someone's age if you are given their height and weight. You call age the dependent variable because it depends on height and weight, and you can height and weight the independent variables. \n\nWhen you fit your model, you get something like this:\n\nAge = Intercept + A * Height * B * Weight\n\nThe model gives you the values of A and B and Intercept, so you can predict the Age of someone if you know their Height and Weight.\n\nThis is multinomial regression where the dependent variable is continuous, that is, it can be any value within reason and it varies.\n\nLogistic regression is used when you want a binary outcome, for example 1 meaning success and 0 failure.\n\nUsing the same example, you might want to predict someone's sex based on their height, weight and age. In this model, you would have three independent variables making it multinomial and your outcome would be 1 = male, 0 = not male, that is female. \n\nThe logistic regression model gives you the probability that, given someone's age, weight and height, that they are male (in this example, the higher the probability, the greater the chance that they person is male).\n\nAgain, you get a number of coefficients, but this time you use them to calculate a LOGIT:\n\nLOGIT = Intercept + A * Height + B * Weight * C * Age\n\nThe LOGIT is also written as:\n\nLOGIT(P) = Log [( P )/ ( 1 - P )]\n\nSo if you fit your model and get your values for A, B, C and the INTERCEPT, you can calculate the LOGIT value from some data and with this, calculate the probability P, that they are male. This probability P is given by:\n\nP = 1 / ( 1 + e**-(LOGIT))\n\nwhere ** is to the power of.\n\nThis is a much better way to model categories or classications, which are often used for understanding risk in banks and marketing, where you want to know the likelihood that someone will buy a product or default on a loan payment.
849282
If I'm not mistaken, try to calculate the prediction equation and use it to draw the regression line.\ny= a + bx
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.
26357
You don't need linear programming for this...\nYou want to maximize the investment in C, then in B, then in A.\n\nHe should in invest $6,000 in bond C.\nThen he can invest only $2,000 in bond B.\nThen he can invest only $8,000 in bond A.\n\nHe'll have $2,000 remaining.\n\n\nEdit to add:\nSorry, my LP skills are a bit rusty.  Here's all I remember:\nMaximize .05*A + .75*B + 1.05*C;\nsubject to   \nA + B + C <=18000\nA + B <=10000\nB + C <=8000\nC <=6000
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